ideal

Chapter 7 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Book on a Table A book weighing 5 N rests on top of a table. Part A A downward force of magnitude 5 N is exerted on the book by the force of ANSWER: Part B An upward force of magnitude _____ is exerted on the _____ by the table. the table gravity inertia . ANSWER: Part C Do the downward force in Part A and the upward force in Part B constitute a 3rd law pair? You did not open hints for this part. ANSWER: Part D The reaction to the force in Part A is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____ . You did not open hints for this part. ANSWER: 6 N / table 5 N / table 5 N / book 6 N / book yes no Part E The reaction to the force in Part B is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____. ANSWER: Part F Which of Newton’s laws dictates that the forces in Parts A and B are equal and opposite? ANSWER: Part G Which of Newton’s laws dictates that the forces in Parts B and E are equal and opposite? ANSWER: 5 N / earth / book / upward 5 N / book / table / upward 5 N / book / earth / upward 5 N / earth / book / downward 5 N / table / book / upward 5 N / table / earth / upward 5 N / book / table / upward 5 N / table / book / downward 5 N / earth / book / downward Newton’s 1st or 2nd law Newton’s 3rd law Blocks in an Elevator Ranking Task Three blocks are stacked on top of each other inside an elevator as shown in the figure. Answer the following questions with reference to the eight forces defined as follows. the force of the 3 block on the 2 block, , the force of the 2 block on the 3 block, , the force of the 3 block on the 1 block, , the force of the 1 block on the 3 block, , the force of the 2 block on the 1 block, , the force of the 1 block on the 2 block, , the force of the 1 block on the floor, , and the force of the floor on the 1 block, . Part A Assume the elevator is at rest. Rank the magnitude of the forces. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Newton’s 1st or 2nd law Newton’s 3rd law kg kg F3 on 2 kg kg F2 on 3 kg kg F3 on 1 kg kg F1 on 3 kg kg F2 on 1 kg kg F1 on 2 kg F1 on floor kg Ffloor on 1 Part B This question will be shown after you complete previous question(s). Newton’s 3rd Law Discussed Learning Goal: To understand Newton’s 3rd law, which states that a physical interaction always generates a pair of forces on the two interacting bodies. In Principia, Newton wrote: To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. (translation by Cajori) The phrase after the colon (often omitted from textbooks) makes it clear that this is a statement about the nature of force. The central idea is that physical interactions (e.g., due to gravity, bodies touching, or electric forces) cause forces to arise between pairs of bodies. Each pairwise interaction produces a pair of opposite forces, one acting on each body. In summary, each physical interaction between two bodies generates a pair of forces. Whatever the physical cause of the interaction, the force on body A from body B is equal in magnitude and opposite in direction to the force on body B from body A. Incidentally, Newton states that the word “action” denotes both (a) the force due to an interaction and (b) the changes in momentum that it imparts to the two interacting bodies. If you haven’t learned about momentum, don’t worry; for now this is just a statement about the origin of forces. Mark each of the following statements as true or false. If a statement refers to “two bodies” interacting via some force, you are not to assume that these two bodies have the same mass. Part A Every force has one and only one 3rd law pair force. ANSWER: Part B The two forces in each pair act in opposite directions. ANSWER: Part C The two forces in each pair can either both act on the same body or they can act on different bodies. ANSWER: true false true false Part D The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be due to friction or electric charge). ANSWER: Part E The two forces of a 3rd law pair always act on different bodies. ANSWER: Part F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.) You did not open hints for this part. ANSWER: true false true false true false Part G According to Newton’s 3rd law, the force on the (smaller) moon due to the (larger) earth is ANSWER: Pulling Three Blocks Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force . The magnitude of the tension in the string between blocks B and C is = 3.00 . Assume that each block has mass = 0.400 . true false greater in magnitude and antiparallel to the force on the earth due to the moon. greater in magnitude and parallel to the force on the earth due to the moon. equal in magnitude but antiparallel to the force on the earth due to the moon. equal in magnitude and parallel to the force on the earth due to the moon. smaller in magnitude and antiparallel to the force on the earth due to the moon. smaller in magnitude and parallel to the force on the earth due to the moon. F T N m kg Part A What is the magnitude of the force? Express your answer numerically in newtons. You did not open hints for this part. ANSWER: Part B What is the tension in the string between block A and block B? Express your answer numerically in newtons You did not open hints for this part. ANSWER: Pulling Two Blocks In the situation shown in the figure, a person is pulling with a constant, nonzero force on string 1, which is attached to block A. Block A is also attached to block B via string 2, as shown. For this problem, assume that neither string stretches and that friction is negligible. Both blocks have finite (nonzero) mass. F F = N TAB TAB = N F Part A Which one of the following statements correctly descibes the relationship between the accelerations of blocks A and B? You did not open hints for this part. ANSWER: Part B How does the magnitude of the tension in string 1, , compare with the tension in string 2, ? You did not open hints for this part. Block A has a larger acceleration than block B. Block B has a larger acceleration than block A. Both blocks have the same acceleration. More information is needed to determine the relationship between the accelerations. T1 T2 ANSWER: Tension in a Massless Rope Learning Goal: To understand the concept of tension and the relationship between tension and force. This problem introduces the concept of tension. The example is a rope, oriented vertically, that is being pulled from both ends. Let and (with u for up and d for down) represent the magnitude of the forces acting on the top and bottom of the rope, respectively. Assume that the rope is massless, so that its weight is negligible compared with the tension. (This is not a ridiculous approximation–modern rope materials such as Kevlar can carry tensions thousands of times greater than the weight of tens of meters of such rope.) Consider the three sections of rope labeled a, b, and c in the figure. At point 1, a downward force of magnitude acts on section a. At point 1, an upward force of magnitude acts on section b. At point 1, the tension in the rope is . At point 2, a downward force of magnitude acts on section b. At point 2, an upward force of magnitude acts on section c. At point 2, the tension in the rope is . Assume, too, that the rope is at equilibrium. Part A What is the magnitude of the downward force on section a? Express your answer in terms of the tension . ANSWER: More information is needed to determine the relationship between and . T1 > T2 T1 = T2 T1 < T2 T1 T2 Fu Fd Fad Fbu T1 Fbd Fcu T2 Fad T1 Part B What is the magnitude of the upward force on section b? Express your answer in terms of the tension . ANSWER: Part C The magnitude of the upward force on c, , and the magnitude of the downward force on b, , are equal because of which of Newton's laws? ANSWER: Part D The magnitude of the force is ____ . ANSWER: Fad = Fbu T1 Fbu = Fcu Fbd 1st 2nd 3rd Fbu Fbd Part E Now consider the forces on the ends of the rope. What is the relationship between the magnitudes of these two forces? You did not open hints for this part. ANSWER: Part F The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straight line. For this situation, which of the following statements are true? Check all that apply. ANSWER: less than greater than equal to Fu > Fd Fu = Fd Fu < Fd The tension in the rope is everywhere the same. The magnitudes of the forces exerted on the two objects by the rope are the same. The forces exerted on the two objects by the rope must be in opposite directions. The forces exerted on the two objects by the rope must be in the direction of the rope. Two Hanging Masses Two blocks with masses and hang one under the other. For this problem, take the positive direction to be upward, and use for the magnitude of the acceleration due to gravity. Case 1: Blocks at rest For Parts A and B assume the blocks are at rest. Part A Find , the tension in the lower rope. Express your answer in terms of some or all of the variables , , and . You did not open hints for this part. ANSWER: M1 M2 g T2 M1 M2 g Part B Find , the tension in the upper rope. Express your answer in terms of some or all of the variables , , and . You did not open hints for this part. ANSWER: Case 2: Accelerating blocks For Parts C and D the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude . Part C Find , the tension in the lower rope. Express your answer in terms of some or all of the variables , , , and . You did not open hints for this part. ANSWER: T2 = T1 M1 M2 g T1 = a T2 M1 M2 a g Part D Find , the tension in the upper rope. Express your answer in terms of some or all of the variables , , , and . You did not open hints for this part. ANSWER: Video Tutor: Suspended Balls: Which String Breaks? First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point. T2 = T1 M1 M2 a g T1 = Part A A heavy crate is attached to the wall by a light rope, as shown in the figure. Another rope hangs off the opposite edge of the box. If you slowly increase the force on the free rope by pulling on it in a horizontal direction, which rope will break? Ignore friction and the mass of the ropes. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. The rope attached to the wall will break. The rope that you are pulling on will break. Both ropes are equally likely to break.

Chapter 7 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Book on a Table A book weighing 5 N rests on top of a table. Part A A downward force of magnitude 5 N is exerted on the book by the force of ANSWER: Part B An upward force of magnitude ___ is exerted on the _ by the table. the table gravity inertia . ANSWER: Part C Do the downward force in Part A and the upward force in Part B constitute a 3rd law pair? You did not open hints for this part. ANSWER: Part D The reaction to the force in Part A is a force of magnitude _, exerted on the _ by the _. Its direction is _ . You did not open hints for this part. ANSWER: 6 N / table 5 N / table 5 N / book 6 N / book yes no Part E The reaction to the force in Part B is a force of magnitude _, exerted on the _ by the _. Its direction is _. ANSWER: Part F Which of Newton’s laws dictates that the forces in Parts A and B are equal and opposite? ANSWER: Part G Which of Newton’s laws dictates that the forces in Parts B and E are equal and opposite? ANSWER: 5 N / earth / book / upward 5 N / book / table / upward 5 N / book / earth / upward 5 N / earth / book / downward 5 N / table / book / upward 5 N / table / earth / upward 5 N / book / table / upward 5 N / table / book / downward 5 N / earth / book / downward Newton’s 1st or 2nd law Newton’s 3rd law Blocks in an Elevator Ranking Task Three blocks are stacked on top of each other inside an elevator as shown in the figure. Answer the following questions with reference to the eight forces defined as follows. the force of the 3 block on the 2 block, , the force of the 2 block on the 3 block, , the force of the 3 block on the 1 block, , the force of the 1 block on the 3 block, , the force of the 2 block on the 1 block, , the force of the 1 block on the 2 block, , the force of the 1 block on the floor, , and the force of the floor on the 1 block, . Part A Assume the elevator is at rest. Rank the magnitude of the forces. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Newton’s 1st or 2nd law Newton’s 3rd law kg kg F3 on 2 kg kg F2 on 3 kg kg F3 on 1 kg kg F1 on 3 kg kg F2 on 1 kg kg F1 on 2 kg F1 on floor kg Ffloor on 1 Part B This question will be shown after you complete previous question(s). Newton’s 3rd Law Discussed Learning Goal: To understand Newton’s 3rd law, which states that a physical interaction always generates a pair of forces on the two interacting bodies. In Principia, Newton wrote: To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. (translation by Cajori) The phrase after the colon (often omitted from textbooks) makes it clear that this is a statement about the nature of force. The central idea is that physical interactions (e.g., due to gravity, bodies touching, or electric forces) cause forces to arise between pairs of bodies. Each pairwise interaction produces a pair of opposite forces, one acting on each body. In summary, each physical interaction between two bodies generates a pair of forces. Whatever the physical cause of the interaction, the force on body A from body B is equal in magnitude and opposite in direction to the force on body B from body A. Incidentally, Newton states that the word “action” denotes both (a) the force due to an interaction and (b) the changes in momentum that it imparts to the two interacting bodies. If you haven’t learned about momentum, don’t worry; for now this is just a statement about the origin of forces. Mark each of the following statements as true or false. If a statement refers to “two bodies” interacting via some force, you are not to assume that these two bodies have the same mass. Part A Every force has one and only one 3rd law pair force. ANSWER: Part B The two forces in each pair act in opposite directions. ANSWER: Part C The two forces in each pair can either both act on the same body or they can act on different bodies. ANSWER: true false true false Part D The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be due to friction or electric charge). ANSWER: Part E The two forces of a 3rd law pair always act on different bodies. ANSWER: Part F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.) You did not open hints for this part. ANSWER: true false true false true false Part G According to Newton’s 3rd law, the force on the (smaller) moon due to the (larger) earth is ANSWER: Pulling Three Blocks Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force . The magnitude of the tension in the string between blocks B and C is = 3.00 . Assume that each block has mass = 0.400 . true false greater in magnitude and antiparallel to the force on the earth due to the moon. greater in magnitude and parallel to the force on the earth due to the moon. equal in magnitude but antiparallel to the force on the earth due to the moon. equal in magnitude and parallel to the force on the earth due to the moon. smaller in magnitude and antiparallel to the force on the earth due to the moon. smaller in magnitude and parallel to the force on the earth due to the moon. F T N m kg Part A What is the magnitude of the force? Express your answer numerically in newtons. You did not open hints for this part. ANSWER: Part B What is the tension in the string between block A and block B? Express your answer numerically in newtons You did not open hints for this part. ANSWER: Pulling Two Blocks In the situation shown in the figure, a person is pulling with a constant, nonzero force on string 1, which is attached to block A. Block A is also attached to block B via string 2, as shown. For this problem, assume that neither string stretches and that friction is negligible. Both blocks have finite (nonzero) mass. F F = N TAB TAB = N F Part A Which one of the following statements correctly descibes the relationship between the accelerations of blocks A and B? You did not open hints for this part. ANSWER: Part B How does the magnitude of the tension in string 1, , compare with the tension in string 2, ? You did not open hints for this part. Block A has a larger acceleration than block B. Block B has a larger acceleration than block A. Both blocks have the same acceleration. More information is needed to determine the relationship between the accelerations. T1 T2 ANSWER: Tension in a Massless Rope Learning Goal: To understand the concept of tension and the relationship between tension and force. This problem introduces the concept of tension. The example is a rope, oriented vertically, that is being pulled from both ends. Let and (with u for up and d for down) represent the magnitude of the forces acting on the top and bottom of the rope, respectively. Assume that the rope is massless, so that its weight is negligible compared with the tension. (This is not a ridiculous approximation–modern rope materials such as Kevlar can carry tensions thousands of times greater than the weight of tens of meters of such rope.) Consider the three sections of rope labeled a, b, and c in the figure. At point 1, a downward force of magnitude acts on section a. At point 1, an upward force of magnitude acts on section b. At point 1, the tension in the rope is . At point 2, a downward force of magnitude acts on section b. At point 2, an upward force of magnitude acts on section c. At point 2, the tension in the rope is . Assume, too, that the rope is at equilibrium. Part A What is the magnitude of the downward force on section a? Express your answer in terms of the tension . ANSWER: More information is needed to determine the relationship between and . T1 > T2 T1 = T2 T1 < T2 T1 T2 Fu Fd Fad Fbu T1 Fbd Fcu T2 Fad T1 Part B What is the magnitude of the upward force on section b? Express your answer in terms of the tension . ANSWER: Part C The magnitude of the upward force on c, , and the magnitude of the downward force on b, , are equal because of which of Newton's laws? ANSWER: Part D The magnitude of the force is __ . ANSWER: Fad = Fbu T1 Fbu = Fcu Fbd 1st 2nd 3rd Fbu Fbd Part E Now consider the forces on the ends of the rope. What is the relationship between the magnitudes of these two forces? You did not open hints for this part. ANSWER: Part F The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straight line. For this situation, which of the following statements are true? Check all that apply. ANSWER: less than greater than equal to Fu > Fd Fu = Fd Fu < Fd The tension in the rope is everywhere the same. The magnitudes of the forces exerted on the two objects by the rope are the same. The forces exerted on the two objects by the rope must be in opposite directions. The forces exerted on the two objects by the rope must be in the direction of the rope. Two Hanging Masses Two blocks with masses and hang one under the other. For this problem, take the positive direction to be upward, and use for the magnitude of the acceleration due to gravity. Case 1: Blocks at rest For Parts A and B assume the blocks are at rest. Part A Find , the tension in the lower rope. Express your answer in terms of some or all of the variables , , and . You did not open hints for this part. ANSWER: M1 M2 g T2 M1 M2 g Part B Find , the tension in the upper rope. Express your answer in terms of some or all of the variables , , and . You did not open hints for this part. ANSWER: Case 2: Accelerating blocks For Parts C and D the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude . Part C Find , the tension in the lower rope. Express your answer in terms of some or all of the variables , , , and . You did not open hints for this part. ANSWER: T2 = T1 M1 M2 g T1 = a T2 M1 M2 a g Part D Find , the tension in the upper rope. Express your answer in terms of some or all of the variables , , , and . You did not open hints for this part. ANSWER: Video Tutor: Suspended Balls: Which String Breaks? First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point. T2 = T1 M1 M2 a g T1 = Part A A heavy crate is attached to the wall by a light rope, as shown in the figure. Another rope hangs off the opposite edge of the box. If you slowly increase the force on the free rope by pulling on it in a horizontal direction, which rope will break? Ignore friction and the mass of the ropes. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. The rope attached to the wall will break. The rope that you are pulling on will break. Both ropes are equally likely to break.

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Assignment 8 Due: 11:59pm on Friday, April 4, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 10.3 Part A If a particle’s speed increases by a factor of 5, by what factor does its kinetic energy change? ANSWER: Correct Conceptual Question 10.11 A spring is compressed 1.5 . Part A How far must you compress a spring with twice the spring constant to store the same amount of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct = 25 K2 K1 cm x = 1.1 cm Problem 10.2 The lowest point in Death Valley is below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 . Part A What is the change in potential energy of an energetic 80 hiker who makes it from the floor of Death Valley to the top of Mt.Whitney? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.3 Part A At what speed does a 1800 compact car have the same kinetic energy as a 1.80×104 truck going 21.0 ? Express your answer with the appropriate units. ANSWER: Correct Problem 10.5 A boy reaches out of a window and tosses a ball straight up with a speed of 13 . The ball is 21 above the ground as he releases it. 85m m kg U = 3.5×106 J kg kg km/hr vc = 66.4 km hr m/s m Part A Use energy to find the ball’s maximum height above the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Use energy to find the ball’s speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Use energy to find the speed of impact on the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Hmax = 30 m v = 13 ms v = 24 ms Problem 10.8 A 59.0 skateboarder wants to just make it to the upper edge of a “quarter pipe,” a track that is one-quarter of a circle with a radius of 2.30 . Part A What speed does he need at the bottom? Express your answer with the appropriate units. ANSWER: Correct Problem 10.12 A 1500 car traveling at 12 suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll. Part A kg m 6.71 ms kg m/s What will be the car’s speed as it coasts into the gas station on the other side of the valley? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object’s kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object’s kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where “i” denotes the “initial” moment and “f” denotes the “final” moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: v = 6.8 ms K = (1/2)mv2 U = mgh Ki + Ui = Kf + Uf v Correct Part B At the top point of the flight, what can be said about the projectile’s kinetic and potential energy? ANSWER: Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system “Earth-ball.” Although we will often talk about “the gravitational potential energy of an elevated object,” it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch __________. ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases. Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum. Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level–but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: Correct You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: is negative is positive is zero depends on the choice of the “zero level” of potential energy U = 0 hmax v g hmax = v2 2g h 0.5v v g h = 3 v2 8g Correct Part F What is the speed of the object at the height of ? Express your answer in terms of and . Use three significant figures in the numeric coefficient. Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. ANSWER: Correct Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball’s flight. Express your answer in terms of , , and . Hint 1. Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of “final” moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. u (1/2)hmax v g h = 0 v (1/2)mv2 h (1/2)hmax (1/4)mv2 h = (1/2)hmax v u = 0.707v v hmax v g Kf Express your answer in terms of , , and . Hint 1. Find the speed at the maximum height The speed of the ball at the maximum height is __________. ANSWER: ANSWER: ANSWER: Correct Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. v m 0 v v cos v sin v tan Kf = 0.5m(vcos())2 hmax = (vsin())2 2g v hmax v g ANSWER: Correct Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: Correct The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Problem 10.14 A 12- -long spring is attached to the ceiling. When a 2.2 mass is hung from it, the spring stretches to a length of 17 . Part A What is the spring constant ? Express your answer to two significant figures and include the appropriate units. hmax = v2 2g v hmax v g hmax = v2 2g Ki + Ui = Kf + Uf cm kg cm k ANSWER: Correct Part B How long is the spring when a 3.0 mass is suspended from it? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.17 A 6.2 mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in . You may want to review ( pages 255 – 257) . For help with math skills, you may want to review: Solving Algebraic Equations = 430 k Nm kg y = 19 cm kg Part A What does the spring scale read just before the mass touches the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass before it touches the scale. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? ANSWER: Correct Part B The scale reads 22 when the lower spring has been compressed by 2.7 . What is the value of the spring constant for the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? Use these to determine the force on the mass by the spring, taking note of the directions from your picture. How is the spring constant related to the force by the spring and the compression of the spring? Check your units. ANSWER: F = 61 N N cm k = 1400 k Nm Correct Part C At what compression length will the scale read zero? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces on the mass. When the scale reads zero, what is the force on the mass due to the scale? What is the gravitational force on the mass? What is the force on the mass by the spring? How is the compression length related to the force by the spring and the spring constant? Check your units. ANSWER: Correct Problem 10.18 Part A How far must you stretch a spring with = 800 to store 180 of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: y = 4.2 cm k N/m J Correct Problem 10.22 A 15 runaway grocery cart runs into a spring with spring constant 230 and compresses it by 57 . Part A What was the speed of the cart just before it hit the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Spring Gun A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measured from the equilibrium position of the spring. s = 0.67 m kg N/m cm v = 2.2 ms H Part A The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to . Hint 1. Potential energy of the spring The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial. Hint 2. Potential energy of the ball At the highest point in the ball’s trajectory, all of the spring’s potential energy has been converted into gravitational potential energy of the ball. ANSWER: Correct A Bullet Is Fired into a Wooden Block A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed . H height = H 4 mb vi mw vf Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of , , and . Hint 1. Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy vi mw mb ptotal vf ptotal = (mw + mb)vf ptotal vi ptotal = mbvi ANSWER: Correct Problem 10.31 Ball 1, with a mass of 150 and traveling at 15.0 , collides head on with ball 2, which has a mass of 340 and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Express your answer with the appropriate units. ANSWER: Correct Part B Express your answer with the appropriate units. ANSWER: Correct Part C vf = mb vi mb+mw g m/s g (vfx) = -5.82 1 ms (vfx) = 9.18 2 ms What are the final velocities of each ball if the collision is perfectly inelastic? Express your answer with the appropriate units. ANSWER: Correct Part D Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.43 A package of mass is released from rest at a warehouse loading dock and slides down the = 2.2 – high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. You may want to review ( pages 265 – 269) . For help with math skills, you may want to review: Solving Algebraic Equations (vfx) = 4.59 1 ms (vfx) = 4.59 2 ms m h m 2m Part A Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are two parts to this problem: the block sliding down the frictionless incline and the collision. What conservation laws are valid in each part? In terms of , what are the kinetic and potential energies of the block at the top of the incline? What is the potential energy of the same block at the bottom just before the collision? What are the kinetic energy and velocity of block just before the collision? What is conserved during the collision? What is the total momentum of the two blocks before the collision? What is the momentum of the two blocks stuck together after the collision? What is the velocity of the two blocks after the collision? ANSWER: Correct Part B Suppose the collision between the packages is perfectly elastic. To what height does the package of mass rebound? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are three parts to this problem: the block sliding down the incline, the collision, and mass going back up the incline. What conservation laws are valid in each part? m m v = 2.2 ms m m What is an elastic collision? For an elastic collision, how are the initial and final velocities related when one of the masses is initially at rest? Using the velocity of just before the collision from Part A, what is the velocity of just after the collision in this case? What are the kinetic and potential energies of mass just after the collision? What is the kinetic energy of mass at its maximum rebound height? Using conservation of energy, what is the potential energy of mass at its maximum height? What is the maximum height? ANSWER: Correct Problem 10.35 A cannon tilted up at a 35.0 angle fires a cannon ball at 79.0 from atop a 21.0 -high fortress wall. Part A What is the ball’s impact speed on the ground below? Express your answer with the appropriate units. ANSWER: Correct Problem 10.45 A 1000 safe is 2.5 above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 . m m m m m h = 24 cm $ m/s m vf = 81.6 ms kg m cm Part A What is the spring constant of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.49 A 100 block on a frictionless table is firmly attached to one end of a spring with = 21 . The other end of the spring is anchored to the wall. A 30 ball is thrown horizontally toward the block with a speed of 6.0 . Part A If the collision is perfectly elastic, what is the ball’s speed immediately after the collision? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the maximum compression of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.5×105 k Nm g k N/m g m/s v = 3.2 ms Correct Part C Repeat part A for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Repeat part B for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.4%. You received 120.28 out of a possible total of 121 points. x = 0.19 m v = 1.4 ms x = 0.11 m

Assignment 8 Due: 11:59pm on Friday, April 4, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 10.3 Part A If a particle’s speed increases by a factor of 5, by what factor does its kinetic energy change? ANSWER: Correct Conceptual Question 10.11 A spring is compressed 1.5 . Part A How far must you compress a spring with twice the spring constant to store the same amount of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct = 25 K2 K1 cm x = 1.1 cm Problem 10.2 The lowest point in Death Valley is below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 . Part A What is the change in potential energy of an energetic 80 hiker who makes it from the floor of Death Valley to the top of Mt.Whitney? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.3 Part A At what speed does a 1800 compact car have the same kinetic energy as a 1.80×104 truck going 21.0 ? Express your answer with the appropriate units. ANSWER: Correct Problem 10.5 A boy reaches out of a window and tosses a ball straight up with a speed of 13 . The ball is 21 above the ground as he releases it. 85m m kg U = 3.5×106 J kg kg km/hr vc = 66.4 km hr m/s m Part A Use energy to find the ball’s maximum height above the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Use energy to find the ball’s speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Use energy to find the speed of impact on the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Hmax = 30 m v = 13 ms v = 24 ms Problem 10.8 A 59.0 skateboarder wants to just make it to the upper edge of a “quarter pipe,” a track that is one-quarter of a circle with a radius of 2.30 . Part A What speed does he need at the bottom? Express your answer with the appropriate units. ANSWER: Correct Problem 10.12 A 1500 car traveling at 12 suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll. Part A kg m 6.71 ms kg m/s What will be the car’s speed as it coasts into the gas station on the other side of the valley? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object’s kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object’s kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where “i” denotes the “initial” moment and “f” denotes the “final” moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: v = 6.8 ms K = (1/2)mv2 U = mgh Ki + Ui = Kf + Uf v Correct Part B At the top point of the flight, what can be said about the projectile’s kinetic and potential energy? ANSWER: Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system “Earth-ball.” Although we will often talk about “the gravitational potential energy of an elevated object,” it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch . ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases. Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum. Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level–but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: Correct You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: is negative is positive is zero depends on the choice of the “zero level” of potential energy U = 0 hmax v g hmax = v2 2g h 0.5v v g h = 3 v2 8g Correct Part F What is the speed of the object at the height of ? Express your answer in terms of and . Use three significant figures in the numeric coefficient. Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. ANSWER: Correct Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball’s flight. Express your answer in terms of , , and . Hint 1. Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of “final” moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. u (1/2)hmax v g h = 0 v (1/2)mv2 h (1/2)hmax (1/4)mv2 h = (1/2)hmax v u = 0.707v v hmax v g Kf Express your answer in terms of , , and . Hint 1. Find the speed at the maximum height The speed of the ball at the maximum height is . ANSWER: ANSWER: ANSWER: Correct Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. v m 0 v v cos v sin v tan Kf = 0.5m(vcos())2 hmax = (vsin())2 2g v hmax v g ANSWER: Correct Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: Correct The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Problem 10.14 A 12- -long spring is attached to the ceiling. When a 2.2 mass is hung from it, the spring stretches to a length of 17 . Part A What is the spring constant ? Express your answer to two significant figures and include the appropriate units. hmax = v2 2g v hmax v g hmax = v2 2g Ki + Ui = Kf + Uf cm kg cm k ANSWER: Correct Part B How long is the spring when a 3.0 mass is suspended from it? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.17 A 6.2 mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in . You may want to review ( pages 255 – 257) . For help with math skills, you may want to review: Solving Algebraic Equations = 430 k Nm kg y = 19 cm kg Part A What does the spring scale read just before the mass touches the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass before it touches the scale. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? ANSWER: Correct Part B The scale reads 22 when the lower spring has been compressed by 2.7 . What is the value of the spring constant for the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? Use these to determine the force on the mass by the spring, taking note of the directions from your picture. How is the spring constant related to the force by the spring and the compression of the spring? Check your units. ANSWER: F = 61 N N cm k = 1400 k Nm Correct Part C At what compression length will the scale read zero? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces on the mass. When the scale reads zero, what is the force on the mass due to the scale? What is the gravitational force on the mass? What is the force on the mass by the spring? How is the compression length related to the force by the spring and the spring constant? Check your units. ANSWER: Correct Problem 10.18 Part A How far must you stretch a spring with = 800 to store 180 of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: y = 4.2 cm k N/m J Correct Problem 10.22 A 15 runaway grocery cart runs into a spring with spring constant 230 and compresses it by 57 . Part A What was the speed of the cart just before it hit the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Spring Gun A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measured from the equilibrium position of the spring. s = 0.67 m kg N/m cm v = 2.2 ms H Part A The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to . Hint 1. Potential energy of the spring The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial. Hint 2. Potential energy of the ball At the highest point in the ball’s trajectory, all of the spring’s potential energy has been converted into gravitational potential energy of the ball. ANSWER: Correct A Bullet Is Fired into a Wooden Block A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed . H height = H 4 mb vi mw vf Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of , , and . Hint 1. Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy vi mw mb ptotal vf ptotal = (mw + mb)vf ptotal vi ptotal = mbvi ANSWER: Correct Problem 10.31 Ball 1, with a mass of 150 and traveling at 15.0 , collides head on with ball 2, which has a mass of 340 and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Express your answer with the appropriate units. ANSWER: Correct Part B Express your answer with the appropriate units. ANSWER: Correct Part C vf = mb vi mb+mw g m/s g (vfx) = -5.82 1 ms (vfx) = 9.18 2 ms What are the final velocities of each ball if the collision is perfectly inelastic? Express your answer with the appropriate units. ANSWER: Correct Part D Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.43 A package of mass is released from rest at a warehouse loading dock and slides down the = 2.2 – high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. You may want to review ( pages 265 – 269) . For help with math skills, you may want to review: Solving Algebraic Equations (vfx) = 4.59 1 ms (vfx) = 4.59 2 ms m h m 2m Part A Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are two parts to this problem: the block sliding down the frictionless incline and the collision. What conservation laws are valid in each part? In terms of , what are the kinetic and potential energies of the block at the top of the incline? What is the potential energy of the same block at the bottom just before the collision? What are the kinetic energy and velocity of block just before the collision? What is conserved during the collision? What is the total momentum of the two blocks before the collision? What is the momentum of the two blocks stuck together after the collision? What is the velocity of the two blocks after the collision? ANSWER: Correct Part B Suppose the collision between the packages is perfectly elastic. To what height does the package of mass rebound? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are three parts to this problem: the block sliding down the incline, the collision, and mass going back up the incline. What conservation laws are valid in each part? m m v = 2.2 ms m m What is an elastic collision? For an elastic collision, how are the initial and final velocities related when one of the masses is initially at rest? Using the velocity of just before the collision from Part A, what is the velocity of just after the collision in this case? What are the kinetic and potential energies of mass just after the collision? What is the kinetic energy of mass at its maximum rebound height? Using conservation of energy, what is the potential energy of mass at its maximum height? What is the maximum height? ANSWER: Correct Problem 10.35 A cannon tilted up at a 35.0 angle fires a cannon ball at 79.0 from atop a 21.0 -high fortress wall. Part A What is the ball’s impact speed on the ground below? Express your answer with the appropriate units. ANSWER: Correct Problem 10.45 A 1000 safe is 2.5 above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 . m m m m m h = 24 cm $ m/s m vf = 81.6 ms kg m cm Part A What is the spring constant of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.49 A 100 block on a frictionless table is firmly attached to one end of a spring with = 21 . The other end of the spring is anchored to the wall. A 30 ball is thrown horizontally toward the block with a speed of 6.0 . Part A If the collision is perfectly elastic, what is the ball’s speed immediately after the collision? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the maximum compression of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.5×105 k Nm g k N/m g m/s v = 3.2 ms Correct Part C Repeat part A for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Repeat part B for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.4%. You received 120.28 out of a possible total of 121 points. x = 0.19 m v = 1.4 ms x = 0.11 m

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Chapter 10 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A One-Dimensional Inelastic Collision Block 1, of mass = 3.70 , moves along a frictionless air track with speed = 15.0 . It collides with block 2, of mass = 19.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: m1 kg v1 m/s m2 kg pi Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: Part C What is the change in the two-block system’s kinetic energy due to the collision? Express your answer numerically in joules. You did not open hints for this part. ANSWER: pi = kg m/s vf vf = m/s K = Kfinal − Kinitial K = J Conservation of Energy Ranking Task Six pendulums of various masses are released from various heights above a tabletop, as shown in the figures below. All the pendulums have the same length and are mounted such that at the vertical position their lowest points are the height of the tabletop and just do not strike the tabletop when released. Assume that the size of each bob is negligible. Part A Rank each pendulum on the basis of its initial gravitational potential energy (before being released) relative to the tabletop. Rank from largest to smallest To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: m h Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Momentum and Kinetic Energy Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed and has mass . Object 2 moves with speed and has mass . Part A Which object has the larger magnitude of its momentum? You did not open hints for this part. ANSWER: Part B Which object has the larger kinetic energy? You did not open hints for this part. ANSWER: v1 = v m1 = 2m v2 = 2v m2 = m Object 1 has the greater magnitude of its momentum. Object 2 has the greater magnitude of its momentum. Both objects have the same magnitude of their momenta. Object 1 has the greater kinetic energy. Object 2 has the greater kinetic energy. The objects have the same kinetic energy. Projectile Motion and Conservation of Energy Ranking Task Part A Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height above the ground. Assume that the effects of air resistance are negligible. Rank these throws according to the speed of the baseball the instant before it hits the ground. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: H PSS 10.1 Conservation of Mechanical Energy Learning Goal: To practice Problem-Solving Strategy 10.1 for conservation of mechanical energy problems. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 that makes an angle of 45 with the vertical, steps off his tree limb, and swings down and then up to Jane’s open arms. When he arrives, his vine makes an angle of 30 with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan’s speed just before he reaches Jane. You can ignore air resistance and the mass of the vine. PROBLEM-SOLVING STRATEGY 10.1 Conservation of mechanical energy MODEL: Choose a system without friction or other losses of mechanical energy. m VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you’re trying to find. SOLVE: The mathematical representation is based on the law of conservation of mechanical energy: . ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The problem does not involve friction, nor are there losses of mechanical energy, so conservation of mechanical energy applies. Model Tarzan and the vine as a pendulum. Visualize Part A Which of the following sketches can be used in drawing a before-and-after pictorial representation? ANSWER: Kf + Uf = Ki + Ui Solve Part B What is Tarzan’s speed just before he reaches Jane? Express your answer in meters per second to two significant figures. You did not open hints for this part. ANSWER: Assess Part C This question will be shown after you complete previous question(s). Bungee Jumping Diagram A Diagram B Diagram C Diagram D vf vf = m/s Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass , and the surface of the bridge is a height above the water. The bungee cord, which has length when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant . Kate doesn’t actually jump but simply steps off the edge of the bridge and falls straight downward. Kate’s height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use for the magnitude of the acceleration due to gravity. Part A How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn’t touch the water. Express the distance in terms of quantities given in the problem introduction. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Spinning Mass on a Spring An object of mass is attached to a spring with spring constant whose unstretched length is , and whose far end is fixed to a shaft that is rotating with angular speed . Neglect gravity and assume that the mass rotates with angular speed as shown. When solving this problem use an inertial coordinate system, as drawn here. m h L k g d = M k L Part A Given the angular speed , find the radius at which the mass rotates without moving toward or away from the origin. Express the radius in terms of , , , and . You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C R( ) k L M R( ) = This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). ± Baby Bounce with a Hooke One of the pioneers of modern science, Sir Robert Hooke (1635-1703), studied the elastic properties of springs and formulated the law that bears his name. Hooke found the relationship among the force a spring exerts, , the distance from equilibrium the end of the spring is displaced, , and a number called the spring constant (or, sometimes, the force constant of the spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or . In its scalar form, this equation is simply . The negative sign indicates that the force that the spring exerts and its displacement have opposite directions. The value of depends on the geometry and the material of the spring; it can be easily determined experimentally using this scalar equation. Toy makers have always been interested in springs for the entertainment value of the motion they produce. One well-known application is a baby bouncer,which consists of a harness seat for a toddler, attached to a spring. The entire contraption hooks onto the top of a doorway. The idea is for the baby to hang in the seat with his or her feet just touching the ground so that a good push up will get the baby bouncing, providing potentially hours of entertainment. F x k F = −kx F = −kx k Part A The following chart and accompanying graph depict an experiment to determine the spring constant for a baby bouncer. Displacement from equilibrium, ( ) Force exerted on the spring, ( ) 0 0 0.005 2.5 0.010 5.0 0.015 7.5 0.020 10 What is the spring constant of the spring being tested for the baby bouncer? Express your answer to two significant figures in newtons per meter. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Shooting a ball into a box Two children are trying to shoot a marble of mass into a small box using a spring-loaded gun that is fixed on a table and shoots horizontally from the edge of the table. The edge of the table is a height above the top of the box (the height of which is negligibly small), and the center of the box is a distance from the edge of the table. x m F N k k = N/m m H d The spring has a spring constant . The first child compresses the spring a distance and finds that the marble falls short of its target by a horizontal distance . Part A By what distance, , should the second child compress the spring so that the marble lands in the middle of the box? (Assume that height of the box is negligible, so that there is no chance that the marble will hit the side of the box before it lands in the bottom.) Express the distance in terms of , , , , and . You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). k x1 d12 x2 m k g H d x2 = Elastic Collision in One Dimension Block 1, of mass , moves across a frictionless surface with speed . It collides elastically with block 2, of mass , which is at rest ( ). After the collision, block 1 moves with speed , while block 2 moves with speed . Assume that , so that after the collision, the two objects move off in the direction of the first object before the collision. Part A This collision is elastic. What quantities, if any, are conserved in this collision? You did not open hints for this part. ANSWER: Part B What is the final speed of block 1? m1 ui m2 vi = 0 uf vf m1 > m2 kinetic energy only momentum only kinetic energy and momentum uf Express in terms of , , and . You did not open hints for this part. ANSWER: Part C What is the final speed of block 2? Express in terms of , , and . You did not open hints for this part. ANSWER: Ballistic Pendulum In a ballistic pendulum an object of mass is fired with an initial speed at a pendulum bob. The bob has a mass , which is suspended by a rod of length and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement as shown . uf m1 m2 ui uf = vf vf m1 m2 ui vf = m v0 M L Part A Find an expression for , the initial speed of the fired object. Express your answer in terms of some or all of the variables , , , and and the acceleration due to gravity, . You did not open hints for this part. ANSWER: Part B An experiment is done to compare the initial speed of bullets fired from different handguns: a 9.0 and a .44 caliber. The guns are fired into a 10- pendulum bob of length . Assume that the 9.0- bullet has a mass of 6.0 and the .44-caliber bullet has a mass of 12 . If the 9.0- bullet causes the pendulum to swing to a maximum angular displacement of 4.3 and the .44-caliber bullet causes a displacement of 10.1 , find the ratio of the initial speed of the 9.0- bullet to the speed of the .44-caliber bullet, . Express your answer numerically. You did not open hints for this part. ANSWER: v0 m M L g v0 = mm kg L mm g g mm mm (v /( 0 )9.0 v0)44 Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. (v0 )9.0/(v0 )44 =

Chapter 10 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A One-Dimensional Inelastic Collision Block 1, of mass = 3.70 , moves along a frictionless air track with speed = 15.0 . It collides with block 2, of mass = 19.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: m1 kg v1 m/s m2 kg pi Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: Part C What is the change in the two-block system’s kinetic energy due to the collision? Express your answer numerically in joules. You did not open hints for this part. ANSWER: pi = kg m/s vf vf = m/s K = Kfinal − Kinitial K = J Conservation of Energy Ranking Task Six pendulums of various masses are released from various heights above a tabletop, as shown in the figures below. All the pendulums have the same length and are mounted such that at the vertical position their lowest points are the height of the tabletop and just do not strike the tabletop when released. Assume that the size of each bob is negligible. Part A Rank each pendulum on the basis of its initial gravitational potential energy (before being released) relative to the tabletop. Rank from largest to smallest To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: m h Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Momentum and Kinetic Energy Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed and has mass . Object 2 moves with speed and has mass . Part A Which object has the larger magnitude of its momentum? You did not open hints for this part. ANSWER: Part B Which object has the larger kinetic energy? You did not open hints for this part. ANSWER: v1 = v m1 = 2m v2 = 2v m2 = m Object 1 has the greater magnitude of its momentum. Object 2 has the greater magnitude of its momentum. Both objects have the same magnitude of their momenta. Object 1 has the greater kinetic energy. Object 2 has the greater kinetic energy. The objects have the same kinetic energy. Projectile Motion and Conservation of Energy Ranking Task Part A Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height above the ground. Assume that the effects of air resistance are negligible. Rank these throws according to the speed of the baseball the instant before it hits the ground. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: H PSS 10.1 Conservation of Mechanical Energy Learning Goal: To practice Problem-Solving Strategy 10.1 for conservation of mechanical energy problems. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 that makes an angle of 45 with the vertical, steps off his tree limb, and swings down and then up to Jane’s open arms. When he arrives, his vine makes an angle of 30 with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan’s speed just before he reaches Jane. You can ignore air resistance and the mass of the vine. PROBLEM-SOLVING STRATEGY 10.1 Conservation of mechanical energy MODEL: Choose a system without friction or other losses of mechanical energy. m VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you’re trying to find. SOLVE: The mathematical representation is based on the law of conservation of mechanical energy: . ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The problem does not involve friction, nor are there losses of mechanical energy, so conservation of mechanical energy applies. Model Tarzan and the vine as a pendulum. Visualize Part A Which of the following sketches can be used in drawing a before-and-after pictorial representation? ANSWER: Kf + Uf = Ki + Ui Solve Part B What is Tarzan’s speed just before he reaches Jane? Express your answer in meters per second to two significant figures. You did not open hints for this part. ANSWER: Assess Part C This question will be shown after you complete previous question(s). Bungee Jumping Diagram A Diagram B Diagram C Diagram D vf vf = m/s Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass , and the surface of the bridge is a height above the water. The bungee cord, which has length when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant . Kate doesn’t actually jump but simply steps off the edge of the bridge and falls straight downward. Kate’s height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use for the magnitude of the acceleration due to gravity. Part A How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn’t touch the water. Express the distance in terms of quantities given in the problem introduction. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Spinning Mass on a Spring An object of mass is attached to a spring with spring constant whose unstretched length is , and whose far end is fixed to a shaft that is rotating with angular speed . Neglect gravity and assume that the mass rotates with angular speed as shown. When solving this problem use an inertial coordinate system, as drawn here. m h L k g d = M k L Part A Given the angular speed , find the radius at which the mass rotates without moving toward or away from the origin. Express the radius in terms of , , , and . You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C R( ) k L M R( ) = This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). ± Baby Bounce with a Hooke One of the pioneers of modern science, Sir Robert Hooke (1635-1703), studied the elastic properties of springs and formulated the law that bears his name. Hooke found the relationship among the force a spring exerts, , the distance from equilibrium the end of the spring is displaced, , and a number called the spring constant (or, sometimes, the force constant of the spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or . In its scalar form, this equation is simply . The negative sign indicates that the force that the spring exerts and its displacement have opposite directions. The value of depends on the geometry and the material of the spring; it can be easily determined experimentally using this scalar equation. Toy makers have always been interested in springs for the entertainment value of the motion they produce. One well-known application is a baby bouncer,which consists of a harness seat for a toddler, attached to a spring. The entire contraption hooks onto the top of a doorway. The idea is for the baby to hang in the seat with his or her feet just touching the ground so that a good push up will get the baby bouncing, providing potentially hours of entertainment. F x k F = −kx F = −kx k Part A The following chart and accompanying graph depict an experiment to determine the spring constant for a baby bouncer. Displacement from equilibrium, ( ) Force exerted on the spring, ( ) 0 0 0.005 2.5 0.010 5.0 0.015 7.5 0.020 10 What is the spring constant of the spring being tested for the baby bouncer? Express your answer to two significant figures in newtons per meter. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Shooting a ball into a box Two children are trying to shoot a marble of mass into a small box using a spring-loaded gun that is fixed on a table and shoots horizontally from the edge of the table. The edge of the table is a height above the top of the box (the height of which is negligibly small), and the center of the box is a distance from the edge of the table. x m F N k k = N/m m H d The spring has a spring constant . The first child compresses the spring a distance and finds that the marble falls short of its target by a horizontal distance . Part A By what distance, , should the second child compress the spring so that the marble lands in the middle of the box? (Assume that height of the box is negligible, so that there is no chance that the marble will hit the side of the box before it lands in the bottom.) Express the distance in terms of , , , , and . You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). k x1 d12 x2 m k g H d x2 = Elastic Collision in One Dimension Block 1, of mass , moves across a frictionless surface with speed . It collides elastically with block 2, of mass , which is at rest ( ). After the collision, block 1 moves with speed , while block 2 moves with speed . Assume that , so that after the collision, the two objects move off in the direction of the first object before the collision. Part A This collision is elastic. What quantities, if any, are conserved in this collision? You did not open hints for this part. ANSWER: Part B What is the final speed of block 1? m1 ui m2 vi = 0 uf vf m1 > m2 kinetic energy only momentum only kinetic energy and momentum uf Express in terms of , , and . You did not open hints for this part. ANSWER: Part C What is the final speed of block 2? Express in terms of , , and . You did not open hints for this part. ANSWER: Ballistic Pendulum In a ballistic pendulum an object of mass is fired with an initial speed at a pendulum bob. The bob has a mass , which is suspended by a rod of length and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement as shown . uf m1 m2 ui uf = vf vf m1 m2 ui vf = m v0 M L Part A Find an expression for , the initial speed of the fired object. Express your answer in terms of some or all of the variables , , , and and the acceleration due to gravity, . You did not open hints for this part. ANSWER: Part B An experiment is done to compare the initial speed of bullets fired from different handguns: a 9.0 and a .44 caliber. The guns are fired into a 10- pendulum bob of length . Assume that the 9.0- bullet has a mass of 6.0 and the .44-caliber bullet has a mass of 12 . If the 9.0- bullet causes the pendulum to swing to a maximum angular displacement of 4.3 and the .44-caliber bullet causes a displacement of 10.1 , find the ratio of the initial speed of the 9.0- bullet to the speed of the .44-caliber bullet, . Express your answer numerically. You did not open hints for this part. ANSWER: v0 m M L g v0 = mm kg L mm g g mm mm (v /( 0 )9.0 v0)44 Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. (v0 )9.0/(v0 )44 =

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Sex, Gender, and Popular Culture Spring 2015 Look through popular magazines, and see if you can find advertisements that objectify women in order to sell a product. Alternately, you may use an advertisement on television (but make sure to provide a link to the ad so I can see it!). Study these images then write a paper about objectification that deals with all or some of the following: • What effect(s), if any, do you think the objectification of women’s bodies has on our culture? • Jean Kilbourne states “turning a human being into a thing is almost always the first step toward justifying violence against that person.” What do you think she means by this? Do you agree with her reasoning? Why or why not? • Some people would argue that depicting a woman’s body as an object is a form of art. What is your opinion of this point of view? Explain your reasoning. • Why do you think that women are objectified more often than men are? • How does sexualization and objectification play out differently across racial lines? • Kilbourne explains that the consequences of being objectified are different – and more serious – for women than for men. Do you agree? How is the world different for women than it is for men? How do objectified images of women interact with those in our culture differently from the way images of men do? Why is it important to look at images in the context of the culture? • What is the difference between sexual objectification and sexual subjectification? (Ros Gill ) • How do ads construct violent white masculinity and how does that vision of masculinity hurt both men and women? Throughout your written analysis, be sure to make clear and specific reference to the images you selected, and please submit these images with your paper. Make sure you engage with and reference to at least 4 of the following authors: Kilbourne, Bordo, Hunter & Soto, Rose, Durham, Gill, Katz, Schuchardt, Ono and Buescher. Guidelines:  Keep your content focused on structural, systemic, institutional factors rather than the individual: BE ANALYTICAL NOT ANECDOTAL.  Avoid using the first person or including personal stories/reactions. You must make sure to actively engage with your readings: these essays need to be informed and framed by the theoretical material you have been reading this semester.  Keep within the 4-6 page limit; use 12-point font, double spacing and 1-inch margins.  Use formal writing conventions (introduction/thesis statement, body, conclusion) and correct grammar. Resources may be cited within the text of your paper, i.e. (Walters, 2013).

Sex, Gender, and Popular Culture Spring 2015 Look through popular magazines, and see if you can find advertisements that objectify women in order to sell a product. Alternately, you may use an advertisement on television (but make sure to provide a link to the ad so I can see it!). Study these images then write a paper about objectification that deals with all or some of the following: • What effect(s), if any, do you think the objectification of women’s bodies has on our culture? • Jean Kilbourne states “turning a human being into a thing is almost always the first step toward justifying violence against that person.” What do you think she means by this? Do you agree with her reasoning? Why or why not? • Some people would argue that depicting a woman’s body as an object is a form of art. What is your opinion of this point of view? Explain your reasoning. • Why do you think that women are objectified more often than men are? • How does sexualization and objectification play out differently across racial lines? • Kilbourne explains that the consequences of being objectified are different – and more serious – for women than for men. Do you agree? How is the world different for women than it is for men? How do objectified images of women interact with those in our culture differently from the way images of men do? Why is it important to look at images in the context of the culture? • What is the difference between sexual objectification and sexual subjectification? (Ros Gill ) • How do ads construct violent white masculinity and how does that vision of masculinity hurt both men and women? Throughout your written analysis, be sure to make clear and specific reference to the images you selected, and please submit these images with your paper. Make sure you engage with and reference to at least 4 of the following authors: Kilbourne, Bordo, Hunter & Soto, Rose, Durham, Gill, Katz, Schuchardt, Ono and Buescher. Guidelines:  Keep your content focused on structural, systemic, institutional factors rather than the individual: BE ANALYTICAL NOT ANECDOTAL.  Avoid using the first person or including personal stories/reactions. You must make sure to actively engage with your readings: these essays need to be informed and framed by the theoretical material you have been reading this semester.  Keep within the 4-6 page limit; use 12-point font, double spacing and 1-inch margins.  Use formal writing conventions (introduction/thesis statement, body, conclusion) and correct grammar. Resources may be cited within the text of your paper, i.e. (Walters, 2013).

The objectification of women has been a very controversial topic … Read More...

Assignment 12 Due: 11:59pm on Friday, May 9, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 15.6 A 0.600 -diameter vat of liquid is 2.30 deep. The pressure at the bottom of the vat is 1.30 . Part A What is the mass of the liquid in the vat? Express your answer with the appropriate units. ANSWER: Correct Problem 15.8 A 90-cm-thick layer of oil floats on a 160-cm-thick layer of water. Part A What is the pressure at the bottom of the water layer? Express your answer with the appropriate units. ANSWER: Correct m m atm 876 kg p = 1.25×105 Pa Problem 15.9 A research submarine has a 40.0 -diameter window 9.00 thick. The manufacturer says the window can withstand forces up to 1.20×106 . What is the submarine’s maximum safe depth? Part A The pressure inside the submarine is maintained at 1.0 atm. Express your answer with the appropriate units. ANSWER: Correct Problem 15.13 Part A What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 12 dog off the floor? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 15.40 The 78 student in the figure balances a 1100 elephant on a hydraulic lift. cm cm N 947 m kg d = 3.8 cm kg kg You may want to review ( pages 415 – 419) . For help with math skills, you may want to review: Rearrangement of Equations Involving Multiplication and Division Part A What is the diameter of the piston the student is standing on? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Given that the height of the fluid on the two sides is the same in the figure, how is the pressure of the fluid on the two sides related? What is the definition of pressure? What is the area of the right cylinder? What is the force exerted by the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? What is the additional pressure above atmospheric pressure under the student in the left cylinder? What is the force exerted by the student on the left cylinder? What is the area of the left cylinder? ANSWER: Correct Part B d = 0.53 m When a second student joins the first, the piston sinks 40 . What is the second student’s mass? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the definition of pressure? How is the height difference between the left and right cylinders related to the pressure difference in the two cylinders? What is the standard value for the density of the oil given in the text? What is the force due to the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? Given the height difference between the two cylinders and the pressure in the right cylinder, what is the pressure above atmospheric pressure in the left cylinder? What is the force due to both students on the left cylinder? What is the sum of the masses of the students? What is the mass of the second student? ANSWER: Correct Enhanced EOC: Problem 15.17 A 6.80 rock whose density is 4900 is suspended by a string such that half of the rock’s volume is under water. You may want to review ( pages 419 – 423) . For help with math skills, you may want to review: Conversion Factors Part A What is the tension in the string? Express your answer with the appropriate units. Hint 1. How to approach the problem cm m = 80 kg kg kg/m3 What are the three forces acting on the rock? Draw a picture indicating the direction of the forces on the rock and an appropriate coordinate system indicating the positive direction. How is volume related to mass and density? What is the volume of the rock? What is the buoyant force on the rock given that half of the rock is underwater? What is the gravitational force on the rock? Given that the rock is suspended, what is the net force on the rock? Now, determine the tension in the string. ANSWER: Correct Problem 15.15 A block floats in water with its long axis vertical. The length of the block above water is 1.0 . Part A What is the block’s mass density? Express your answer with the appropriate units. ANSWER: Correct Crown of Gold? According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crown that he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when 59.8 N 2.0 cm × 2.0 cm × 7.0 cm cm 857 kg m3 Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold. A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, , and the apparent weight of the crown when it is submerged in water, . See whether you can follow in Archimedes’ footsteps. The figure shows what is meant by weighing the crown while it is submerged in water. Part A Take the density of the crown to be . What is the ratio of the crown’s apparent weight (in water) to its actual weight ? Express your answer in terms of the density of the crown and the density of water . Hint 1. Find an expression for the actual weight of the crown Assume that the crown has volume . Find the actual weight of the crown. Express in terms of , (the magnitude of the acceleration due to gravity), and . ANSWER: Wactual Wapparent c Wapparent Wactual c w V Wactual Wactual V g c Wactual = cV g Hint 2. Find an expression for the apparent weight of the crown Assume that the crown has volume , and take the density of water to be . Find the apparent weight of the crown submerged in water. Express your answer in terms of , (the magnitude of the acceleration due to gravity), , and . Hint 1. How to approach the problem The apparent weight of the crown when it is submerged in water will be less than its actual weight (weight in air) due to the buoyant force, which opposes gravity. Hint 2. Find an algebraic expression for the buoyant force. Find the magnitude of the buoyant force on the crown when it is completely submerged in water. Express your answer in terms of , , and the gravitational acceleration . ANSWER: ANSWER: ANSWER: Correct Part B Imagine that the apparent weight of the crown in water is , and the actual weight is . Is the crown made of pure (100%) gold? The density of water is V w Wapparent V g w c Fbuoyant w V g Fbuoyant = wV g Wapparent = (c − w)gV = Wapparent Wactual 1 − w c Wapparent = 4.50 N Wactual = 5.00 N grams per cubic centimeter. The density of gold is grams per cubic centimeter. Hint 1. Find the ratio of weights for a crown of pure gold Given the expression you obtained for , what should the ratio of weights be if the crown is made of pure gold? Express your answer numerically, to two decimal places. ANSWER: ANSWER: Correct For the values given, , whereas for pure gold, . Thus, you can conclude that the the crown is not pure gold but contains some less-dense metal. The goldsmith made sure that the crown’s (true) weight was the same as that of the amount of gold he was allocated, but in so doing was forced to make the volume of the crown slightly larger than it would otherwise have been. Problem 15.23 A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 9.0 . Part A What is the speed in the second segment? Express your answer with the appropriate units. w = 1.00 g = 19.32 Wapparent Wactual = 0.95 Wapparent Wactual Yes No = 4.50/5.00 = 0.90 Wapparent Wactual = 1 − = 0.95 Wapparent Wactual w g m/s ANSWER: Correct Part B What is the speed in the third segment? Express your answer with the appropriate units. ANSWER: Correct Part C What is the volume flow rate through the pipe? Express your answer with the appropriate units. ANSWER: Correct Understanding Bernoulli’s Equation Bernoulli’s equation is a simple relation that can give useful insight into the balance among fluid pressure, flow speed, and elevation. It applies exclusively to ideal fluids with steady flow, that is, fluids with a constant density and no internal friction forces, whose flow patterns do not change with time. Despite its limitations, however, Bernoulli’s equation is an essential tool in understanding the behavior of fluids in many practical applications, from plumbing systems to the flight of airplanes. 2.25 ms 36.0 ms 7.07×10−4 m3 s For a fluid element of density that flows along a streamline, Bernoulli’s equation states that , where is the pressure, is the flow speed, is the height, is the acceleration due to gravity, and subscripts 1 and 2 refer to any two points along the streamline. The physical interpretation of Bernoulli’s equation becomes clearer if we rearrange the terms of the equation as follows: . The term on the left-hand side represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from point 1 to point 2. The two terms on the right-hand side represent, respectively, the change in potential energy, , and the change in kinetic energy, , of the unit volume during its flow from point 1 to point 2. In other words, Bernoulli’s equation states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occurs during the flow. This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing along a streamline. Part A Consider the portion of a flow tube shown in the figure. Point 1 and point 2 are at the same height. An ideal fluid enters the flow tube at point 1 and moves steadily toward point 2. If the cross section of the flow tube at point 1 is greater than that at point 2, what can you say about the pressure at point 2? Hint 1. How to approach the problem Apply Bernoulli’s equation to point 1 and to point 2. Since the points are both at the same height, their elevations cancel out in the equation and you are left with a relation between pressure and flow speeds. Even though the problem does not give direct information on the flow speed along the flow tube, it does tell you that the cross section of the flow tube decreases as the fluid flows toward point 2. Apply the continuity equation to points 1 and 2 and determine whether the flow speed at point 2 is greater than or smaller than the flow speed at point 1. With that information and Bernoulli’s equation, you will be able to determine the pressure at point 2 with respect to the pressure at point 1. Hint 2. Apply Bernoulli’s equation p1 +gh1 + = +g + 1 2 v21 p2 h2 1 2 v22 p v h g p1 − p2 = g(h2 −h1)+ ( − ) 1 2 v22 v21 p1 − p2 g(h2 − h1) 1 ( − ) 2 v22 v21 Apply Bernoulli’s equation to point 1 and to point 2 to complete the expression below. Here and are the pressure and flow speed, respectively, and subscripts 1 and 2 refer to point 1 and point 2. Also, use for elevation with the appropriate subscript, and use for the density of the fluid. Express your answer in terms of some or all of the variables , , , , , , and . Hint 1. Flow along a horizontal streamline Along a horizontal streamline, the change in potential energy of the flowing fluid is zero. In other words, when applying Bernoulli’s equation to any two points of the streamline, and they cancel out. ANSWER: Hint 3. Determine with respect to By applying the continuity equation, determine which of the following is true. Hint 1. The continuity equation The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube. It says that the amount of fluid flowing through a cross section of the tube in a time interval must be the same for all cross sections, or . Therefore, the flow speed must increase when the cross section of the flow tube decreases, and vice versa. ANSWER: p v h p1 v1 h1 p2 v2 h2 h1 = h2 p1 + = 1 2 v21 p2 + v2 2 2 v2 v1 $V A $t $V = = $t A1v1 A2v2 v2 > v1 v2 = v1 v2 < v1 ANSWER: Correct Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of an ideal fluid.When the cross section of the flow tube decreases, the flow speed increases, and therefore the pressure decreases. In other words, if , then and . Part B As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows through a flow tube with decreasing cross section moves toward a region of lower pressure. Physically, the pressure drop experienced by the fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to __________. Hint 1. Effects from conservation of mass Recall that, if the cross section of the flow tube varies, the flow speed must change to conserve mass. This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decrease speed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger cross section. ANSWER: Correct Part C Now assume that point 2 is at height with respect to point 1, as shown in the figure. The ends of the flow tube have the same areas as The pressure at point 2 is lower than the pressure at point 1. equal to the pressure at point 1. higher than the pressure at point 1. A2 < A1 v2 > v1 p2 < p1 A v decrease in speed increase in speed remain in equilibrium h the ends of the horizontal flow tube shown in Part A. Since the cross section of the flow tube is decreasing, Bernoulli's equation tells us that a fluid element flowing toward point 2 from point 1 moves toward a region of lower pressure. In this case, what is the pressure drop experienced by the fluid element? Hint 1. How to approach the problem Apply Bernoulli's equation to point 1 and to point 2, as you did in Part A. Note that this time you must take into account the difference in elevation between points 1 and 2. Do you need to add this additional term to the other term representing the pressure drop between the two ends of the flow tube or do you subtract it? ANSWER: Correct Part D From a physical point of view, how do you explain the fact that the pressure drop at the ends of the elevated flow tube from Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A? Hint 1. Physical meaning of the pressure drop in a tube As explained in the introduction, the difference in pressure between the ends of a flow tube represents the total work done on a unit volume of fluid by the pressure forces of the The pressure drop is smaller than the pressure drop occurring in a purely horizontal flow. equal to the pressure drop occurring in a purely horizontal flow. larger than the pressure drop occurring in a purely horizontal flow. p1 − p2 surrounding fluid to move that volume of fluid from one end to the other end of the flow tube. ANSWER: Correct In the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube had to balance only the increase in kinetic energy resulting from the acceleration of the fluid. In an elevated flow tube, the difference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure is needed for the flow to occur. Venturi Meter with Two Tubes A pair of vertical, open-ended glass tubes inserted into a horizontal pipe are often used together to measure flow velocity in the pipe, a configuration called a Venturi meter. Consider such an arrangement with a horizontal pipe carrying fluid of density . The fluid rises to heights and in the two open-ended tubes (see figure). The cross-sectional area of the pipe is at the position of tube 1, and at the position of tube 2. A greater amount of work is needed to balance the increase in potential energy from the elevation change. decrease in potential energy from the elevation change. larger increase in kinetic energy. larger decrease in kinetic energy. h1 h2 A1 A2 Part A Find , the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and , the magnitude of the acceleration due to gravity. Hint 1. How to approach the problem Use Bernoulli's law to compute the difference in pressure between the top and bottom of tube 1. The pressure at the top of the tube is defined to be atmospheric pressure. Note: Inside the tube, since the fluid is not flowing, the terms involving velocity in Bernoulli's equation can be ignored. Thus, Bernoulli's equation reduces to the formula for pressure as a function of depth in a fluid of uniform density. Hint 2. Simplified Bernoulli's equation For a fluid of uniform density that is not flowing, the pressure at a depth below the surface is given by , where is the pressure at the surface and is the magnitude of the acceleration due to gravity. ANSWER: Correct The fluid is pushed up tube 1 by the pressure of the fluid at the base of the tube, and not by its kinetic energy, since there is no streamline around the sharp edge of the tube. Thus energy is not conserved (there is turbulence at the edge of the tube) at the entrance of the tube. Since Bernoulli's law is essentially a statement of energy conservation, it must be applied separately to the fluid in the tube and the fluid flowing in the main pipe. However, the pressure in the fluid is the same just inside and just outside the tube. Part B Find , the speed of the fluid in the left end of the main pipe. Express your answer in terms of , , , and either and or , which is equal to . Hint 1. How to approach the problem Energy is conserved along the streamlines in the main flow. This means that Bernoulli's law can be applied to obtain a relationship between the fluid pressure and velocity at the bottom of p1 g p h p = p0 + gh p0 g p1 = gh1 v1 h1 h2 g A1 A2 A1 A2 tube 1, and the fluid pressure and velocity at the bottom of tube 2. Hint 2. Find in terms of What is , the pressure at the bottom of tube 2? Express your answer in terms of , , and any other given quantities. Hint 1. Recall Part A Obtain the solution for this part in the same way that you found an expression for in terms of in Part A of this problem. ANSWER: Hint 3. Find in terms of given quantities Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , , , and . Hint 1. Find the pressure at the bottom of tube 2 Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , and . ANSWER: Hint 2. Find in terms of The fluid is incompressible, so you can use the continuity equation to relate the fluid velocities and in terms of the geometry of the pipe. Find , the fluid velocity at the bottom p2 h2 p2 h2 g p1 h1 p2 = gh2 p2 p2 p1 v1 A1 A2 p2 p1 v1 v2 p2 = p1 + ( − ) 1 2 v1 2 v2 2 v2 v1 v1 v2 v2 of tube 2, in terms of . Your answer may include and , the cross-sectional areas of the pipe. ANSWER: ANSWER: ANSWER: Correct Note that this result depends on the difference between the heights of the fluid in the tubes, a quantity that is more easily measured than the heights themselves. Problem 15.39 The container shown in the figure is filled with oil. It is open to the atmosphere on the left. v1 A1 A2 v2 = A1 A2 v1 p2 = p1 + ( )(1 − ) 1 2 v1 2 ( ) A1 A2 2 v1 = 2g h1−h2 ( ) −1 A1 A2 2 −−−−−−−−−−−−−− Part A What is the pressure at point A? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the pressure difference between points A and B? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the pressure difference between points A and C? PA = 106 kPa PB − PA = 4.4 kPa Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 15.48 You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 32.4 . If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17 . Part A What is the density? Express your answer with the appropriate units. ANSWER: Correct Problem 15.60 Water flows from the pipe shown in the figure with a speed of 7.0 . PC − PA = 4.4 kPa N N statue = 2100 kg m3 m/s Part A What is the water pressure as it exits into the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the height of the standing column of water? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.9%. You received 93.92 out of a possible total of 94 points. P = 1.0×105 Pa h h = 5.9 m

Assignment 12 Due: 11:59pm on Friday, May 9, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 15.6 A 0.600 -diameter vat of liquid is 2.30 deep. The pressure at the bottom of the vat is 1.30 . Part A What is the mass of the liquid in the vat? Express your answer with the appropriate units. ANSWER: Correct Problem 15.8 A 90-cm-thick layer of oil floats on a 160-cm-thick layer of water. Part A What is the pressure at the bottom of the water layer? Express your answer with the appropriate units. ANSWER: Correct m m atm 876 kg p = 1.25×105 Pa Problem 15.9 A research submarine has a 40.0 -diameter window 9.00 thick. The manufacturer says the window can withstand forces up to 1.20×106 . What is the submarine’s maximum safe depth? Part A The pressure inside the submarine is maintained at 1.0 atm. Express your answer with the appropriate units. ANSWER: Correct Problem 15.13 Part A What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 12 dog off the floor? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 15.40 The 78 student in the figure balances a 1100 elephant on a hydraulic lift. cm cm N 947 m kg d = 3.8 cm kg kg You may want to review ( pages 415 – 419) . For help with math skills, you may want to review: Rearrangement of Equations Involving Multiplication and Division Part A What is the diameter of the piston the student is standing on? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Given that the height of the fluid on the two sides is the same in the figure, how is the pressure of the fluid on the two sides related? What is the definition of pressure? What is the area of the right cylinder? What is the force exerted by the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? What is the additional pressure above atmospheric pressure under the student in the left cylinder? What is the force exerted by the student on the left cylinder? What is the area of the left cylinder? ANSWER: Correct Part B d = 0.53 m When a second student joins the first, the piston sinks 40 . What is the second student’s mass? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the definition of pressure? How is the height difference between the left and right cylinders related to the pressure difference in the two cylinders? What is the standard value for the density of the oil given in the text? What is the force due to the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? Given the height difference between the two cylinders and the pressure in the right cylinder, what is the pressure above atmospheric pressure in the left cylinder? What is the force due to both students on the left cylinder? What is the sum of the masses of the students? What is the mass of the second student? ANSWER: Correct Enhanced EOC: Problem 15.17 A 6.80 rock whose density is 4900 is suspended by a string such that half of the rock’s volume is under water. You may want to review ( pages 419 – 423) . For help with math skills, you may want to review: Conversion Factors Part A What is the tension in the string? Express your answer with the appropriate units. Hint 1. How to approach the problem cm m = 80 kg kg kg/m3 What are the three forces acting on the rock? Draw a picture indicating the direction of the forces on the rock and an appropriate coordinate system indicating the positive direction. How is volume related to mass and density? What is the volume of the rock? What is the buoyant force on the rock given that half of the rock is underwater? What is the gravitational force on the rock? Given that the rock is suspended, what is the net force on the rock? Now, determine the tension in the string. ANSWER: Correct Problem 15.15 A block floats in water with its long axis vertical. The length of the block above water is 1.0 . Part A What is the block’s mass density? Express your answer with the appropriate units. ANSWER: Correct Crown of Gold? According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crown that he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when 59.8 N 2.0 cm × 2.0 cm × 7.0 cm cm 857 kg m3 Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold. A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, , and the apparent weight of the crown when it is submerged in water, . See whether you can follow in Archimedes’ footsteps. The figure shows what is meant by weighing the crown while it is submerged in water. Part A Take the density of the crown to be . What is the ratio of the crown’s apparent weight (in water) to its actual weight ? Express your answer in terms of the density of the crown and the density of water . Hint 1. Find an expression for the actual weight of the crown Assume that the crown has volume . Find the actual weight of the crown. Express in terms of , (the magnitude of the acceleration due to gravity), and . ANSWER: Wactual Wapparent c Wapparent Wactual c w V Wactual Wactual V g c Wactual = cV g Hint 2. Find an expression for the apparent weight of the crown Assume that the crown has volume , and take the density of water to be . Find the apparent weight of the crown submerged in water. Express your answer in terms of , (the magnitude of the acceleration due to gravity), , and . Hint 1. How to approach the problem The apparent weight of the crown when it is submerged in water will be less than its actual weight (weight in air) due to the buoyant force, which opposes gravity. Hint 2. Find an algebraic expression for the buoyant force. Find the magnitude of the buoyant force on the crown when it is completely submerged in water. Express your answer in terms of , , and the gravitational acceleration . ANSWER: ANSWER: ANSWER: Correct Part B Imagine that the apparent weight of the crown in water is , and the actual weight is . Is the crown made of pure (100%) gold? The density of water is V w Wapparent V g w c Fbuoyant w V g Fbuoyant = wV g Wapparent = (c − w)gV = Wapparent Wactual 1 − w c Wapparent = 4.50 N Wactual = 5.00 N grams per cubic centimeter. The density of gold is grams per cubic centimeter. Hint 1. Find the ratio of weights for a crown of pure gold Given the expression you obtained for , what should the ratio of weights be if the crown is made of pure gold? Express your answer numerically, to two decimal places. ANSWER: ANSWER: Correct For the values given, , whereas for pure gold, . Thus, you can conclude that the the crown is not pure gold but contains some less-dense metal. The goldsmith made sure that the crown’s (true) weight was the same as that of the amount of gold he was allocated, but in so doing was forced to make the volume of the crown slightly larger than it would otherwise have been. Problem 15.23 A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 9.0 . Part A What is the speed in the second segment? Express your answer with the appropriate units. w = 1.00 g = 19.32 Wapparent Wactual = 0.95 Wapparent Wactual Yes No = 4.50/5.00 = 0.90 Wapparent Wactual = 1 − = 0.95 Wapparent Wactual w g m/s ANSWER: Correct Part B What is the speed in the third segment? Express your answer with the appropriate units. ANSWER: Correct Part C What is the volume flow rate through the pipe? Express your answer with the appropriate units. ANSWER: Correct Understanding Bernoulli’s Equation Bernoulli’s equation is a simple relation that can give useful insight into the balance among fluid pressure, flow speed, and elevation. It applies exclusively to ideal fluids with steady flow, that is, fluids with a constant density and no internal friction forces, whose flow patterns do not change with time. Despite its limitations, however, Bernoulli’s equation is an essential tool in understanding the behavior of fluids in many practical applications, from plumbing systems to the flight of airplanes. 2.25 ms 36.0 ms 7.07×10−4 m3 s For a fluid element of density that flows along a streamline, Bernoulli’s equation states that , where is the pressure, is the flow speed, is the height, is the acceleration due to gravity, and subscripts 1 and 2 refer to any two points along the streamline. The physical interpretation of Bernoulli’s equation becomes clearer if we rearrange the terms of the equation as follows: . The term on the left-hand side represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from point 1 to point 2. The two terms on the right-hand side represent, respectively, the change in potential energy, , and the change in kinetic energy, , of the unit volume during its flow from point 1 to point 2. In other words, Bernoulli’s equation states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occurs during the flow. This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing along a streamline. Part A Consider the portion of a flow tube shown in the figure. Point 1 and point 2 are at the same height. An ideal fluid enters the flow tube at point 1 and moves steadily toward point 2. If the cross section of the flow tube at point 1 is greater than that at point 2, what can you say about the pressure at point 2? Hint 1. How to approach the problem Apply Bernoulli’s equation to point 1 and to point 2. Since the points are both at the same height, their elevations cancel out in the equation and you are left with a relation between pressure and flow speeds. Even though the problem does not give direct information on the flow speed along the flow tube, it does tell you that the cross section of the flow tube decreases as the fluid flows toward point 2. Apply the continuity equation to points 1 and 2 and determine whether the flow speed at point 2 is greater than or smaller than the flow speed at point 1. With that information and Bernoulli’s equation, you will be able to determine the pressure at point 2 with respect to the pressure at point 1. Hint 2. Apply Bernoulli’s equation p1 +gh1 + = +g + 1 2 v21 p2 h2 1 2 v22 p v h g p1 − p2 = g(h2 −h1)+ ( − ) 1 2 v22 v21 p1 − p2 g(h2 − h1) 1 ( − ) 2 v22 v21 Apply Bernoulli’s equation to point 1 and to point 2 to complete the expression below. Here and are the pressure and flow speed, respectively, and subscripts 1 and 2 refer to point 1 and point 2. Also, use for elevation with the appropriate subscript, and use for the density of the fluid. Express your answer in terms of some or all of the variables , , , , , , and . Hint 1. Flow along a horizontal streamline Along a horizontal streamline, the change in potential energy of the flowing fluid is zero. In other words, when applying Bernoulli’s equation to any two points of the streamline, and they cancel out. ANSWER: Hint 3. Determine with respect to By applying the continuity equation, determine which of the following is true. Hint 1. The continuity equation The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube. It says that the amount of fluid flowing through a cross section of the tube in a time interval must be the same for all cross sections, or . Therefore, the flow speed must increase when the cross section of the flow tube decreases, and vice versa. ANSWER: p v h p1 v1 h1 p2 v2 h2 h1 = h2 p1 + = 1 2 v21 p2 + v2 2 2 v2 v1 $V A $t $V = = $t A1v1 A2v2 v2 > v1 v2 = v1 v2 < v1 ANSWER: Correct Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of an ideal fluid.When the cross section of the flow tube decreases, the flow speed increases, and therefore the pressure decreases. In other words, if , then and . Part B As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows through a flow tube with decreasing cross section moves toward a region of lower pressure. Physically, the pressure drop experienced by the fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to __________. Hint 1. Effects from conservation of mass Recall that, if the cross section of the flow tube varies, the flow speed must change to conserve mass. This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decrease speed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger cross section. ANSWER: Correct Part C Now assume that point 2 is at height with respect to point 1, as shown in the figure. The ends of the flow tube have the same areas as The pressure at point 2 is lower than the pressure at point 1. equal to the pressure at point 1. higher than the pressure at point 1. A2 < A1 v2 > v1 p2 < p1 A v decrease in speed increase in speed remain in equilibrium h the ends of the horizontal flow tube shown in Part A. Since the cross section of the flow tube is decreasing, Bernoulli's equation tells us that a fluid element flowing toward point 2 from point 1 moves toward a region of lower pressure. In this case, what is the pressure drop experienced by the fluid element? Hint 1. How to approach the problem Apply Bernoulli's equation to point 1 and to point 2, as you did in Part A. Note that this time you must take into account the difference in elevation between points 1 and 2. Do you need to add this additional term to the other term representing the pressure drop between the two ends of the flow tube or do you subtract it? ANSWER: Correct Part D From a physical point of view, how do you explain the fact that the pressure drop at the ends of the elevated flow tube from Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A? Hint 1. Physical meaning of the pressure drop in a tube As explained in the introduction, the difference in pressure between the ends of a flow tube represents the total work done on a unit volume of fluid by the pressure forces of the The pressure drop is smaller than the pressure drop occurring in a purely horizontal flow. equal to the pressure drop occurring in a purely horizontal flow. larger than the pressure drop occurring in a purely horizontal flow. p1 − p2 surrounding fluid to move that volume of fluid from one end to the other end of the flow tube. ANSWER: Correct In the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube had to balance only the increase in kinetic energy resulting from the acceleration of the fluid. In an elevated flow tube, the difference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure is needed for the flow to occur. Venturi Meter with Two Tubes A pair of vertical, open-ended glass tubes inserted into a horizontal pipe are often used together to measure flow velocity in the pipe, a configuration called a Venturi meter. Consider such an arrangement with a horizontal pipe carrying fluid of density . The fluid rises to heights and in the two open-ended tubes (see figure). The cross-sectional area of the pipe is at the position of tube 1, and at the position of tube 2. A greater amount of work is needed to balance the increase in potential energy from the elevation change. decrease in potential energy from the elevation change. larger increase in kinetic energy. larger decrease in kinetic energy. h1 h2 A1 A2 Part A Find , the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and , the magnitude of the acceleration due to gravity. Hint 1. How to approach the problem Use Bernoulli's law to compute the difference in pressure between the top and bottom of tube 1. The pressure at the top of the tube is defined to be atmospheric pressure. Note: Inside the tube, since the fluid is not flowing, the terms involving velocity in Bernoulli's equation can be ignored. Thus, Bernoulli's equation reduces to the formula for pressure as a function of depth in a fluid of uniform density. Hint 2. Simplified Bernoulli's equation For a fluid of uniform density that is not flowing, the pressure at a depth below the surface is given by , where is the pressure at the surface and is the magnitude of the acceleration due to gravity. ANSWER: Correct The fluid is pushed up tube 1 by the pressure of the fluid at the base of the tube, and not by its kinetic energy, since there is no streamline around the sharp edge of the tube. Thus energy is not conserved (there is turbulence at the edge of the tube) at the entrance of the tube. Since Bernoulli's law is essentially a statement of energy conservation, it must be applied separately to the fluid in the tube and the fluid flowing in the main pipe. However, the pressure in the fluid is the same just inside and just outside the tube. Part B Find , the speed of the fluid in the left end of the main pipe. Express your answer in terms of , , , and either and or , which is equal to . Hint 1. How to approach the problem Energy is conserved along the streamlines in the main flow. This means that Bernoulli's law can be applied to obtain a relationship between the fluid pressure and velocity at the bottom of p1 g p h p = p0 + gh p0 g p1 = gh1 v1 h1 h2 g A1 A2 A1 A2 tube 1, and the fluid pressure and velocity at the bottom of tube 2. Hint 2. Find in terms of What is , the pressure at the bottom of tube 2? Express your answer in terms of , , and any other given quantities. Hint 1. Recall Part A Obtain the solution for this part in the same way that you found an expression for in terms of in Part A of this problem. ANSWER: Hint 3. Find in terms of given quantities Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , , , and . Hint 1. Find the pressure at the bottom of tube 2 Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , and . ANSWER: Hint 2. Find in terms of The fluid is incompressible, so you can use the continuity equation to relate the fluid velocities and in terms of the geometry of the pipe. Find , the fluid velocity at the bottom p2 h2 p2 h2 g p1 h1 p2 = gh2 p2 p2 p1 v1 A1 A2 p2 p1 v1 v2 p2 = p1 + ( − ) 1 2 v1 2 v2 2 v2 v1 v1 v2 v2 of tube 2, in terms of . Your answer may include and , the cross-sectional areas of the pipe. ANSWER: ANSWER: ANSWER: Correct Note that this result depends on the difference between the heights of the fluid in the tubes, a quantity that is more easily measured than the heights themselves. Problem 15.39 The container shown in the figure is filled with oil. It is open to the atmosphere on the left. v1 A1 A2 v2 = A1 A2 v1 p2 = p1 + ( )(1 − ) 1 2 v1 2 ( ) A1 A2 2 v1 = 2g h1−h2 ( ) −1 A1 A2 2 −−−−−−−−−−−−−− Part A What is the pressure at point A? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the pressure difference between points A and B? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the pressure difference between points A and C? PA = 106 kPa PB − PA = 4.4 kPa Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 15.48 You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 32.4 . If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17 . Part A What is the density? Express your answer with the appropriate units. ANSWER: Correct Problem 15.60 Water flows from the pipe shown in the figure with a speed of 7.0 . PC − PA = 4.4 kPa N N statue = 2100 kg m3 m/s Part A What is the water pressure as it exits into the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the height of the standing column of water? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.9%. You received 93.92 out of a possible total of 94 points. P = 1.0×105 Pa h h = 5.9 m

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The “deadman” lever controls the augers of a snowthrower. If force F = 30 N is required to begin rotating the lever about the ideal pivot at O, determine the corresponding tension T in the control cable AB and the magnitude of the pin reaction at O. Neglect the weight of the lever.

The “deadman” lever controls the augers of a snowthrower. If force F = 30 N is required to begin rotating the lever about the ideal pivot at O, determine the corresponding tension T in the control cable AB and the magnitude of the pin reaction at O. Neglect the weight of the lever.

The “deadman” lever controls the augers of a snowthrower. If … Read More...

Chapter 12 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, May 16, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Spinning Grinding Wheel At time a grinding wheel has an angular velocity of 26.0 . It has a constant angular acceleration of 33.0 until a circuit breaker trips at time = 1.80 . From then on, the wheel turns through an angle of 432 as it coasts to a stop at constant angular deceleration. Part A Through what total angle did the wheel turn between and the time it stopped? Express your answer in radians. You did not open hints for this part. ANSWER: Part B At what time does the wheel stop? Express your answer in seconds. You did not open hints for this part. ANSWER: t = 0 rad/s rad/s2 t s rad t = 0 rad Part C What was the wheel’s angular acceleration as it slowed down? Express your answer in radians per second per second. You did not open hints for this part. ANSWER: An Exhausted Bicyclist An exhausted bicyclist pedals somewhat erratically when exercising on a static bicycle. The angular velocity of the wheels follows the equation , where represents time (measured in seconds), = 0.500 , = 0.250 and = 2.00 . Part A There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time to be at angle radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel’s rotation. What angular displacement has the spot of paint undergone between time 0 and 2 seconds? Express your answer in radians using three significant figures. s rad/s2 (t) = at − bsin(ct) for t 0 t a rad/s2 b rad/s c rad/s t = 0 = 0 Typesetting math: 29% You did not open hints for this part. ANSWER: Part B Express the angular displacement undergone by the spot of paint at seconds in degrees. Remember to use the unrounded value from Part A, should you need it. Express your answer in degrees using three significant figures. You did not open hints for this part. ANSWER: Part C What distance has the spot of paint moved in 2 seconds if the radius of the wheel is 50 centimeters? Express your answer in centimeters using three significant figures. You did not open hints for this part. ANSWER: = rad t = 2 = d Typesetting math: 29% Part D Which one of the following statements describes the motion of the spot of paint at seconds? You did not open hints for this part. ANSWER: Flywheel Kinematics A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration . The flywheel is assumed to be at rest at time in Parts A and B of this problem. Part A Find the time it takes to accelerate the flywheel to if the angular acceleration is . Express your answer in terms of and . d = cm t = 2.0 The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is increasing. t = 0 t1 1 1 Typesetting math: 29% You did not open hints for this part. ANSWER: Part B Find the angle through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity . Express your answer in terms of some or all of the following: , , and . You did not open hints for this part. ANSWER: Part C Assume that the motor has accelerated the wheel up to an angular velocity with angular acceleration in time . At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of . Find , the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity). Express your answer in terms of some or all of the following: , \texttip{\alpha }{alpha}, and \texttip{t_{\rm 1}}{t_1}. You did not open hints for this part. t1 = 1 1 1 t1 1 = 1 t1 −5 t2 1 Typesetting math: 29% ANSWER: Surprising Exploding Firework A mortar fires a shell of mass \texttip{m}{m} at speed \texttip{v_{\rm 0}}{v_0}. The shell explodes at the top of its trajectory (shown by a star in the figure) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass \large{\frac15m} and a larger piece of mass \large{\frac45m}. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance \texttip{d}{d} from the mortar. If there had been no explosion, the shell would have landed a distance \texttip{r}{r} from the mortar. Assume that air resistance and the mass of the shell’s explosive charge are negligible. Part A Find the distance \texttip{d}{d} from the mortar at which the larger piece of the shell lands. Express \texttip{d}{d} in terms of \texttip{r}{r}. You did not open hints for this part. \texttip{t_{\rm 2}}{t_2} = s Typesetting math: 29% ANSWER: Kinetic Energy of a Dumbbell This problem illustrates the two contributions to the kinetic energy of an extended object: rotational kinetic energy and translational kinetic energy. You are to find the total kinetic energy \texttip{K_{\rm total}}{K_total} of a dumbbell of mass \texttip{m}{m} when it is rotating with angular speed \texttip{\omega }{omega} and its center of mass is moving translationally with speed \texttip{v}{v}. Denote the dumbbell’s moment of inertia about its center of mass by \texttip{I_{\rm cm}}{I_cm}. Note that if you approximate the spheres as point masses of mass m/2 each located a distance \texttip{r}{r} from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by I_{\rm cm} = mr^2, but this fact will not be necessary for this problem. Part A Find the total kinetic energy \texttip{K_{\rm tot}}{K_tot} of the dumbbell. Express your answer in terms of \texttip{m}{m}, \texttip{v}{v}, \texttip{I_{\rm cm}}{I_cm}, and \texttip{\omega }{omega}. You did not open hints for this part. \texttip{d}{d} = Typesetting math: 29% ANSWER: Part B This question will be shown after you complete previous question(s). Unwinding Cylinder A cylinder with moment of inertia \texttip{I}{I} about its center of mass, mass \texttip{m}{m}, and radius \texttip{r}{r} has a string wrapped around it which is tied to the ceiling . The cylinder’s vertical position as a function of time is y(t). At time t = 0 the cylinder is released from rest at a height \texttip{h}{h} above the ground. Part A The string constrains the rotational and translational motion of the cylinder. What is the relationship between the angular rotation rate \texttip{\omega }{omega} and \texttip{v}{v}, the velocity of the center of mass of the cylinder? \texttip{K_{\rm tot}}{K_tot} = Typesetting math: 29% Remember that upward motion corresponds to positive linear velocity, and counterclockwise rotation corresponds to positive angular velocity. Express \texttip{\omega }{omega} in terms of \texttip{v}{v} and other given quantities. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C Suppose that at a certain instant the velocity of the cylinder is \texttip{v}{v}. What is its total kinetic energy, \texttip{K_{\rm total}}{K_total}, at that instant? Express \texttip{K_{\rm total}}{K_total} in terms of \texttip{m}{m}, \texttip{r}{r}, \texttip{I}{I}, and \texttip{v}{v}. You did not open hints for this part. ANSWER: Part D \texttip{\omega }{omega} = \texttip{K_{\rm total}}{K_total} = Typesetting math: 29% Find \texttip{v_{\rm f \hspace{1 pt}}}{v_f}, the cylinder’s vertical velocity when it hits the ground. Express \texttip{v_{\rm f \hspace{1 pt}}}{v_f}, in terms of \texttip{g}{g}, \texttip{h}{h}, \texttip{I}{I}, \texttip{m}{m}, and \texttip{r}{r}. You did not open hints for this part. ANSWER: Kinetic Energy of a Rotating Wheel A simple wheel has the form of a solid cylinder of radius \texttip{r}{r} with a mass \texttip{m}{m} uniformly distributed throughout its volume. The wheel is pivoted on a stationary axle through the axis of the cylinder and rotates about the axle at a constant angular speed. The wheel rotates \texttip{n}{n} full revolutions in a time interval \texttip{t}{t}. Part A What is the kinetic energy \texttip{K}{K} of the rotating wheel? Express your answer in terms of \texttip{m}{m}, \texttip{r}{r}, \texttip{n}{n}, \texttip{t}{t} and, \texttip{\pi }{pi}. You did not open hints for this part. ANSWER: Finding Torque \texttip{v_{\rm f \hspace{1 pt}}}{v_f} = \texttip{K}{K} = Typesetting math: 29% A force \texttip{\vec{F}}{F_vec} of magnitude \texttip{F}{F} making an angle \texttip{\theta }{theta} with the x axis is applied to a particle located along axis of rotation A, at Cartesian coordinates (0, 0) in the figure. The vector \texttip{\vec{F}}{F_vec} lies in the xy plane, and the four axes of rotation A, B, C, and D all lie perpendicular to the xy plane. A particle is located at a vector position \texttip{\vec{r}}{r_vec} with respect to an axis of rotation (thus \texttip{\vec{r}}{r_vec} points from the axis to the point at which the particle is located). The magnitude of the torque \texttip{\tau }{tau} about this axis due to a force \texttip{\vec{F}}{F_vec} acting on the particle is given by \tau = r F \sin(\alpha) = ({\rm moment \; arm}) \cdot F = rF_{\perp}, where \texttip{\alpha }{alpha} is the angle between \texttip{\vec{r}}{r_vec} and \texttip{\vec{F}}{F_vec}, \texttip{r}{r} is the magnitude of \texttip{\vec{r}}{r_vec}, \texttip{F}{F} is the magnitude of \texttip{\vec{F}}{F_vec}, the component of \texttip{\vec{r}}{r_vec} that is perpendicualr to \texttip{\vec{F}}{F_vec} is the moment arm, and \texttip{F_{\rm \perp}}{F_\perp} is the component of the force that is perpendicular to \texttip{\vec{r}}{r_vec}. Sign convention: You will need to determine the sign by analyzing the direction of the rotation that the torque would tend to produce. Recall that negative torque about an axis corresponds to clockwise rotation. In this problem, you must express the angle \texttip{\alpha }{alpha} in the above equation in terms of \texttip{\theta }{theta}, \texttip{\phi }{phi}, and/or \texttip{\pi }{pi} when entering your answers. Keep in mind that \pi = 180\;\rm degrees and (\pi/2) = 90\;\rm degrees . Part A What is the torque \texttip{\tau_{\rm A}}{tau_A} about axis A due to the force \texttip{\vec{F}}{F_vec}? Express the torque about axis A at Cartesian coordinates (0, 0). You did not open hints for this part. Typesetting math: 29% ANSWER: Part B What is the torque \texttip{\tau_{\rm B}}{tau_B} about axis B due to the force \texttip{\vec{F}}{F_vec}? (B is the point at Cartesian coordinates (0, b), located a distance \texttip{b}{b} from the origin along the y axis.) Express the torque about axis B in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. You did not open hints for this part. ANSWER: Part C What is the torque \texttip{\tau_{\rm C}}{tau_C} about axis C due to \texttip{\vec{F}}{F_vec}? (C is the point at Cartesian coordinates (c, 0), a distance \texttip{c}{c} along the x axis.) Express the torque about axis C in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. You did not open hints for this part. ANSWER: \texttip{\tau_{\rm A}}{tau_A} = \texttip{\tau_{\rm B}}{tau_B} = Typesetting math: 29% Part D What is the torque \texttip{\tau_{\rm D}}{tau_D} about axis D due to \texttip{\vec{F}}{F_vec}? (D is the point located at a distance \texttip{d}{d} from the origin and making an angle \texttip{\phi }{phi} with the x axis.) Express the torque about axis D in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. ANSWER: Torque Magnitude Ranking Task The wrench in the figure has six forces of equal magnitude acting on it. \texttip{\tau_{\rm C}}{tau_C} = \texttip{\tau_{\rm D}}{tau_D} = Typesetting math: 29% Part A Rank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench, measured about an axis centered on the bolt. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: The Parallel-Axis Theorem Typesetting math: 29% Learning Goal: To understand the parallel-axis theorem and its applications To solve many problems about rotational motion, it is important to know the moment of inertia of each object involved. Calculating the moments of inertia of various objects, even highly symmetrical ones, may be a lengthy and tedious process. While it is important to be able to calculate moments of inertia from the definition (I=\sum m_ir_i^2), in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of frequently occurring shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily available from any mechanics text, including your textbook. However, one must take into account that an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of tranlsational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moments of inertia that you can find in the textbooks are usually calculated with respect to an axis passing through the center of mass of the object. However, in many problems the axis of rotation does not pass through the center of mass. Does that mean that one has to go through the lengthy process of finding the moment of inertia from scratch? It turns out that in many cases, calculating the moment of inertia can be done rather easily if one uses the parallel-axis theorem. Mathematically, it can be expressed as I=I_{\rm cm}+md^2, where \texttip{I_{\rm cm}}{I_cm} is the moment of inertia about an axis passing through the center of mass, \texttip{m}{m} is the total mass of the object, and \texttip{I}{I} is the moment of inertia about another axis, parallel to the one for which \texttip{I_{\rm cm}}{I_cm} is calculated and located a distance \texttip{d}{d} from the center of mass. In this problem you will show that the theorem does indeed work for at least one object: a dumbbell of length \texttip{2r}{2r} made of two small spheres of mass \texttip{m}{m} each connected by a light rod (see the figure). NOTE: Unless otherwise noted, all axes considered are perpendicular to the plane of the page. Part A Using the definition of moment of inertia, calculate I_{\rm cm}, the moment of inertia about the center of mass, for this object. Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B Using the definition of moment of inertia, calculate I_{\rm B}, the moment of inertia about an axis through point B, for this object. Point B coincides with (the center of) one of the spheres (see the figure). Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. ANSWER: Part C Now calculate I_{\rm B} for this object using the parallel-axis theorem. Express your answer in terms of \texttip{I_{\rm cm}}{I_cm}, \texttip{m}{m}, and \texttip{r}{r}. ANSWER: I_{\rm cm} = I_{\rm B} = I_{\rm B} = Typesetting math: 29% Part D Using the definition of moment of inertia, calculate I_{\rm C}, the moment of inertia about an axis through point C, for this object. Point C is located a distance \texttip{r}{r} from the center of mass (see the figure). Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. ANSWER: Part E Now calculate I_{\rm C} for this object using the parallel-axis theorem. Express your answer in terms of \texttip{I_{\rm cm}}{I_cm}, \texttip{m}{m}, and \texttip{r}{r}. ANSWER: Consider an irregular object of mass \texttip{m}{m}. Its moment of inertia measured with respect to axis A (parallel to the plane of the page), which passes through the center of mass (see the second diagram), is given by I_{\rm A}=0.64mr^2. Axes B, C, D, and E are parallel to axis A; their separations from axis A are shown in the diagram. In the subsequent questions, the subscript indicates the axis with respect to which the moment of inertia is measured: for instance, I_{\rm C} is the moment of inertia about axis C. I_{\rm C} = I_{\rm C} = Typesetting math: 29% Part F Which moment of inertia is the smallest? ANSWER: Part G Which moment of inertia is the largest? ANSWER: I_{\rm A} I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} I_{\rm A} I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} Typesetting math: 29% Part H Which moments of inertia are equal? ANSWER: Part I Which moment of inertia equals 4.64mr^2? ANSWER: Part J Axis X, not shown in the diagram, is parallel to the axes shown. It is known that I_{\rm X}=6mr^2. Which of the following is a possible location for axis X? ANSWER: I_{\rm A} and I_{\rm D} I_{\rm B} and I_{\rm C} I_{\rm C} and I_{\rm E} No two moments of inertia are equal. I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} between axes A and C between axes C and D between axes D and E to the right of axis E Typesetting math: 29% Torque and Angular Acceleration Learning Goal: To understand and apply the formula \tau=I\alpha to rigid objects rotating about a fixed axis. To find the acceleration \texttip{a}{a} of a particle of mass \texttip{m}{m}, we use Newton’s second law: \vec {F}_{\rm net}=m\vec{a}, where \texttip{\vec{F}_{\rm net}}{F_vec_net} is the net force acting on the particle. To find the angular acceleration \texttip{\alpha }{alpha} of a rigid object rotating about a fixed axis, we can use a similar formula: \tau_{\rm net}=I\alpha, where \tau_{\rm net}=\sum \tau is the net torque acting on the object and \texttip{I}{I} is its moment of inertia. In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses \texttip{m_{\rm 1}}{m_1} and \texttip{m_{\rm 2}}{m_2} are attached to a seesaw. The seesaw is made of a bar that has length \texttip{l}{l} and is pivoted so that it is free to rotate in the vertical plane without friction. You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m_1>m_2, and that counterclockwise is considered the positive rotational direction. Part A The seesaw is pivoted in the middle, and the mass of the swing bar is negligible. Find the angular acceleration \texttip{\alpha }{alpha} of the seesaw. Express your answer in terms of some or all of the quantities \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{l}{l}, as well as the acceleration due to gravity \texttip{g}{g}. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B In what direction will the seesaw rotate, and what will the sign of the angular acceleration be? ANSWER: Part C This question will be shown after you complete previous question(s). \texttip{\alpha }{alpha} = The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Typesetting math: 29% Part D In what direction will the seesaw rotate and what will the sign of the angular acceleration be? ANSWER: Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Pivoted Rod with Unequal Masses The figure shows a simple model of a seesaw. These consist of a plank/rod of mass \texttip{m_{\rm r}}{m_r} and length 2x allowed to pivot freely about its center (or central axis), as shown in the diagram. A small sphere of mass \texttip{m_{\rm 1}}{m_1} is attached to the left end of the rod, and a small sphere of mass \texttip{m_{\rm 2}}{m_2} is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to \texttip{g}{g}. The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Typesetting math: 29% Part A What is the moment of inertia \texttip{I}{I} of this assembly about the axis through which it is pivoted? Express the moment of inertia in terms of \texttip{m_{\rm r}}{m_r}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{x}{x}. Keep in mind that the length of the rod is 2x, not \texttip{x}{x}. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Weight and Wheel Consider a bicycle wheel that initially is not rotating. A block of mass \texttip{m}{m} is attached to the wheel and is allowed to fall a distance \texttip{h}{h}. Assume that the wheel has a moment of inertia \texttip{I}{I} about its rotation axis. Part A Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius \texttip{r_{\mit A}}{r_A} \texttip{I}{I} = Typesetting math: 29% . Find \texttip{\omega _{\mit A}}{omega_A}, the angular speed of the wheel after the block has fallen a distance \texttip{h}{h}, for this case. Express \texttip{\omega _{\mit A}}{omega_A} in terms of \texttip{m}{m}, \texttip{g}{g}, \texttip{h}{h}, \texttip{r_{\mit A}}{r_A}, and \texttip{I}{I}. You did not open hints for this part. ANSWER: Part B Now consider the case that the string tied to the block is wrapped around a smaller inside axle of the wheel of radius \texttip{r_{\mit B}}{r_B} . Find \texttip{\omega _{\mit B}}{omega_B}, the angular speed of the wheel after the block has fallen a distance \texttip{h}{h}, for this case. Express \texttip{\omega _{\mit B}}{omega_B} in terms of \texttip{m}{m}, \texttip{g}{g}, \texttip{h}{h}, \texttip{r_{\mit B}}{r_B}, and \texttip{I}{I}. \texttip{\omega _{\mit A}}{omega_A} = Typesetting math: 29% You did not open hints for this part. ANSWER: Part C Which of the following describes the relationship between \texttip{\omega _{\mit A}}{omega_A} and \texttip{\omega _{\mit B}}{omega_B}? You did not open hints for this part. ANSWER: A Bar Suspended by Two Vertical Strings A rigid, uniform, horizontal bar of mass \texttip{m_{\rm 1}}{m_1} and length \texttip{L}{L} is supported by two identical massless strings. Both strings are vertical. String A is attached at a distance d < L/2 from the left end of the bar and is connected to the ceiling; string B is attached to \texttip{\omega _{\mit B}}{omega_B} = \omega_A > \omega_B \omega_B > \omega_A \omega_A = \omega_B Typesetting math: 29% the left end of the bar and is connected to the floor. A small block of mass \texttip{m_{\rm 2}}{m_2} is supported against gravity by the bar at a distance \texttip{x}{x} from the left end of the bar, as shown in the figure. Throughout this problem positive torque is that which spins an object counterclockwise. Use \texttip{g}{g} for the magnitude of the acceleration due to gravity. Part A Find \texttip{T_{\mit A}}{T_A}, the tension in string A. Express the tension in string A in terms of \texttip{g}{g}, \texttip{m_{\rm 1}}{m_1}, \texttip{L}{L}, \texttip{d}{d}, \texttip{m_{\rm 2}}{m_2}, and \texttip{x}{x}. You did not open hints for this part. ANSWER: Part B Find \texttip{T_{\mit B}}{T_B}, the magnitude of the tension in string B. Express the magnitude of the tension in string B in terms of \texttip{T_{\mit A}}{T_A}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{g}{g}. \texttip{T_{\mit A}}{T_A} = Typesetting math: 29% You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Part D If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of \texttip{x}{x} such that the bar remains stable (call it \texttip{x_{\rm critical}}{x_critical})? Express your answer for \texttip{x_{\rm critical}}{x_critical} in terms of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{d}{d}, and \texttip{L}{L}. You did not open hints for this part. ANSWER: Part E \texttip{T_{\mit B}}{T_B} = \texttip{x_{\rm critical}}{x_critical} = Typesetting math: 29% This question will be shown after you complete previous question(s). A Tale of Two Nutcrackers This problem explores the ways that torque can be used in everyday life. Case 1 To crack a nut a force of magnitude \texttip{F_{\rm n}}{F_n} (or greater) must be applied on both sides, as shown in the figure. One can see that a nutcracker only applies this force at the point in which it contacts the nut (at a distance \texttip{d}{d} from the nutcracker pivot). In this problem the nut is placed in a nutcracker and equal forces of magnitude \texttip{F}{F} are applied to each end, directed perpendicular to the handle, at a distance \texttip{D}{D} from the pivot. The frictional forces between the nut and the nutcracker are equal and large enough that the nut doesn’t shoot out of the nutcracker. Part A Find \texttip{F}{F}, the magnitude of the force applied to each side of the nutcracker required to crack the nut. Express the force in terms of \texttip{F_{\rm n}}{F_n}, \texttip{d}{d}, and \texttip{D}{D}. You did not open hints for this part. ANSWER: Typesetting math: 29% Case 2 The nut is now placed in a nutcracker with only one lever, as shown, and again friction keeps the nut from slipping. The top “jaw” (in black) is fixed to a stationary frame so that a person just has to apply a force to the bottom lever. Assume that \texttip{F_{\rm 2}}{F_2} is directed perpendicular to the handle. Part B Find the magnitude of the force \texttip{F_{\rm 2}}{F_2} required to crack the nut. Express your answer in terms of \texttip{F_{\rm n}}{F_n}, \texttip{d}{d}, and \texttip{D}{D}. You did not open hints for this part. ANSWER: \texttip{F}{F} = \texttip{F_{\rm 2}}{F_2} = Typesetting math: 29% Part C This question will be shown after you complete previous question(s). Precarious Lunch A uniform steel beam of length \texttip{L}{L} and mass \texttip{m_{\rm 1}}{m_1} is attached via a hinge to the side of a building. The beam is supported by a steel cable attached to the end of the beam at an angle \texttip{\theta }{theta}, as shown. Through the hinge, the wall exerts an unknown force, \texttip{F}{F}, on the beam. A workman of mass \texttip{m_{\rm 2}}{m_2} sits eating lunch a distance \texttip{d}{d} from the building. Part A Find \texttip{T}{T}, the tension in the cable. Remember to account for all the forces in the problem. Express your answer in terms of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{L}{L}, \texttip{d}{d}, \texttip{\theta }{theta}, and \texttip{g}{g}, the magnitude of the acceleration due to gravity. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B Find \texttip{F_{\mit x}}{F_x}, the \texttip{x}{x}-component of the force exerted by the wall on the beam ( \texttip{F}{F}), using the axis shown. Remember to pay attention to the direction that the wall exerts the force. Express your answer in terms of \texttip{T}{T} and other given quantities. You did not open hints for this part. ANSWER: Part C Find \texttip{F_{\mit y}}{F_y}, the y-component of force that the wall exerts on the beam ( \texttip{F}{F}), using the axis shown. Remember to pay attention to the direction that the wall exerts the force. Express your answer in terms of \texttip{T}{T}, \texttip{\theta }{theta}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{g}{g}. ANSWER: \texttip{T}{T} = \texttip{F_{\mit x}}{F_x} = \texttip{F_{\mit y}}{F_y} = Typesetting math: 29% Pulling Out a Nail A nail is hammered into a board so that it would take a force \texttip{F_{\rm nail}}{F_nail}, applied straight upward on the head of the nail, to pull it out. (Take an upward force to be positive.) A carpenter uses a crowbar to try to pry it out. The length of the handle of the crowbar is \texttip{L_{\rm h}}{L_h}, and the length of the forked portion of the crowbar (which fits around the nail) is \texttip{L_{\rm n}}{L_n}. Assume that the forked portion of the crowbar is perfectly horizontal. The handle of the crowbar makes an angle \texttip{\theta }{theta} with the horizontal, and the carpenter pulls directly along the horizontal. Typesetting math: 29% Part A With what force \texttip{F_{\rm pull}}{F_pull} must the carpenter pull on the crowbar to remove the nail? Express the force in terms of \texttip{F_{\rm nail}}{F_nail}, \texttip{L_{\rm h}}{L_h}, \texttip{L_{\rm n}}{L_n}, and \texttip{\theta }{theta}. You did not open hints for this part. ANSWER: Now, imagine that \texttip{F_{\rm pull}}{F_pull} is not large enough to dislodge the nail. In other words, the nail stays in place, and, if the surface below the crowbar weren’t present, the crowbar would rotate around the point of contact with the nail. This makes it natural to take the pivot point to be the point where the crowbar is in contact with the nail. (But you are always free to choose the pivot point to be any fixed point, even one some distance from the object.) Part B What is the magnitude of the normal force that the surface exerts on the crowbar, \texttip{F_{\rm bar}}{F_bar}? Express your answer for the normal force in terms of \texttip{F_{\rm pull}}{F_pull}, \texttip{\theta }{theta}, \texttip{L_{\rm n}}{L_n}, and \texttip{L_{\rm h}}{L_h}. Take the upward direction to be positive. You did not open hints for this part. ANSWER: Three bars are shown in the figure. Both bars A and B have \texttip{F_{\rm pull}}{F_pull} acting on them in the horizontal direction. Bar C has \texttip{F_{\rm pull}}{F_pull} = \texttip{F_{\rm bar}}{F_bar} = Typesetting math: 29% \texttip{F_{\rm pull}}{F_pull} strictly perpendicular to the bar. \texttip{L_{\rm h}}{L_h}, \texttip{L_{\rm n}}{L_n}, and \texttip{\theta }{theta} are the same quantities in each case. Part C Let the magnitude of the torque about the bend in the crowbars be denoted \texttip{\tau _{\mit A}}{tau_A}, \texttip{\tau _{\mit B}}{tau_B} and \texttip{\tau _{\mit C}}{tau_C} for each of the three cases shown. Which of the following is the correct relationship between the magnitude of of the torques? You did not open hints for this part. ANSWER: Tipping Crane \tau_A > \tau_B > \tau_C \tau_A > \tau_B = \tau_C \tau_A = \tau_B = \tau_C \tau_A < \tau_B = \tau_C \tau_A < \tau_B < \tau_C \tau_A = \tau_B > \tau_C \tau_A = \tau_B < \tau_C Typesetting math: 29% Learning Goal: To step through the application of \Sigma \vec{\tau} = 0 to prevent a crane from tipping over. A crane of weight \texttip{W}{W} has a length (wheelbase) \texttip{c}{c}, and its center of mass is midway between the wheels (i.e., the mass of the lifting arm is negligible). The arm extending from the front of the crane has a length \texttip{b}{b} and makes an angle \texttip{\theta }{theta} with the horizontal. The crane contacts the ground only at its front and rear tires. Part A While watching the crane in operation, an observer mentions to you that for a given load there is a maximum angle \texttip{\theta _{\rm max}}{theta_max} between 0 \degree and 90 \degree that the crane arm can make with the horizontal without tipping the crane over. Is this correct? ANSWER: Part B Later that week, while watching the same crane in operation, a different observer mentions to you that there is a maximum load the crane can lift without tipping, and you can find that maximum load by observing the minimum angle \texttip{\theta _{\rm min}}{theta_min} that the crane arm makes with the horizontal. Is this correct? ANSWER: yes no Typesetting math: 29% Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H yes no Typesetting math: 29% Notice that we have the weight of the crane exerting a torque about the front wheels of the same crane. To create a torque, a force must be present, so it would seem that somehow the weight of the crane is exerting a force upon its front wheels. However, the crane is one object, and it follows from Newton's laws that an object cannot exert a net force upon itself. This crane seems to be defying Newton's laws. What's going on here? ANSWER: Part I Assume you get a summer job as a crane operator. On the first day you are lifting a heavy piece of machinery. Even though you have the arm at 70^\circ above the horizontal, the crane begins to tip slowly forward. Consider the following possible actions: Release the brake on the lifting cable so that the load accelerates 1. downward. 2. Release the lifting arm so that \texttip{\theta }{theta} decreases rapidly and the load accelerates downward. 3. Increase \texttip{\theta }{theta} while simultaneously letting out the lifting cable so that the load accelerates downward. 4. Put the crane wheels in gear and accelerate the crane forward. None of these solutions is ideal, but which will have the short-term effect of restoring contact of the crane's rear wheels with the ground? ANSWER: Spinning Situations Suppose you are standing on the center of a merry-go-round that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis is pointing upward. The wheel is rotating counterclockwise when observed from above. Newton's laws don't apply to torques. The rear wheels exert a downward force on the front wheels. The crane is not accelerating so forces don't matter. The earth exerts forces on the crane and the load. all but 1 all but 2 all but 3 all but 4 all of them Typesetting math: 29% For this problem, neglect any air resistance or friction between the merry-go-round and its foundation. Part A Suppose you now grab the edge of the wheel with your hand, stopping it from spinning. What happens to the merry-go-round? You did not open hints for this part. ANSWER: Twirling a Baton A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120{\rm kg} and length 80.0{\rm cm} . Part A Initially, the baton is spinning about a line through its center at angular velocity 3.00{\rm rad/s} . What is its angular momentum? Express your answer in kilogram meters squared per second. It remains at rest. It begins to rotate counterclockwise (as observed from above). It begins to rotate clockwise (as observed from above). Typesetting math: 29% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \rm kg \cdot m^2/s Typesetting math: 29%

Chapter 12 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, May 16, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Spinning Grinding Wheel At time a grinding wheel has an angular velocity of 26.0 . It has a constant angular acceleration of 33.0 until a circuit breaker trips at time = 1.80 . From then on, the wheel turns through an angle of 432 as it coasts to a stop at constant angular deceleration. Part A Through what total angle did the wheel turn between and the time it stopped? Express your answer in radians. You did not open hints for this part. ANSWER: Part B At what time does the wheel stop? Express your answer in seconds. You did not open hints for this part. ANSWER: t = 0 rad/s rad/s2 t s rad t = 0 rad Part C What was the wheel’s angular acceleration as it slowed down? Express your answer in radians per second per second. You did not open hints for this part. ANSWER: An Exhausted Bicyclist An exhausted bicyclist pedals somewhat erratically when exercising on a static bicycle. The angular velocity of the wheels follows the equation , where represents time (measured in seconds), = 0.500 , = 0.250 and = 2.00 . Part A There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time to be at angle radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel’s rotation. What angular displacement has the spot of paint undergone between time 0 and 2 seconds? Express your answer in radians using three significant figures. s rad/s2 (t) = at − bsin(ct) for t 0 t a rad/s2 b rad/s c rad/s t = 0 = 0 Typesetting math: 29% You did not open hints for this part. ANSWER: Part B Express the angular displacement undergone by the spot of paint at seconds in degrees. Remember to use the unrounded value from Part A, should you need it. Express your answer in degrees using three significant figures. You did not open hints for this part. ANSWER: Part C What distance has the spot of paint moved in 2 seconds if the radius of the wheel is 50 centimeters? Express your answer in centimeters using three significant figures. You did not open hints for this part. ANSWER: = rad t = 2 = d Typesetting math: 29% Part D Which one of the following statements describes the motion of the spot of paint at seconds? You did not open hints for this part. ANSWER: Flywheel Kinematics A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration . The flywheel is assumed to be at rest at time in Parts A and B of this problem. Part A Find the time it takes to accelerate the flywheel to if the angular acceleration is . Express your answer in terms of and . d = cm t = 2.0 The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is increasing. t = 0 t1 1 1 Typesetting math: 29% You did not open hints for this part. ANSWER: Part B Find the angle through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity . Express your answer in terms of some or all of the following: , , and . You did not open hints for this part. ANSWER: Part C Assume that the motor has accelerated the wheel up to an angular velocity with angular acceleration in time . At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of . Find , the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity). Express your answer in terms of some or all of the following: , \texttip{\alpha }{alpha}, and \texttip{t_{\rm 1}}{t_1}. You did not open hints for this part. t1 = 1 1 1 t1 1 = 1 t1 −5 t2 1 Typesetting math: 29% ANSWER: Surprising Exploding Firework A mortar fires a shell of mass \texttip{m}{m} at speed \texttip{v_{\rm 0}}{v_0}. The shell explodes at the top of its trajectory (shown by a star in the figure) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass \large{\frac15m} and a larger piece of mass \large{\frac45m}. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance \texttip{d}{d} from the mortar. If there had been no explosion, the shell would have landed a distance \texttip{r}{r} from the mortar. Assume that air resistance and the mass of the shell’s explosive charge are negligible. Part A Find the distance \texttip{d}{d} from the mortar at which the larger piece of the shell lands. Express \texttip{d}{d} in terms of \texttip{r}{r}. You did not open hints for this part. \texttip{t_{\rm 2}}{t_2} = s Typesetting math: 29% ANSWER: Kinetic Energy of a Dumbbell This problem illustrates the two contributions to the kinetic energy of an extended object: rotational kinetic energy and translational kinetic energy. You are to find the total kinetic energy \texttip{K_{\rm total}}{K_total} of a dumbbell of mass \texttip{m}{m} when it is rotating with angular speed \texttip{\omega }{omega} and its center of mass is moving translationally with speed \texttip{v}{v}. Denote the dumbbell’s moment of inertia about its center of mass by \texttip{I_{\rm cm}}{I_cm}. Note that if you approximate the spheres as point masses of mass m/2 each located a distance \texttip{r}{r} from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by I_{\rm cm} = mr^2, but this fact will not be necessary for this problem. Part A Find the total kinetic energy \texttip{K_{\rm tot}}{K_tot} of the dumbbell. Express your answer in terms of \texttip{m}{m}, \texttip{v}{v}, \texttip{I_{\rm cm}}{I_cm}, and \texttip{\omega }{omega}. You did not open hints for this part. \texttip{d}{d} = Typesetting math: 29% ANSWER: Part B This question will be shown after you complete previous question(s). Unwinding Cylinder A cylinder with moment of inertia \texttip{I}{I} about its center of mass, mass \texttip{m}{m}, and radius \texttip{r}{r} has a string wrapped around it which is tied to the ceiling . The cylinder’s vertical position as a function of time is y(t). At time t = 0 the cylinder is released from rest at a height \texttip{h}{h} above the ground. Part A The string constrains the rotational and translational motion of the cylinder. What is the relationship between the angular rotation rate \texttip{\omega }{omega} and \texttip{v}{v}, the velocity of the center of mass of the cylinder? \texttip{K_{\rm tot}}{K_tot} = Typesetting math: 29% Remember that upward motion corresponds to positive linear velocity, and counterclockwise rotation corresponds to positive angular velocity. Express \texttip{\omega }{omega} in terms of \texttip{v}{v} and other given quantities. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C Suppose that at a certain instant the velocity of the cylinder is \texttip{v}{v}. What is its total kinetic energy, \texttip{K_{\rm total}}{K_total}, at that instant? Express \texttip{K_{\rm total}}{K_total} in terms of \texttip{m}{m}, \texttip{r}{r}, \texttip{I}{I}, and \texttip{v}{v}. You did not open hints for this part. ANSWER: Part D \texttip{\omega }{omega} = \texttip{K_{\rm total}}{K_total} = Typesetting math: 29% Find \texttip{v_{\rm f \hspace{1 pt}}}{v_f}, the cylinder’s vertical velocity when it hits the ground. Express \texttip{v_{\rm f \hspace{1 pt}}}{v_f}, in terms of \texttip{g}{g}, \texttip{h}{h}, \texttip{I}{I}, \texttip{m}{m}, and \texttip{r}{r}. You did not open hints for this part. ANSWER: Kinetic Energy of a Rotating Wheel A simple wheel has the form of a solid cylinder of radius \texttip{r}{r} with a mass \texttip{m}{m} uniformly distributed throughout its volume. The wheel is pivoted on a stationary axle through the axis of the cylinder and rotates about the axle at a constant angular speed. The wheel rotates \texttip{n}{n} full revolutions in a time interval \texttip{t}{t}. Part A What is the kinetic energy \texttip{K}{K} of the rotating wheel? Express your answer in terms of \texttip{m}{m}, \texttip{r}{r}, \texttip{n}{n}, \texttip{t}{t} and, \texttip{\pi }{pi}. You did not open hints for this part. ANSWER: Finding Torque \texttip{v_{\rm f \hspace{1 pt}}}{v_f} = \texttip{K}{K} = Typesetting math: 29% A force \texttip{\vec{F}}{F_vec} of magnitude \texttip{F}{F} making an angle \texttip{\theta }{theta} with the x axis is applied to a particle located along axis of rotation A, at Cartesian coordinates (0, 0) in the figure. The vector \texttip{\vec{F}}{F_vec} lies in the xy plane, and the four axes of rotation A, B, C, and D all lie perpendicular to the xy plane. A particle is located at a vector position \texttip{\vec{r}}{r_vec} with respect to an axis of rotation (thus \texttip{\vec{r}}{r_vec} points from the axis to the point at which the particle is located). The magnitude of the torque \texttip{\tau }{tau} about this axis due to a force \texttip{\vec{F}}{F_vec} acting on the particle is given by \tau = r F \sin(\alpha) = ({\rm moment \; arm}) \cdot F = rF_{\perp}, where \texttip{\alpha }{alpha} is the angle between \texttip{\vec{r}}{r_vec} and \texttip{\vec{F}}{F_vec}, \texttip{r}{r} is the magnitude of \texttip{\vec{r}}{r_vec}, \texttip{F}{F} is the magnitude of \texttip{\vec{F}}{F_vec}, the component of \texttip{\vec{r}}{r_vec} that is perpendicualr to \texttip{\vec{F}}{F_vec} is the moment arm, and \texttip{F_{\rm \perp}}{F_\perp} is the component of the force that is perpendicular to \texttip{\vec{r}}{r_vec}. Sign convention: You will need to determine the sign by analyzing the direction of the rotation that the torque would tend to produce. Recall that negative torque about an axis corresponds to clockwise rotation. In this problem, you must express the angle \texttip{\alpha }{alpha} in the above equation in terms of \texttip{\theta }{theta}, \texttip{\phi }{phi}, and/or \texttip{\pi }{pi} when entering your answers. Keep in mind that \pi = 180\;\rm degrees and (\pi/2) = 90\;\rm degrees . Part A What is the torque \texttip{\tau_{\rm A}}{tau_A} about axis A due to the force \texttip{\vec{F}}{F_vec}? Express the torque about axis A at Cartesian coordinates (0, 0). You did not open hints for this part. Typesetting math: 29% ANSWER: Part B What is the torque \texttip{\tau_{\rm B}}{tau_B} about axis B due to the force \texttip{\vec{F}}{F_vec}? (B is the point at Cartesian coordinates (0, b), located a distance \texttip{b}{b} from the origin along the y axis.) Express the torque about axis B in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. You did not open hints for this part. ANSWER: Part C What is the torque \texttip{\tau_{\rm C}}{tau_C} about axis C due to \texttip{\vec{F}}{F_vec}? (C is the point at Cartesian coordinates (c, 0), a distance \texttip{c}{c} along the x axis.) Express the torque about axis C in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. You did not open hints for this part. ANSWER: \texttip{\tau_{\rm A}}{tau_A} = \texttip{\tau_{\rm B}}{tau_B} = Typesetting math: 29% Part D What is the torque \texttip{\tau_{\rm D}}{tau_D} about axis D due to \texttip{\vec{F}}{F_vec}? (D is the point located at a distance \texttip{d}{d} from the origin and making an angle \texttip{\phi }{phi} with the x axis.) Express the torque about axis D in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. ANSWER: Torque Magnitude Ranking Task The wrench in the figure has six forces of equal magnitude acting on it. \texttip{\tau_{\rm C}}{tau_C} = \texttip{\tau_{\rm D}}{tau_D} = Typesetting math: 29% Part A Rank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench, measured about an axis centered on the bolt. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: The Parallel-Axis Theorem Typesetting math: 29% Learning Goal: To understand the parallel-axis theorem and its applications To solve many problems about rotational motion, it is important to know the moment of inertia of each object involved. Calculating the moments of inertia of various objects, even highly symmetrical ones, may be a lengthy and tedious process. While it is important to be able to calculate moments of inertia from the definition (I=\sum m_ir_i^2), in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of frequently occurring shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily available from any mechanics text, including your textbook. However, one must take into account that an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of tranlsational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moments of inertia that you can find in the textbooks are usually calculated with respect to an axis passing through the center of mass of the object. However, in many problems the axis of rotation does not pass through the center of mass. Does that mean that one has to go through the lengthy process of finding the moment of inertia from scratch? It turns out that in many cases, calculating the moment of inertia can be done rather easily if one uses the parallel-axis theorem. Mathematically, it can be expressed as I=I_{\rm cm}+md^2, where \texttip{I_{\rm cm}}{I_cm} is the moment of inertia about an axis passing through the center of mass, \texttip{m}{m} is the total mass of the object, and \texttip{I}{I} is the moment of inertia about another axis, parallel to the one for which \texttip{I_{\rm cm}}{I_cm} is calculated and located a distance \texttip{d}{d} from the center of mass. In this problem you will show that the theorem does indeed work for at least one object: a dumbbell of length \texttip{2r}{2r} made of two small spheres of mass \texttip{m}{m} each connected by a light rod (see the figure). NOTE: Unless otherwise noted, all axes considered are perpendicular to the plane of the page. Part A Using the definition of moment of inertia, calculate I_{\rm cm}, the moment of inertia about the center of mass, for this object. Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B Using the definition of moment of inertia, calculate I_{\rm B}, the moment of inertia about an axis through point B, for this object. Point B coincides with (the center of) one of the spheres (see the figure). Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. ANSWER: Part C Now calculate I_{\rm B} for this object using the parallel-axis theorem. Express your answer in terms of \texttip{I_{\rm cm}}{I_cm}, \texttip{m}{m}, and \texttip{r}{r}. ANSWER: I_{\rm cm} = I_{\rm B} = I_{\rm B} = Typesetting math: 29% Part D Using the definition of moment of inertia, calculate I_{\rm C}, the moment of inertia about an axis through point C, for this object. Point C is located a distance \texttip{r}{r} from the center of mass (see the figure). Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. ANSWER: Part E Now calculate I_{\rm C} for this object using the parallel-axis theorem. Express your answer in terms of \texttip{I_{\rm cm}}{I_cm}, \texttip{m}{m}, and \texttip{r}{r}. ANSWER: Consider an irregular object of mass \texttip{m}{m}. Its moment of inertia measured with respect to axis A (parallel to the plane of the page), which passes through the center of mass (see the second diagram), is given by I_{\rm A}=0.64mr^2. Axes B, C, D, and E are parallel to axis A; their separations from axis A are shown in the diagram. In the subsequent questions, the subscript indicates the axis with respect to which the moment of inertia is measured: for instance, I_{\rm C} is the moment of inertia about axis C. I_{\rm C} = I_{\rm C} = Typesetting math: 29% Part F Which moment of inertia is the smallest? ANSWER: Part G Which moment of inertia is the largest? ANSWER: I_{\rm A} I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} I_{\rm A} I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} Typesetting math: 29% Part H Which moments of inertia are equal? ANSWER: Part I Which moment of inertia equals 4.64mr^2? ANSWER: Part J Axis X, not shown in the diagram, is parallel to the axes shown. It is known that I_{\rm X}=6mr^2. Which of the following is a possible location for axis X? ANSWER: I_{\rm A} and I_{\rm D} I_{\rm B} and I_{\rm C} I_{\rm C} and I_{\rm E} No two moments of inertia are equal. I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} between axes A and C between axes C and D between axes D and E to the right of axis E Typesetting math: 29% Torque and Angular Acceleration Learning Goal: To understand and apply the formula \tau=I\alpha to rigid objects rotating about a fixed axis. To find the acceleration \texttip{a}{a} of a particle of mass \texttip{m}{m}, we use Newton’s second law: \vec {F}_{\rm net}=m\vec{a}, where \texttip{\vec{F}_{\rm net}}{F_vec_net} is the net force acting on the particle. To find the angular acceleration \texttip{\alpha }{alpha} of a rigid object rotating about a fixed axis, we can use a similar formula: \tau_{\rm net}=I\alpha, where \tau_{\rm net}=\sum \tau is the net torque acting on the object and \texttip{I}{I} is its moment of inertia. In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses \texttip{m_{\rm 1}}{m_1} and \texttip{m_{\rm 2}}{m_2} are attached to a seesaw. The seesaw is made of a bar that has length \texttip{l}{l} and is pivoted so that it is free to rotate in the vertical plane without friction. You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m_1>m_2, and that counterclockwise is considered the positive rotational direction. Part A The seesaw is pivoted in the middle, and the mass of the swing bar is negligible. Find the angular acceleration \texttip{\alpha }{alpha} of the seesaw. Express your answer in terms of some or all of the quantities \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{l}{l}, as well as the acceleration due to gravity \texttip{g}{g}. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B In what direction will the seesaw rotate, and what will the sign of the angular acceleration be? ANSWER: Part C This question will be shown after you complete previous question(s). \texttip{\alpha }{alpha} = The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Typesetting math: 29% Part D In what direction will the seesaw rotate and what will the sign of the angular acceleration be? ANSWER: Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Pivoted Rod with Unequal Masses The figure shows a simple model of a seesaw. These consist of a plank/rod of mass \texttip{m_{\rm r}}{m_r} and length 2x allowed to pivot freely about its center (or central axis), as shown in the diagram. A small sphere of mass \texttip{m_{\rm 1}}{m_1} is attached to the left end of the rod, and a small sphere of mass \texttip{m_{\rm 2}}{m_2} is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to \texttip{g}{g}. The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Typesetting math: 29% Part A What is the moment of inertia \texttip{I}{I} of this assembly about the axis through which it is pivoted? Express the moment of inertia in terms of \texttip{m_{\rm r}}{m_r}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{x}{x}. Keep in mind that the length of the rod is 2x, not \texttip{x}{x}. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Weight and Wheel Consider a bicycle wheel that initially is not rotating. A block of mass \texttip{m}{m} is attached to the wheel and is allowed to fall a distance \texttip{h}{h}. Assume that the wheel has a moment of inertia \texttip{I}{I} about its rotation axis. Part A Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius \texttip{r_{\mit A}}{r_A} \texttip{I}{I} = Typesetting math: 29% . Find \texttip{\omega _{\mit A}}{omega_A}, the angular speed of the wheel after the block has fallen a distance \texttip{h}{h}, for this case. Express \texttip{\omega _{\mit A}}{omega_A} in terms of \texttip{m}{m}, \texttip{g}{g}, \texttip{h}{h}, \texttip{r_{\mit A}}{r_A}, and \texttip{I}{I}. You did not open hints for this part. ANSWER: Part B Now consider the case that the string tied to the block is wrapped around a smaller inside axle of the wheel of radius \texttip{r_{\mit B}}{r_B} . Find \texttip{\omega _{\mit B}}{omega_B}, the angular speed of the wheel after the block has fallen a distance \texttip{h}{h}, for this case. Express \texttip{\omega _{\mit B}}{omega_B} in terms of \texttip{m}{m}, \texttip{g}{g}, \texttip{h}{h}, \texttip{r_{\mit B}}{r_B}, and \texttip{I}{I}. \texttip{\omega _{\mit A}}{omega_A} = Typesetting math: 29% You did not open hints for this part. ANSWER: Part C Which of the following describes the relationship between \texttip{\omega _{\mit A}}{omega_A} and \texttip{\omega _{\mit B}}{omega_B}? You did not open hints for this part. ANSWER: A Bar Suspended by Two Vertical Strings A rigid, uniform, horizontal bar of mass \texttip{m_{\rm 1}}{m_1} and length \texttip{L}{L} is supported by two identical massless strings. Both strings are vertical. String A is attached at a distance d < L/2 from the left end of the bar and is connected to the ceiling; string B is attached to \texttip{\omega _{\mit B}}{omega_B} = \omega_A > \omega_B \omega_B > \omega_A \omega_A = \omega_B Typesetting math: 29% the left end of the bar and is connected to the floor. A small block of mass \texttip{m_{\rm 2}}{m_2} is supported against gravity by the bar at a distance \texttip{x}{x} from the left end of the bar, as shown in the figure. Throughout this problem positive torque is that which spins an object counterclockwise. Use \texttip{g}{g} for the magnitude of the acceleration due to gravity. Part A Find \texttip{T_{\mit A}}{T_A}, the tension in string A. Express the tension in string A in terms of \texttip{g}{g}, \texttip{m_{\rm 1}}{m_1}, \texttip{L}{L}, \texttip{d}{d}, \texttip{m_{\rm 2}}{m_2}, and \texttip{x}{x}. You did not open hints for this part. ANSWER: Part B Find \texttip{T_{\mit B}}{T_B}, the magnitude of the tension in string B. Express the magnitude of the tension in string B in terms of \texttip{T_{\mit A}}{T_A}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{g}{g}. \texttip{T_{\mit A}}{T_A} = Typesetting math: 29% You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Part D If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of \texttip{x}{x} such that the bar remains stable (call it \texttip{x_{\rm critical}}{x_critical})? Express your answer for \texttip{x_{\rm critical}}{x_critical} in terms of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{d}{d}, and \texttip{L}{L}. You did not open hints for this part. ANSWER: Part E \texttip{T_{\mit B}}{T_B} = \texttip{x_{\rm critical}}{x_critical} = Typesetting math: 29% This question will be shown after you complete previous question(s). A Tale of Two Nutcrackers This problem explores the ways that torque can be used in everyday life. Case 1 To crack a nut a force of magnitude \texttip{F_{\rm n}}{F_n} (or greater) must be applied on both sides, as shown in the figure. One can see that a nutcracker only applies this force at the point in which it contacts the nut (at a distance \texttip{d}{d} from the nutcracker pivot). In this problem the nut is placed in a nutcracker and equal forces of magnitude \texttip{F}{F} are applied to each end, directed perpendicular to the handle, at a distance \texttip{D}{D} from the pivot. The frictional forces between the nut and the nutcracker are equal and large enough that the nut doesn’t shoot out of the nutcracker. Part A Find \texttip{F}{F}, the magnitude of the force applied to each side of the nutcracker required to crack the nut. Express the force in terms of \texttip{F_{\rm n}}{F_n}, \texttip{d}{d}, and \texttip{D}{D}. You did not open hints for this part. ANSWER: Typesetting math: 29% Case 2 The nut is now placed in a nutcracker with only one lever, as shown, and again friction keeps the nut from slipping. The top “jaw” (in black) is fixed to a stationary frame so that a person just has to apply a force to the bottom lever. Assume that \texttip{F_{\rm 2}}{F_2} is directed perpendicular to the handle. Part B Find the magnitude of the force \texttip{F_{\rm 2}}{F_2} required to crack the nut. Express your answer in terms of \texttip{F_{\rm n}}{F_n}, \texttip{d}{d}, and \texttip{D}{D}. You did not open hints for this part. ANSWER: \texttip{F}{F} = \texttip{F_{\rm 2}}{F_2} = Typesetting math: 29% Part C This question will be shown after you complete previous question(s). Precarious Lunch A uniform steel beam of length \texttip{L}{L} and mass \texttip{m_{\rm 1}}{m_1} is attached via a hinge to the side of a building. The beam is supported by a steel cable attached to the end of the beam at an angle \texttip{\theta }{theta}, as shown. Through the hinge, the wall exerts an unknown force, \texttip{F}{F}, on the beam. A workman of mass \texttip{m_{\rm 2}}{m_2} sits eating lunch a distance \texttip{d}{d} from the building. Part A Find \texttip{T}{T}, the tension in the cable. Remember to account for all the forces in the problem. Express your answer in terms of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{L}{L}, \texttip{d}{d}, \texttip{\theta }{theta}, and \texttip{g}{g}, the magnitude of the acceleration due to gravity. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B Find \texttip{F_{\mit x}}{F_x}, the \texttip{x}{x}-component of the force exerted by the wall on the beam ( \texttip{F}{F}), using the axis shown. Remember to pay attention to the direction that the wall exerts the force. Express your answer in terms of \texttip{T}{T} and other given quantities. You did not open hints for this part. ANSWER: Part C Find \texttip{F_{\mit y}}{F_y}, the y-component of force that the wall exerts on the beam ( \texttip{F}{F}), using the axis shown. Remember to pay attention to the direction that the wall exerts the force. Express your answer in terms of \texttip{T}{T}, \texttip{\theta }{theta}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{g}{g}. ANSWER: \texttip{T}{T} = \texttip{F_{\mit x}}{F_x} = \texttip{F_{\mit y}}{F_y} = Typesetting math: 29% Pulling Out a Nail A nail is hammered into a board so that it would take a force \texttip{F_{\rm nail}}{F_nail}, applied straight upward on the head of the nail, to pull it out. (Take an upward force to be positive.) A carpenter uses a crowbar to try to pry it out. The length of the handle of the crowbar is \texttip{L_{\rm h}}{L_h}, and the length of the forked portion of the crowbar (which fits around the nail) is \texttip{L_{\rm n}}{L_n}. Assume that the forked portion of the crowbar is perfectly horizontal. The handle of the crowbar makes an angle \texttip{\theta }{theta} with the horizontal, and the carpenter pulls directly along the horizontal. Typesetting math: 29% Part A With what force \texttip{F_{\rm pull}}{F_pull} must the carpenter pull on the crowbar to remove the nail? Express the force in terms of \texttip{F_{\rm nail}}{F_nail}, \texttip{L_{\rm h}}{L_h}, \texttip{L_{\rm n}}{L_n}, and \texttip{\theta }{theta}. You did not open hints for this part. ANSWER: Now, imagine that \texttip{F_{\rm pull}}{F_pull} is not large enough to dislodge the nail. In other words, the nail stays in place, and, if the surface below the crowbar weren’t present, the crowbar would rotate around the point of contact with the nail. This makes it natural to take the pivot point to be the point where the crowbar is in contact with the nail. (But you are always free to choose the pivot point to be any fixed point, even one some distance from the object.) Part B What is the magnitude of the normal force that the surface exerts on the crowbar, \texttip{F_{\rm bar}}{F_bar}? Express your answer for the normal force in terms of \texttip{F_{\rm pull}}{F_pull}, \texttip{\theta }{theta}, \texttip{L_{\rm n}}{L_n}, and \texttip{L_{\rm h}}{L_h}. Take the upward direction to be positive. You did not open hints for this part. ANSWER: Three bars are shown in the figure. Both bars A and B have \texttip{F_{\rm pull}}{F_pull} acting on them in the horizontal direction. Bar C has \texttip{F_{\rm pull}}{F_pull} = \texttip{F_{\rm bar}}{F_bar} = Typesetting math: 29% \texttip{F_{\rm pull}}{F_pull} strictly perpendicular to the bar. \texttip{L_{\rm h}}{L_h}, \texttip{L_{\rm n}}{L_n}, and \texttip{\theta }{theta} are the same quantities in each case. Part C Let the magnitude of the torque about the bend in the crowbars be denoted \texttip{\tau _{\mit A}}{tau_A}, \texttip{\tau _{\mit B}}{tau_B} and \texttip{\tau _{\mit C}}{tau_C} for each of the three cases shown. Which of the following is the correct relationship between the magnitude of of the torques? You did not open hints for this part. ANSWER: Tipping Crane \tau_A > \tau_B > \tau_C \tau_A > \tau_B = \tau_C \tau_A = \tau_B = \tau_C \tau_A < \tau_B = \tau_C \tau_A < \tau_B < \tau_C \tau_A = \tau_B > \tau_C \tau_A = \tau_B < \tau_C Typesetting math: 29% Learning Goal: To step through the application of \Sigma \vec{\tau} = 0 to prevent a crane from tipping over. A crane of weight \texttip{W}{W} has a length (wheelbase) \texttip{c}{c}, and its center of mass is midway between the wheels (i.e., the mass of the lifting arm is negligible). The arm extending from the front of the crane has a length \texttip{b}{b} and makes an angle \texttip{\theta }{theta} with the horizontal. The crane contacts the ground only at its front and rear tires. Part A While watching the crane in operation, an observer mentions to you that for a given load there is a maximum angle \texttip{\theta _{\rm max}}{theta_max} between 0 \degree and 90 \degree that the crane arm can make with the horizontal without tipping the crane over. Is this correct? ANSWER: Part B Later that week, while watching the same crane in operation, a different observer mentions to you that there is a maximum load the crane can lift without tipping, and you can find that maximum load by observing the minimum angle \texttip{\theta _{\rm min}}{theta_min} that the crane arm makes with the horizontal. Is this correct? ANSWER: yes no Typesetting math: 29% Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H yes no Typesetting math: 29% Notice that we have the weight of the crane exerting a torque about the front wheels of the same crane. To create a torque, a force must be present, so it would seem that somehow the weight of the crane is exerting a force upon its front wheels. However, the crane is one object, and it follows from Newton's laws that an object cannot exert a net force upon itself. This crane seems to be defying Newton's laws. What's going on here? ANSWER: Part I Assume you get a summer job as a crane operator. On the first day you are lifting a heavy piece of machinery. Even though you have the arm at 70^\circ above the horizontal, the crane begins to tip slowly forward. Consider the following possible actions: Release the brake on the lifting cable so that the load accelerates 1. downward. 2. Release the lifting arm so that \texttip{\theta }{theta} decreases rapidly and the load accelerates downward. 3. Increase \texttip{\theta }{theta} while simultaneously letting out the lifting cable so that the load accelerates downward. 4. Put the crane wheels in gear and accelerate the crane forward. None of these solutions is ideal, but which will have the short-term effect of restoring contact of the crane's rear wheels with the ground? ANSWER: Spinning Situations Suppose you are standing on the center of a merry-go-round that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis is pointing upward. The wheel is rotating counterclockwise when observed from above. Newton's laws don't apply to torques. The rear wheels exert a downward force on the front wheels. The crane is not accelerating so forces don't matter. The earth exerts forces on the crane and the load. all but 1 all but 2 all but 3 all but 4 all of them Typesetting math: 29% For this problem, neglect any air resistance or friction between the merry-go-round and its foundation. Part A Suppose you now grab the edge of the wheel with your hand, stopping it from spinning. What happens to the merry-go-round? You did not open hints for this part. ANSWER: Twirling a Baton A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120{\rm kg} and length 80.0{\rm cm} . Part A Initially, the baton is spinning about a line through its center at angular velocity 3.00{\rm rad/s} . What is its angular momentum? Express your answer in kilogram meters squared per second. It remains at rest. It begins to rotate counterclockwise (as observed from above). It begins to rotate clockwise (as observed from above). Typesetting math: 29% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \rm kg \cdot m^2/s Typesetting math: 29%

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Consider an ideal transformer with 100 turns of wire in the primary and 500 turns of wire in the secondary. If the primary coil is fed with 110 V, rms, what is the power consumed by the device connected to the secondary that behaves as a resistor.

Consider an ideal transformer with 100 turns of wire in the primary and 500 turns of wire in the secondary. If the primary coil is fed with 110 V, rms, what is the power consumed by the device connected to the secondary that behaves as a resistor.

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Chapter 11 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Understanding Work and Kinetic Energy Learning Goal: To learn about the Work-Energy Theorem and its basic applications. In this problem, you will learn about the relationship between the work done on an object and the kinetic energy of that object. The kinetic energy of an object of mass moving at a speed is defined as . It seems reasonable to say that the speed of an object–and, therefore, its kinetic energy–can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object. First, let us consider a sled of mass being pulled by a constant, horizontal force of magnitude along a rough, horizontal surface. The sled is speeding up. Part A How many forces are acting on the sled? ANSWER: Part B This question will be shown after you complete previous question(s). Part C K m v K = (1/2)mv2 m F one two three four This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I Typesetting math: 91% This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Work-Energy Theorem Reviewed Learning Goal: Review the work-energy theorem and apply it to a simple problem. If you push a particle of mass in the direction in which it is already moving, you expect the particle’s speed to increase. If you push with a constant force , then the particle will accelerate with acceleration (from Newton’s 2nd law). Part A Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied for a fixed interval of time , then the _____ of the particle will increase by an amount . You did not open hints for this part. ANSWER: M F a = F/M t at Typesetting math: 91% Part B Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied over a given distance , along the path of the particle, then the _____ of the particle will increase by . ANSWER: Part C If the initial kinetic energy of the particle is , and its final kinetic energy is , express in terms of and the work done on the particle. ANSWER: Part D In general, the work done by a force is written as . Now, consider whether the following statements are true or false: The dot product assures that the integrand is always nonnegative. The dot product indicates that only the component of the force perpendicular to the path contributes to the integral. The dot product indicates that only the component of the force parallel to the path contributes to the integral. Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,t). ANSWER: D FD Ki Kf Kf Ki W Kf = F W = ( ) d f i F r r Typesetting math: 91% Part E Assume that the particle has initial speed . Find its final kinetic energy in terms of , , , and . You did not open hints for this part. ANSWER: Part F What is the final speed of the particle? Express your answer in terms of and . ANSWER: ± The Work Done in Pulling a Supertanker Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 , one at an angle 10.0 west of north, and the other at an angle 10.0 east of north, as they pull the tanker a distance 0.660 toward the north. Part A What is the total work done by the two tugboats on the supertanker? Express your answer in joules, to three significant figures. vi Kf vi M F D Kf = Kf M vf = N km Typesetting math: 91% You did not open hints for this part. ANSWER: Energy Required to Lift a Heavy Box As you are trying to move a heavy box of mass , you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap around the pulley. You pull up on the rope to lift the box. Use for the magnitude of the acceleration due to gravity and neglect friction forces. Part A Once you have pulled hard enough to start the box moving upward, what is the magnitude of the upward force you must apply to the rope to start raising the box with constant velocity? Express the magnitude of the force in terms of , the mass of the box. J m g F m Typesetting math: 91% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Pulling a Block on an Incline with Friction A block of weight sits on an inclined plane as shown. A force of magnitude is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is . Part A F = mg F μ Typesetting math: 91% What is the total work done on the block by the force of friction as the block moves a distance up the incline? Express the work done by friction in terms of any or all of the variables , , , , , and . You did not open hints for this part. ANSWER: Part B What is the total work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed. Wfric L μ m g L F Wfric = WF F L μ m g L F WF = Typesetting math: 91% Part C What is the total work done on the block by the force of friction as the block moves a distance down the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Part D What is the total work done on the box by the appled force in this case? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: When Push Comes to Shove Two forces, of magnitudes = 75.0 and = 25.0 , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position = -1.00 . At some later time, the block has moved to the right, and its center is at a new position, = 1.00 . Wfric L μ m g L F Wfric = WF μ m g L F WF = F1 N F2 N xi cm xf cm Typesetting math: 91% Part A Find the work done on the block by the force of magnitude = 75.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Part B Find the work done by the force of magnitude = 25.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: W1 F1 N xi cm xf cm W1 = J W2 F2 N xi cm xf cm Typesetting math: 91% Part C What is the net work done on the block by the two forces? Express your answer numerically, in joules. ANSWER: Part D Determine the change in the kinetic energy of the block as it moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Work from a Constant Force Learning Goal: W2 = J Wnet Wnet = J Kf − Ki xi cm xf cm Kf − Ki = J Typesetting math: 91% To understand how to compute the work done by a constant force acting on a particle that moves in a straight line. In this problem, you will calculate the work done by a constant force. A force is considered constant if is independent of . This is the most frequently encountered situation in elementary Newtonian mechanics. Part A Consider a particle moving in a straight line from initial point B to final point A, acted upon by a constant force . The force (think of it as a field, having a magnitude and direction at every position ) is indicated by a series of identical vectors pointing to the left, parallel to the horizontal axis. The vectors are all identical only because the force is constant along the path. The magnitude of the force is , and the displacement vector from point B to point A is (of magnitude , making and angle (radians) with the positive x axis). Find , the work that the force performs on the particle as it moves from point B to point A. Express the work in terms of , , and . Remember to use radians, not degrees, for any angles that appear in your answer. You did not open hints for this part. ANSWER: Part B Now consider the same force acting on a particle that travels from point A to point B. The displacement vector now points in the opposite direction as it did in Part A. Find the work done by in this case. Express your answer in terms of , , and . F( r) r F r F L L WBA F L F WBA = F L WAB F Typesetting math: 91% L F You did not open hints for this part. ANSWER: ± Vector Dot Product Let vectors , , and . Calculate the following: Part A You did not open hints for this part. ANSWER: WAB = A = (2, 1,−4) B = (−3, 0, 1) C = (−1,−1, 2) Typesetting math: 91% Part B What is the angle between and ? Express your answer using one significant figure. You did not open hints for this part. ANSWER: Part C ANSWER: Part D ANSWER: A B = AB A B AB = radians 2B 3C = Typesetting math: 91% Part E Which of the following can be computed? You did not open hints for this part. ANSWER: and are different vectors with lengths and respectively. Find the following: Part F Express your answer in terms of You did not open hints for this part. ANSWER: 2(B 3C) = A B C A (B C) A (B + C) 3 A V 1 V 2 V1 V2 V1 Typesetting math: 91% Part G If and are perpendicular, You did not open hints for this part. ANSWER: Part H If and are parallel, Express your answer in terms of and . You did not open hints for this part. ANSWER: ± Tactics Box 11.1 Calculating the Work Done by a Constant Force V = 1 V 1 V 1 V 2 V = 1 V 2 V 1 V 2 V1 V2 V = 1 V 2 Typesetting math: 91% Learning Goal: To practice Tactics Box 11.1 Calculating the Work Done by a Constant Force. Recall that the work done by a constant force at an angle to the displacement is . The vector magnitudes and are always positive, so the sign of is determined entirely by the angle between the force and the displacement. W F d W = Fd cos F d W Typesetting math: 91% TACTICS BOX 11.1 Calculating the work done by a constant force Force and displacement Work Sign of Energy transfer Energy is transferred into the system. The particle speeds up. increases. No energy is transferred. Speed and are constant. Energy is transferred out of the system. The particle slows down. decreases. A box has weight of magnitude = 2.00 accelerates down a rough plane that is inclined at an angle = 30.0 above the horizontal, as shown at left. The normal force acting on the box has a magnitude = 1.732 , the coefficient of kinetic friction between the box and the plane is = 0.300, and the displacement of the box is 1.80 down the inclined plane. W W 0 F(“r) + K < 90 F("r) cos + 90 0 0 K > 90 F(“r) cos − K 180 −F(“r) − FG N n N μk d m Typesetting math: 91% Part A What is the work done on the box by gravity? Express your answers in joules to two significant figures. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Wgrav Wgrav = J Typesetting math: 91% Work and Potential Energy on a Sliding Block with Friction A block of weight sits on a plane inclined at an angle as shown. The coefficient of kinetic friction between the plane and the block is . A force is applied to push the block up the incline at constant speed. Part A What is the work done on the block by the force of friction as the block moves a distance up the incline? Express your answer in terms of some or all of the following: , , , . You did not open hints for this part. ANSWER: w μ F Wf L μ w L Wf = Typesetting math: 91% Part B What is the work done by the applied force of magnitude ? Express your answer in terms of some or all of the following: , , , . ANSWER: Part C What is the change in the potential energy of the block, , after it has been pushed a distance up the incline? Express your answer in terms of some or all of the following: , , , . ANSWER: Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). W F μ w L W = “U L μ w L “U = Typesetting math: 91% Part F This question will be shown after you complete previous question(s). Where’s the Energy? Learning Goal: To understand how to apply the law of conservation of energy to situations with and without nonconservative forces acting. The law of conservation of energy states the following: In an isolated system the total energy remains constant. If the objects within the system interact through gravitational and elastic forces only, then the total mechanical energy is conserved. The mechanical energy of a system is defined as the sum of kinetic energy and potential energy . For such systems where no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as , where the quantities with subscript “i” refer to the “initial” moment and those with subscript “f” refer to the final moment. A wise choice of initial and final moments, which is not always obvious, may significantly simplify the solution. The kinetic energy of an object that has mass \texttip{m}{m} and velocity \texttip{v}{v} is given by \large{K=\frac{1}{2}mv^2}. Potential energy, instead, has many forms. The two forms that you will be dealing with most often in this chapter are the gravitational and elastic potential energy. Gravitational potential energy is the energy possessed by elevated objects. For small heights, it can be found as U_{\rm g}=mgh, where \texttip{m}{m} is the mass of the object, \texttip{g}{g} is the acceleration due to gravity, and \texttip{h}{h} is the elevation of the object above the zero level. The zero level is the elevation at which the gravitational potential energy is assumed to be (you guessed it) zero. The choice of the zero level is dictated by convenience; typically (but not necessarily), it is selected to coincide with the lowest position of the object during the motion explored in the problem. Elastic potential energy is associated with stretched or compressed elastic objects such as springs. For a spring with a force constant \texttip{k}{k}, stretched or compressed a distance \texttip{x}{x}, the associated elastic potential energy is \large{U_{\rm e}=\frac{1}{2}kx^2}. When all three types of energy change, the law of conservation of energy for an object of mass \texttip{m}{m} can be written as K U Ki + Ui = Kf + Uf Typesetting math: 91% \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}. The gravitational force and the elastic force are two examples of conservative forces. What if nonconservative forces, such as friction, also act within the system? In that case, the total mechanical energy would change. The law of conservation of energy is then written as \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2+W_{\rm nc}=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}, where \texttip{W_{\rm nc}}{W_nc} represents the work done by the nonconservative forces acting on the object between the initial and the final moments. The work \texttip{W_{\rm nc}}{W_nc} is usually negative; that is, the nonconservative forces tend to decrease, or dissipate, the mechanical energy of the system. In this problem, we will consider the following situation as depicted in the diagram : A block of mass \texttip{m}{m} slides at a speed \texttip{v}{v} along a horizontal, smooth table. It next slides down a smooth ramp, descending a height \texttip{h}{h}, and then slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it does not lose contact with the supporting surfaces (table, ramp, or floor). You will analyze the motion of the block at different moments using the law of conservation of energy. Part A Which word in the statement of this problem allows you to assume that the table is frictionless? ANSWER: Part B straight smooth horizontal Typesetting math: 91% This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H Typesetting math: 91% This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Sliding In Socks Suppose that the coefficient of kinetic friction between Zak’s feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor. Part A If Zak’s speed is 3.00 \rm m/s when he starts to slide, what distance \texttip{d}{d} will he slide before stopping? Express your answer in meters. ANSWER: Typesetting math: 91% Part B This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \rm m Typesetting math: 91%

Chapter 11 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Understanding Work and Kinetic Energy Learning Goal: To learn about the Work-Energy Theorem and its basic applications. In this problem, you will learn about the relationship between the work done on an object and the kinetic energy of that object. The kinetic energy of an object of mass moving at a speed is defined as . It seems reasonable to say that the speed of an object–and, therefore, its kinetic energy–can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object. First, let us consider a sled of mass being pulled by a constant, horizontal force of magnitude along a rough, horizontal surface. The sled is speeding up. Part A How many forces are acting on the sled? ANSWER: Part B This question will be shown after you complete previous question(s). Part C K m v K = (1/2)mv2 m F one two three four This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I Typesetting math: 91% This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Work-Energy Theorem Reviewed Learning Goal: Review the work-energy theorem and apply it to a simple problem. If you push a particle of mass in the direction in which it is already moving, you expect the particle’s speed to increase. If you push with a constant force , then the particle will accelerate with acceleration (from Newton’s 2nd law). Part A Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied for a fixed interval of time , then the _ of the particle will increase by an amount . You did not open hints for this part. ANSWER: M F a = F/M t at Typesetting math: 91% Part B Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied over a given distance , along the path of the particle, then the _ of the particle will increase by . ANSWER: Part C If the initial kinetic energy of the particle is , and its final kinetic energy is , express in terms of and the work done on the particle. ANSWER: Part D In general, the work done by a force is written as . Now, consider whether the following statements are true or false: The dot product assures that the integrand is always nonnegative. The dot product indicates that only the component of the force perpendicular to the path contributes to the integral. The dot product indicates that only the component of the force parallel to the path contributes to the integral. Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,t). ANSWER: D FD Ki Kf Kf Ki W Kf = F W = ( ) d f i F r r Typesetting math: 91% Part E Assume that the particle has initial speed . Find its final kinetic energy in terms of , , , and . You did not open hints for this part. ANSWER: Part F What is the final speed of the particle? Express your answer in terms of and . ANSWER: ± The Work Done in Pulling a Supertanker Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 , one at an angle 10.0 west of north, and the other at an angle 10.0 east of north, as they pull the tanker a distance 0.660 toward the north. Part A What is the total work done by the two tugboats on the supertanker? Express your answer in joules, to three significant figures. vi Kf vi M F D Kf = Kf M vf = N km Typesetting math: 91% You did not open hints for this part. ANSWER: Energy Required to Lift a Heavy Box As you are trying to move a heavy box of mass , you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap around the pulley. You pull up on the rope to lift the box. Use for the magnitude of the acceleration due to gravity and neglect friction forces. Part A Once you have pulled hard enough to start the box moving upward, what is the magnitude of the upward force you must apply to the rope to start raising the box with constant velocity? Express the magnitude of the force in terms of , the mass of the box. J m g F m Typesetting math: 91% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Pulling a Block on an Incline with Friction A block of weight sits on an inclined plane as shown. A force of magnitude is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is . Part A F = mg F μ Typesetting math: 91% What is the total work done on the block by the force of friction as the block moves a distance up the incline? Express the work done by friction in terms of any or all of the variables , , , , , and . You did not open hints for this part. ANSWER: Part B What is the total work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed. Wfric L μ m g L F Wfric = WF F L μ m g L F WF = Typesetting math: 91% Part C What is the total work done on the block by the force of friction as the block moves a distance down the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Part D What is the total work done on the box by the appled force in this case? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: When Push Comes to Shove Two forces, of magnitudes = 75.0 and = 25.0 , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position = -1.00 . At some later time, the block has moved to the right, and its center is at a new position, = 1.00 . Wfric L μ m g L F Wfric = WF μ m g L F WF = F1 N F2 N xi cm xf cm Typesetting math: 91% Part A Find the work done on the block by the force of magnitude = 75.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Part B Find the work done by the force of magnitude = 25.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: W1 F1 N xi cm xf cm W1 = J W2 F2 N xi cm xf cm Typesetting math: 91% Part C What is the net work done on the block by the two forces? Express your answer numerically, in joules. ANSWER: Part D Determine the change in the kinetic energy of the block as it moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Work from a Constant Force Learning Goal: W2 = J Wnet Wnet = J Kf − Ki xi cm xf cm Kf − Ki = J Typesetting math: 91% To understand how to compute the work done by a constant force acting on a particle that moves in a straight line. In this problem, you will calculate the work done by a constant force. A force is considered constant if is independent of . This is the most frequently encountered situation in elementary Newtonian mechanics. Part A Consider a particle moving in a straight line from initial point B to final point A, acted upon by a constant force . The force (think of it as a field, having a magnitude and direction at every position ) is indicated by a series of identical vectors pointing to the left, parallel to the horizontal axis. The vectors are all identical only because the force is constant along the path. The magnitude of the force is , and the displacement vector from point B to point A is (of magnitude , making and angle (radians) with the positive x axis). Find , the work that the force performs on the particle as it moves from point B to point A. Express the work in terms of , , and . Remember to use radians, not degrees, for any angles that appear in your answer. You did not open hints for this part. ANSWER: Part B Now consider the same force acting on a particle that travels from point A to point B. The displacement vector now points in the opposite direction as it did in Part A. Find the work done by in this case. Express your answer in terms of , , and . F( r) r F r F L L WBA F L F WBA = F L WAB F Typesetting math: 91% L F You did not open hints for this part. ANSWER: ± Vector Dot Product Let vectors , , and . Calculate the following: Part A You did not open hints for this part. ANSWER: WAB = A = (2, 1,−4) B = (−3, 0, 1) C = (−1,−1, 2) Typesetting math: 91% Part B What is the angle between and ? Express your answer using one significant figure. You did not open hints for this part. ANSWER: Part C ANSWER: Part D ANSWER: A B = AB A B AB = radians 2B 3C = Typesetting math: 91% Part E Which of the following can be computed? You did not open hints for this part. ANSWER: and are different vectors with lengths and respectively. Find the following: Part F Express your answer in terms of You did not open hints for this part. ANSWER: 2(B 3C) = A B C A (B C) A (B + C) 3 A V 1 V 2 V1 V2 V1 Typesetting math: 91% Part G If and are perpendicular, You did not open hints for this part. ANSWER: Part H If and are parallel, Express your answer in terms of and . You did not open hints for this part. ANSWER: ± Tactics Box 11.1 Calculating the Work Done by a Constant Force V = 1 V 1 V 1 V 2 V = 1 V 2 V 1 V 2 V1 V2 V = 1 V 2 Typesetting math: 91% Learning Goal: To practice Tactics Box 11.1 Calculating the Work Done by a Constant Force. Recall that the work done by a constant force at an angle to the displacement is . The vector magnitudes and are always positive, so the sign of is determined entirely by the angle between the force and the displacement. W F d W = Fd cos F d W Typesetting math: 91% TACTICS BOX 11.1 Calculating the work done by a constant force Force and displacement Work Sign of Energy transfer Energy is transferred into the system. The particle speeds up. increases. No energy is transferred. Speed and are constant. Energy is transferred out of the system. The particle slows down. decreases. A box has weight of magnitude = 2.00 accelerates down a rough plane that is inclined at an angle = 30.0 above the horizontal, as shown at left. The normal force acting on the box has a magnitude = 1.732 , the coefficient of kinetic friction between the box and the plane is = 0.300, and the displacement of the box is 1.80 down the inclined plane. W W 0 F(“r) + K < 90 F("r) cos + 90 0 0 K > 90 F(“r) cos − K 180 −F(“r) − FG N n N μk d m Typesetting math: 91% Part A What is the work done on the box by gravity? Express your answers in joules to two significant figures. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Wgrav Wgrav = J Typesetting math: 91% Work and Potential Energy on a Sliding Block with Friction A block of weight sits on a plane inclined at an angle as shown. The coefficient of kinetic friction between the plane and the block is . A force is applied to push the block up the incline at constant speed. Part A What is the work done on the block by the force of friction as the block moves a distance up the incline? Express your answer in terms of some or all of the following: , , , . You did not open hints for this part. ANSWER: w μ F Wf L μ w L Wf = Typesetting math: 91% Part B What is the work done by the applied force of magnitude ? Express your answer in terms of some or all of the following: , , , . ANSWER: Part C What is the change in the potential energy of the block, , after it has been pushed a distance up the incline? Express your answer in terms of some or all of the following: , , , . ANSWER: Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). W F μ w L W = “U L μ w L “U = Typesetting math: 91% Part F This question will be shown after you complete previous question(s). Where’s the Energy? Learning Goal: To understand how to apply the law of conservation of energy to situations with and without nonconservative forces acting. The law of conservation of energy states the following: In an isolated system the total energy remains constant. If the objects within the system interact through gravitational and elastic forces only, then the total mechanical energy is conserved. The mechanical energy of a system is defined as the sum of kinetic energy and potential energy . For such systems where no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as , where the quantities with subscript “i” refer to the “initial” moment and those with subscript “f” refer to the final moment. A wise choice of initial and final moments, which is not always obvious, may significantly simplify the solution. The kinetic energy of an object that has mass \texttip{m}{m} and velocity \texttip{v}{v} is given by \large{K=\frac{1}{2}mv^2}. Potential energy, instead, has many forms. The two forms that you will be dealing with most often in this chapter are the gravitational and elastic potential energy. Gravitational potential energy is the energy possessed by elevated objects. For small heights, it can be found as U_{\rm g}=mgh, where \texttip{m}{m} is the mass of the object, \texttip{g}{g} is the acceleration due to gravity, and \texttip{h}{h} is the elevation of the object above the zero level. The zero level is the elevation at which the gravitational potential energy is assumed to be (you guessed it) zero. The choice of the zero level is dictated by convenience; typically (but not necessarily), it is selected to coincide with the lowest position of the object during the motion explored in the problem. Elastic potential energy is associated with stretched or compressed elastic objects such as springs. For a spring with a force constant \texttip{k}{k}, stretched or compressed a distance \texttip{x}{x}, the associated elastic potential energy is \large{U_{\rm e}=\frac{1}{2}kx^2}. When all three types of energy change, the law of conservation of energy for an object of mass \texttip{m}{m} can be written as K U Ki + Ui = Kf + Uf Typesetting math: 91% \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}. The gravitational force and the elastic force are two examples of conservative forces. What if nonconservative forces, such as friction, also act within the system? In that case, the total mechanical energy would change. The law of conservation of energy is then written as \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2+W_{\rm nc}=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}, where \texttip{W_{\rm nc}}{W_nc} represents the work done by the nonconservative forces acting on the object between the initial and the final moments. The work \texttip{W_{\rm nc}}{W_nc} is usually negative; that is, the nonconservative forces tend to decrease, or dissipate, the mechanical energy of the system. In this problem, we will consider the following situation as depicted in the diagram : A block of mass \texttip{m}{m} slides at a speed \texttip{v}{v} along a horizontal, smooth table. It next slides down a smooth ramp, descending a height \texttip{h}{h}, and then slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it does not lose contact with the supporting surfaces (table, ramp, or floor). You will analyze the motion of the block at different moments using the law of conservation of energy. Part A Which word in the statement of this problem allows you to assume that the table is frictionless? ANSWER: Part B straight smooth horizontal Typesetting math: 91% This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H Typesetting math: 91% This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Sliding In Socks Suppose that the coefficient of kinetic friction between Zak’s feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor. Part A If Zak’s speed is 3.00 \rm m/s when he starts to slide, what distance \texttip{d}{d} will he slide before stopping? Express your answer in meters. ANSWER: Typesetting math: 91% Part B This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \rm m Typesetting math: 91%

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The “deadman” lever controls the augers of a snowthrower. If force F = 30 N is required to begin rotating the lever about the ideal pivot at O, determine the corresponding tension T in the control cable AB and the magnitude of the pin reaction at O. Neglect the weight of the lever.

Consider an ideal transformer with 100 turns of wire in the primary and 500 turns of wire in the secondary. If the primary coil is fed with 110 V, rms, what is the power consumed by the device connected to the secondary that behaves as a resistor.