Assignment 10 Due: 11:59pm on Wednesday, April 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 12.3 Part A The figure shows three rotating disks, all of equal mass. Rank in order, from largest to smallest, their rotational kinetic energies to . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: Ka Kc Correct Conceptual Question 12.6 You have two steel solid spheres. Sphere 2 has twice the radius of sphere 1. Part A By what factor does the moment of inertia of sphere 2 exceed the moment of inertia of sphere 1? ANSWER: I2 I1 Correct Problem 12.2 A high-speed drill reaches 2500 in 0.59 . Part A What is the drill’s angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Through how many revolutions does it turn during this first 0.59 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct I2/I1 = 32 rpm s  = 440 rad s2 s  = 12 rev Constant Angular Acceleration in the Kitchen Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. Part A What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in degrees per second per second. Hint 1. How to approach the problem Recall from your study of kinematics the three equations of motion derived for systems undergoing constant linear acceleration. You are now studying systems undergoing constant angular acceleration and will need to work with the three analogous equations of motion. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find the angular acceleration . Hint 2. Find the angular velocity of the salad spinner while Dario is spinning it What is the angular velocity of the salad spinner as Dario is spinning it? Express your answer numerically in degrees per second. Hint 1. Converting rotations to degrees When the salad spinner spins through one revolution, it turns through 360 degrees. ANSWER: Hint 3. Find the angular distance the salad spinner travels as it comes to rest Through how many degrees does the salad spinner rotate as it comes to rest? Express your answer numerically in degrees. Hint 1. Converting rotations to degrees  0 = 1440 degrees/s  =  − 0 One revolution is equivalent to 360 degrees. ANSWER: Hint 4. Determine which equation to use You know the initial and final velocities of the system and the angular distance through which the spinner rotates as it comes to a stop. Which equation should be used to solve for the unknown constant angular acceleration ? ANSWER: ANSWER: Correct Part B How long does it take for the salad spinner to come to rest? Express your answer numerically in seconds.  = 2160 degrees   = 0 + 0t+  1 2 t2  = 0 + t = + 2( − ) 2 20 0  = -480 degrees/s2 Hint 1. How to approach the problem Again, you will need the equations of rotational kinematics that apply to situations of constant angular acceleration. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find . Hint 2. Determine which equation to use You have the initial and final velocities of the system and the angular acceleration, which you found in the previous part. Which is the best equation to use to solve for the unknown time ? ANSWER: ANSWER: Correct ± A Spinning Electric Fan An electric fan is turned off, and its angular velocity decreases uniformly from 540 to 250 in a time interval of length 4.40 . Part A Find the angular acceleration in revolutions per second per second. Hint 1. Average acceleration Recall that if the angular velocity decreases uniformly, the angular acceleration will remain constant. Therefore, the angular acceleration is just the total change in angular velocity divided by t t  = 0 + 0t+  1 2 t2  = 0 + t = + 2( − ) 2 20 0 t = 3.00 s rev/min rev/min s  the total change in time. Be careful of the sign of the angular acceleration. ANSWER: Correct Part B Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A. Hint 1. Determine the correct kinematic equation Which of the following kinematic equations is best suited to this problem? Here and are the initial and final angular velocities, is the elapsed time, is the constant angular acceleration, and and are the initial and final angular displacements. Hint 1. How to chose the right equation Notice that you were given in the problem introduction the initial and final speeds, as well as the length of time between them. In this problem, you are asked to find the number of revolutions (which here is the change in angular displacement, ). If you already found the angular acceleration in Part A, you could use that as well, but you would end up using a more complex equation. Also, in general, it is somewhat favorable to use given quantities instead of quantities that you have calculated. ANSWER:  = -1.10 rev/s2 0  t  0   − 0  = 0 + t  = 0 + t+  1 2 t2 = + 2( − ) 2 20 0 − 0 = (+ )t 1 2 0 ANSWER: Correct Part C How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A? Hint 1. Finding the total time for spin down To find the total time for spin down, just calculate when the velocity will equal zero. This is accomplished by setting the initial velocity plus the acceleration multipled by the time equal to zero and then solving for the time. One can then just subtract the time it took to reach 250 from the total time. Be careful of your signs when you set up the equation. ANSWER: Correct Problem 12.8 A 100 ball and a 230 ball are connected by a 34- -long, massless, rigid rod. The balls rotate about their center of mass at 130 . Part A What is the speed of the 100 ball? Express your answer to two significant figures and include the appropriate units. ANSWER: 29.0 rev rev/min 3.79 s g g cm rpm g Correct Problem 12.10 A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through its center with 0.200 of kinetic energy. Part A What is the speed of a point on the rim? Express your answer with the appropriate units. ANSWER: Correct Problem 12.12 A drum major twirls a 95- -long, 470 baton about its center of mass at 150 . Part A What is the baton’s rotational kinetic energy? Express your answer to two significant figures and include the appropriate units. ANSWER: v = 3.2 ms g cm J 3.65 ms cm g rpm K = 4.4 J Correct Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord’s weight can be ignored. Part A Which of the following statements correctly describes the system shown in the figure? Check all that apply. Hint 1. Conditions for equilibrium If the blocks had the same mass, the system would be in equilibrium. The blocks would have zero acceleration and the tension in each part of the cord would equal the weight of each block. Both parts of the cord would then pull with equal force on the pulley, resulting in a zero net torque and no rotation of the pulley. Is this still the case in the current situation where block B has twice the mass of block A? Hint 2. Rotational analogue of Newton’s second law The net torque of all the forces acting on a rigid body is proportional to the angular acceleration of the body net  and is given by , where is the moment of inertia of the body. Hint 3. Relation between linear and angular acceleration A particle that rotates with angular acceleration has linear acceleration equal to , where is the distance of the particle from the axis of rotation. In the present case, where there is no slipping or stretching of the cord, the cord and the pulley must move together at the same speed. Therefore, if the cord moves with linear acceleration , the pulley must rotate with angular acceleration , where is the radius of the pulley. ANSWER: Correct Part B What happens when block B moves downward? Hint 1. How to approach the problem To determine whether the tensions in both parts of the cord are equal, it is convenient to write a mathematical expression for the net torque on the pulley. This will allow you to relate the tensions in the cord to the pulley’s angular acceleration. Hint 2. Find the net torque on the pulley Let’s assume that the tensions in both parts of the cord are different. Let be the tension in the right cord and the tension in the left cord. If is the radius of the pulley, what is the net torque acting on the pulley? Take the positive sense of rotation to be counterclockwise. Express your answer in terms of , , and . net = I I  a a = R R a  = a R R The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero. T1 T2 R net T1 T2 R Hint 1. Torque The torque of a force with respect to a point is defined as the product of the magnitude times the perpendicular distance between the line of action of and the point . In other words, . ANSWER: ANSWER: Correct Note that if the pulley were stationary (as in many systems where only linear motion is studied), then the tensions in both parts of the cord would be equal. However, if the pulley rotates with a certain angular acceleration, as in the present situation, the tensions must be different. If they were equal, the pulley could not have an angular acceleration. Problem 12.18 Part A In the figure , what is the magnitude of net torque about the axle? Express your answer to two significant figures and include the appropriate units.  F  O F l F  O  = Fl net = R(T2 − T1 ) The left cord pulls on the pulley with greater force than the right cord. The left and right cord pull with equal force on the pulley. The right cord pulls on the pulley with greater force than the left cord. ANSWER: Correct Part B What is the direction of net torque about the axle? ANSWER: Correct Problem 12.22 An athlete at the gym holds a 3.5 steel ball in his hand. His arm is 78 long and has a mass of 3.6 . Assume the center of mass of the arm is at the geometrical center of the arm. Part A What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Express your answer to two significant figures and include the appropriate units.  = 0.20 Nm Clockwise Counterclockwise kg cm kg ANSWER: Correct Part B What is the magnitude of the torque about his shoulder if he holds his arm straight, but below horizontal? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Parallel Axis Theorem The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center of mass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that passes through point p, and is the mass of the object. Part A Suppose a uniform slender rod has length and mass . The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by . Find , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the center of mass Find the distance appropriate to this problem. That is, find the perpendicular distance from the center of mass of the rod to the axis passing through one end of the rod.  = 41 Nm 45  = 29 Nm Icm Ip Ip = Icm + Md2 d M L m Icm = m 1 12 L2 Iend Iend m L d ANSWER: ANSWER: Correct Part B Now consider a cube of mass with edges of length . The moment of inertia of the cube about an axis through its center of mass and perpendicular to one of its faces is given by . Find , the moment of inertia about an axis p through one of the edges of the cube Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the axis Find the perpendicular distance from the center of mass axis to the new edge axis (axis labeled p in the figure). ANSWER: d = L 2 Iend = mL2 3 m a Icm Icm = m 1 6 a2 Iedge Iedge m a o p d ANSWER: Correct Problem 12.26 Starting from rest, a 12- -diameter compact disk takes 2.9 to reach its operating angular velocity of 2000 . Assume that the angular acceleration is constant. The disk’s moment of inertia is . Part A How much torque is applied to the disk? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How many revolutions does it make before reaching full speed? Express your answer using two significant figures. ANSWER: d = a 2 Iedge = 2ma2 3 cm s rpm 2.5 × 10−5 kg m2 = 1.8×10−3  Nm Correct Problem 12.23 An object’s moment of inertia is 2.20 . Its angular velocity is increasing at the rate of 3.70 . Part A What is the total torque on the object? ANSWER: Correct Problem 12.31 A 5.1 cat and a 2.5 bowl of tuna fish are at opposite ends of the 4.0- -long seesaw. N = 48 rev kgm2 rad/s2 8.14 N  m kg kg m Part A How far to the left of the pivot must a 3.8 cat stand to keep the seesaw balanced? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Static Equilibrium of the Arm You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass , length and its center of mass is a distance from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance from the scapula. The deltoid muscle makes an angle of with the horizontal, as shown. Use throughout the problem. Part A kg d = 1.4 m M1 = 3.6 kg L = 0.66 m L1 = 0.33 m L2 = 0.15 m  = 17 g = 9.8 m/s2 Find the tension in the deltoid muscle. Express the tension in newtons, to the nearest integer. Hint 1. Nature of the problem Remember that this is a statics problem, so all forces and torques are balanced (their sums equal zero). Hint 2. Origin of torque Calculate the torque about the point at which the arm attaches to the rest of the body. This allows one to balance the torques without having to worry about the undefined forces at this point. Hint 3. Adding up the torques Add up the torques about the point in which the humerus attaches to the body. Answer in terms of , , , , , and . Remember that counterclockwise torque is positive. ANSWER: ANSWER: Correct Part B Using the conditions for static equilibrium, find the magnitude of the vertical component of the force exerted by the scapula on the humerus (where the humerus attaches to the rest of the body). Express your answer in newtons, to the nearest integer. T L1 L2 M1 g T  total = 0 = L1M1g − Tsin()L2 T = 265 N Fy Hint 1. Total forces involved Recall that there are three vertical forces in this problem: the force of gravity acting on the bone, the force from the vertical component of the muscle tension, and the force exerted by the scapula on the humerus (where it attaches to the rest of the body). ANSWER: Correct Part C Now find the magnitude of the horizontal component of the force exerted by the scapula on the humerus. Express your answer in newtons, to the nearest integer. ANSWER: Correct ± Moments around a Rod A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored for this problem. There are three forces that are applied to the rod at different points and angles: , , and . Note that the dimensions of the bent rod are in centimeters in the figure, although the answers are requested in SI units (kilograms, meters, seconds). |Fy| = 42 N Fx |Fx| = 254 N F 1 F  2 F  3 Part A If and , what does the magnitude of have to be for there to be rotational equilibrium? Answer numerically in newtons to two significant figures. Hint 1. Finding torque about pivot from What is the magnitude of the torque | | provided by around the pivot point? Give your answer numerically in newton-meters to two significant figures. ANSWER: ANSWER: Correct Part B If the L-shaped rod has a moment of inertia , , , and again , how long a time would it take for the object to move through ( /4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures. F3 = 0 F1 = 12 N F 2 F 1   1 F  1 |  1 | = 0.36 N  m F2 = 4.5 N I = 9 kg m2 F1 = 12 N F2 = 27 N F3 = 0 t 45  Hint 1. Find the net torque about the pivot What is the magnitude of the total torque around the pivot point? Answer numerically in newton-meters to two significant figures. ANSWER: Hint 2. Calculate Given the total torque around the pivot point, what is , the magnitude of the angular acceleration? Express your answer numerically in radians per second squared to two significant figures. Hint 1. Equation for If you know the magnitude of the total torque ( ) and the rotational inertia ( ), you can then find the rotational acceleration ( ) from ANSWER: Hint 3. Description of angular kinematics Now that you know the angular acceleration, this is a problem in rotational kinematics; find the time needed to go through a given angle . For constant acceleration ( ) and starting with (where is angular speed) the relation is given by which is analogous to the expression for linear displacement ( ) with constant acceleration ( ) starting from rest, | p ivot| | p ivot| = 1.8 N  m    vot Ivot  pivot = Ipivot.  = 0.20 radians/s2    = 0   = 1  , 2 t2 x a . ANSWER: Correct Part C Now consider the situation in which and , but now a force with nonzero magnitude is acting on the rod. What does have to be to obtain equilibrium? Give a numerical answer, without trigonometric functions, in newtons, to two significant figures. Hint 1. Find the required component of Only the tangential (perpendicular) component of (call it ) provides a torque. What is ? Answer in terms of . You will need to evaluate any trigonometric functions. ANSWER: ANSWER: Correct x = 1 a 2 t2 t = 2.8 s F1 = 12 N F2 = 0 F3 F3 F 3 F  3 F3t F3t F3 F3t = 1 2 F3 F3 = 9.0 N Problem 12.32 A car tire is 55.0 in diameter. The car is traveling at a speed of 24.0 . Part A What is the tire’s rotation frequency, in rpm? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part C What is the speed of a point at the bottom edge of the tire? Express your answer as an integer and include the appropriate units. ANSWER: cm m/s 833 rpm 48.0 ms 0 ms Correct Problem 12.33 A 460 , 8.00-cm-diameter solid cylinder rolls across the floor at 1.30 . Part A What is the can’s kinetic energy? Express your answer with the appropriate units. ANSWER: Correct Problem 12.45 Part A What is the magnitude of the angular momentum of the 780 rotating bar in the figure ? g m/s 0.583 J g ANSWER: Correct Part B What is the direction of the angular momentum of the bar ? ANSWER: Correct Problem 12.46 Part A What is the magnitude of the angular momentum of the 2.20 , 4.60-cm-diameter rotating disk in the figure ? 3.27 kgm2/s into the page out of the page kg ANSWER: Correct Part B What is its direction? ANSWER: Correct Problem 12.60 A 3.0- -long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.46. 3.66×10−2 kgm /s 2 x direction -x direction y direction -y direction z direction -z direction m Part A What is the minimum angle the ladder can make with the floor without slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.61 The 3.0- -long, 90 rigid beam in the following figure is supported at each end. An 70 student stands 2.0 from support 1.  = 47 m kg kg m Part A How much upward force does the support 1 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much upward force does the support 2 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 12.63 A 44 , 5.5- -long beam is supported, but not attached to, the two posts in the figure . A 22 boy starts walking along the beam. You may want to review ( pages 330 – 334) . For help with math skills, you may want to review: F1 = 670 N F2 = 900 N kg m kg The Vector Cross Product Part A How close can he get to the right end of the beam without it falling over? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture of the four forces acting on the beam, indicating both their direction and the place on the beam that the forces are acting. Choose a coordinate system with a direction for the axis along the beam, and indicate the position of the boy. What is the net force on the beam if it is stationary? Just before the beam tips, the force of the left support on the beam is zero. Using the zero net force condition, what is the force due to the right support just before the beam tips? For the beam to remain stationary, what must be zero besides the net force on the beam? Choose a point on the beam, and compute the net torque on the beam about that point. Be sure to choose a positive direction for the rotation axis and therefore the torques. Using the zero torque condition, what is the position of the boy on the beam just prior to tipping? How far is this position from the right edge of the beam? ANSWER: Correct d = 2.0 m Problem 12.68 Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel’s energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.6 diameter and a mass of 270 . Its maximum angular velocity is 1500 . Part A A motor spins up the flywheel with a constant torque of 54 . How long does it take the flywheel to reach top speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much energy is stored in the flywheel? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.2 . What is the average power delivered to the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: m kg rpm N  m t = 250 s = 1.1×106 E J s Correct Part D How much torque does the flywheel exert on the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.71 The 3.30 , 40.0-cm-diameter disk in the figure is spinning at 350 . Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.10 ? P = 2.4×105 W  = 1800 Nm kg rpm s Express your answer with the appropriate units. ANSWER: Correct Problem 12.74 A 5.0 , 60- -diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. Part A What is the magnitude of the cylinder’s initial angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: 5.76 N kg cm  = 22 rad s2 Correct Part B What is the magnitude of the cylinder’s angular velocity when it is directly below the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.82 A 45 figure skater is spinning on the toes of her skates at 0.90 . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 , 20 average diameter, 160 tall) plus two rod-like arms (2.5 each, 67 long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 , 20- -diameter, 200- -tall cylinder. Part A What is her new rotation frequency, in revolutions per second? Express your answer to two significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Score Summary:  = 6.6 rad s kg rev/s kg cm cm kg cm kg cm cm 2 = Your score on this assignment is 95.7%. You received 189.42 out of a possible total of 198 points.

Assignment 10 Due: 11:59pm on Wednesday, April 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 12.3 Part A The figure shows three rotating disks, all of equal mass. Rank in order, from largest to smallest, their rotational kinetic energies to . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: Ka Kc Correct Conceptual Question 12.6 You have two steel solid spheres. Sphere 2 has twice the radius of sphere 1. Part A By what factor does the moment of inertia of sphere 2 exceed the moment of inertia of sphere 1? ANSWER: I2 I1 Correct Problem 12.2 A high-speed drill reaches 2500 in 0.59 . Part A What is the drill’s angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Through how many revolutions does it turn during this first 0.59 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct I2/I1 = 32 rpm s  = 440 rad s2 s  = 12 rev Constant Angular Acceleration in the Kitchen Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. Part A What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in degrees per second per second. Hint 1. How to approach the problem Recall from your study of kinematics the three equations of motion derived for systems undergoing constant linear acceleration. You are now studying systems undergoing constant angular acceleration and will need to work with the three analogous equations of motion. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find the angular acceleration . Hint 2. Find the angular velocity of the salad spinner while Dario is spinning it What is the angular velocity of the salad spinner as Dario is spinning it? Express your answer numerically in degrees per second. Hint 1. Converting rotations to degrees When the salad spinner spins through one revolution, it turns through 360 degrees. ANSWER: Hint 3. Find the angular distance the salad spinner travels as it comes to rest Through how many degrees does the salad spinner rotate as it comes to rest? Express your answer numerically in degrees. Hint 1. Converting rotations to degrees  0 = 1440 degrees/s  =  − 0 One revolution is equivalent to 360 degrees. ANSWER: Hint 4. Determine which equation to use You know the initial and final velocities of the system and the angular distance through which the spinner rotates as it comes to a stop. Which equation should be used to solve for the unknown constant angular acceleration ? ANSWER: ANSWER: Correct Part B How long does it take for the salad spinner to come to rest? Express your answer numerically in seconds.  = 2160 degrees   = 0 + 0t+  1 2 t2  = 0 + t = + 2( − ) 2 20 0  = -480 degrees/s2 Hint 1. How to approach the problem Again, you will need the equations of rotational kinematics that apply to situations of constant angular acceleration. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find . Hint 2. Determine which equation to use You have the initial and final velocities of the system and the angular acceleration, which you found in the previous part. Which is the best equation to use to solve for the unknown time ? ANSWER: ANSWER: Correct ± A Spinning Electric Fan An electric fan is turned off, and its angular velocity decreases uniformly from 540 to 250 in a time interval of length 4.40 . Part A Find the angular acceleration in revolutions per second per second. Hint 1. Average acceleration Recall that if the angular velocity decreases uniformly, the angular acceleration will remain constant. Therefore, the angular acceleration is just the total change in angular velocity divided by t t  = 0 + 0t+  1 2 t2  = 0 + t = + 2( − ) 2 20 0 t = 3.00 s rev/min rev/min s  the total change in time. Be careful of the sign of the angular acceleration. ANSWER: Correct Part B Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A. Hint 1. Determine the correct kinematic equation Which of the following kinematic equations is best suited to this problem? Here and are the initial and final angular velocities, is the elapsed time, is the constant angular acceleration, and and are the initial and final angular displacements. Hint 1. How to chose the right equation Notice that you were given in the problem introduction the initial and final speeds, as well as the length of time between them. In this problem, you are asked to find the number of revolutions (which here is the change in angular displacement, ). If you already found the angular acceleration in Part A, you could use that as well, but you would end up using a more complex equation. Also, in general, it is somewhat favorable to use given quantities instead of quantities that you have calculated. ANSWER:  = -1.10 rev/s2 0  t  0   − 0  = 0 + t  = 0 + t+  1 2 t2 = + 2( − ) 2 20 0 − 0 = (+ )t 1 2 0 ANSWER: Correct Part C How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A? Hint 1. Finding the total time for spin down To find the total time for spin down, just calculate when the velocity will equal zero. This is accomplished by setting the initial velocity plus the acceleration multipled by the time equal to zero and then solving for the time. One can then just subtract the time it took to reach 250 from the total time. Be careful of your signs when you set up the equation. ANSWER: Correct Problem 12.8 A 100 ball and a 230 ball are connected by a 34- -long, massless, rigid rod. The balls rotate about their center of mass at 130 . Part A What is the speed of the 100 ball? Express your answer to two significant figures and include the appropriate units. ANSWER: 29.0 rev rev/min 3.79 s g g cm rpm g Correct Problem 12.10 A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through its center with 0.200 of kinetic energy. Part A What is the speed of a point on the rim? Express your answer with the appropriate units. ANSWER: Correct Problem 12.12 A drum major twirls a 95- -long, 470 baton about its center of mass at 150 . Part A What is the baton’s rotational kinetic energy? Express your answer to two significant figures and include the appropriate units. ANSWER: v = 3.2 ms g cm J 3.65 ms cm g rpm K = 4.4 J Correct Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord’s weight can be ignored. Part A Which of the following statements correctly describes the system shown in the figure? Check all that apply. Hint 1. Conditions for equilibrium If the blocks had the same mass, the system would be in equilibrium. The blocks would have zero acceleration and the tension in each part of the cord would equal the weight of each block. Both parts of the cord would then pull with equal force on the pulley, resulting in a zero net torque and no rotation of the pulley. Is this still the case in the current situation where block B has twice the mass of block A? Hint 2. Rotational analogue of Newton’s second law The net torque of all the forces acting on a rigid body is proportional to the angular acceleration of the body net  and is given by , where is the moment of inertia of the body. Hint 3. Relation between linear and angular acceleration A particle that rotates with angular acceleration has linear acceleration equal to , where is the distance of the particle from the axis of rotation. In the present case, where there is no slipping or stretching of the cord, the cord and the pulley must move together at the same speed. Therefore, if the cord moves with linear acceleration , the pulley must rotate with angular acceleration , where is the radius of the pulley. ANSWER: Correct Part B What happens when block B moves downward? Hint 1. How to approach the problem To determine whether the tensions in both parts of the cord are equal, it is convenient to write a mathematical expression for the net torque on the pulley. This will allow you to relate the tensions in the cord to the pulley’s angular acceleration. Hint 2. Find the net torque on the pulley Let’s assume that the tensions in both parts of the cord are different. Let be the tension in the right cord and the tension in the left cord. If is the radius of the pulley, what is the net torque acting on the pulley? Take the positive sense of rotation to be counterclockwise. Express your answer in terms of , , and . net = I I  a a = R R a  = a R R The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero. T1 T2 R net T1 T2 R Hint 1. Torque The torque of a force with respect to a point is defined as the product of the magnitude times the perpendicular distance between the line of action of and the point . In other words, . ANSWER: ANSWER: Correct Note that if the pulley were stationary (as in many systems where only linear motion is studied), then the tensions in both parts of the cord would be equal. However, if the pulley rotates with a certain angular acceleration, as in the present situation, the tensions must be different. If they were equal, the pulley could not have an angular acceleration. Problem 12.18 Part A In the figure , what is the magnitude of net torque about the axle? Express your answer to two significant figures and include the appropriate units.  F  O F l F  O  = Fl net = R(T2 − T1 ) The left cord pulls on the pulley with greater force than the right cord. The left and right cord pull with equal force on the pulley. The right cord pulls on the pulley with greater force than the left cord. ANSWER: Correct Part B What is the direction of net torque about the axle? ANSWER: Correct Problem 12.22 An athlete at the gym holds a 3.5 steel ball in his hand. His arm is 78 long and has a mass of 3.6 . Assume the center of mass of the arm is at the geometrical center of the arm. Part A What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Express your answer to two significant figures and include the appropriate units.  = 0.20 Nm Clockwise Counterclockwise kg cm kg ANSWER: Correct Part B What is the magnitude of the torque about his shoulder if he holds his arm straight, but below horizontal? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Parallel Axis Theorem The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center of mass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that passes through point p, and is the mass of the object. Part A Suppose a uniform slender rod has length and mass . The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by . Find , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the center of mass Find the distance appropriate to this problem. That is, find the perpendicular distance from the center of mass of the rod to the axis passing through one end of the rod.  = 41 Nm 45  = 29 Nm Icm Ip Ip = Icm + Md2 d M L m Icm = m 1 12 L2 Iend Iend m L d ANSWER: ANSWER: Correct Part B Now consider a cube of mass with edges of length . The moment of inertia of the cube about an axis through its center of mass and perpendicular to one of its faces is given by . Find , the moment of inertia about an axis p through one of the edges of the cube Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the axis Find the perpendicular distance from the center of mass axis to the new edge axis (axis labeled p in the figure). ANSWER: d = L 2 Iend = mL2 3 m a Icm Icm = m 1 6 a2 Iedge Iedge m a o p d ANSWER: Correct Problem 12.26 Starting from rest, a 12- -diameter compact disk takes 2.9 to reach its operating angular velocity of 2000 . Assume that the angular acceleration is constant. The disk’s moment of inertia is . Part A How much torque is applied to the disk? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How many revolutions does it make before reaching full speed? Express your answer using two significant figures. ANSWER: d = a 2 Iedge = 2ma2 3 cm s rpm 2.5 × 10−5 kg m2 = 1.8×10−3  Nm Correct Problem 12.23 An object’s moment of inertia is 2.20 . Its angular velocity is increasing at the rate of 3.70 . Part A What is the total torque on the object? ANSWER: Correct Problem 12.31 A 5.1 cat and a 2.5 bowl of tuna fish are at opposite ends of the 4.0- -long seesaw. N = 48 rev kgm2 rad/s2 8.14 N  m kg kg m Part A How far to the left of the pivot must a 3.8 cat stand to keep the seesaw balanced? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Static Equilibrium of the Arm You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass , length and its center of mass is a distance from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance from the scapula. The deltoid muscle makes an angle of with the horizontal, as shown. Use throughout the problem. Part A kg d = 1.4 m M1 = 3.6 kg L = 0.66 m L1 = 0.33 m L2 = 0.15 m  = 17 g = 9.8 m/s2 Find the tension in the deltoid muscle. Express the tension in newtons, to the nearest integer. Hint 1. Nature of the problem Remember that this is a statics problem, so all forces and torques are balanced (their sums equal zero). Hint 2. Origin of torque Calculate the torque about the point at which the arm attaches to the rest of the body. This allows one to balance the torques without having to worry about the undefined forces at this point. Hint 3. Adding up the torques Add up the torques about the point in which the humerus attaches to the body. Answer in terms of , , , , , and . Remember that counterclockwise torque is positive. ANSWER: ANSWER: Correct Part B Using the conditions for static equilibrium, find the magnitude of the vertical component of the force exerted by the scapula on the humerus (where the humerus attaches to the rest of the body). Express your answer in newtons, to the nearest integer. T L1 L2 M1 g T  total = 0 = L1M1g − Tsin()L2 T = 265 N Fy Hint 1. Total forces involved Recall that there are three vertical forces in this problem: the force of gravity acting on the bone, the force from the vertical component of the muscle tension, and the force exerted by the scapula on the humerus (where it attaches to the rest of the body). ANSWER: Correct Part C Now find the magnitude of the horizontal component of the force exerted by the scapula on the humerus. Express your answer in newtons, to the nearest integer. ANSWER: Correct ± Moments around a Rod A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored for this problem. There are three forces that are applied to the rod at different points and angles: , , and . Note that the dimensions of the bent rod are in centimeters in the figure, although the answers are requested in SI units (kilograms, meters, seconds). |Fy| = 42 N Fx |Fx| = 254 N F 1 F  2 F  3 Part A If and , what does the magnitude of have to be for there to be rotational equilibrium? Answer numerically in newtons to two significant figures. Hint 1. Finding torque about pivot from What is the magnitude of the torque | | provided by around the pivot point? Give your answer numerically in newton-meters to two significant figures. ANSWER: ANSWER: Correct Part B If the L-shaped rod has a moment of inertia , , , and again , how long a time would it take for the object to move through ( /4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures. F3 = 0 F1 = 12 N F 2 F 1   1 F  1 |  1 | = 0.36 N  m F2 = 4.5 N I = 9 kg m2 F1 = 12 N F2 = 27 N F3 = 0 t 45  Hint 1. Find the net torque about the pivot What is the magnitude of the total torque around the pivot point? Answer numerically in newton-meters to two significant figures. ANSWER: Hint 2. Calculate Given the total torque around the pivot point, what is , the magnitude of the angular acceleration? Express your answer numerically in radians per second squared to two significant figures. Hint 1. Equation for If you know the magnitude of the total torque ( ) and the rotational inertia ( ), you can then find the rotational acceleration ( ) from ANSWER: Hint 3. Description of angular kinematics Now that you know the angular acceleration, this is a problem in rotational kinematics; find the time needed to go through a given angle . For constant acceleration ( ) and starting with (where is angular speed) the relation is given by which is analogous to the expression for linear displacement ( ) with constant acceleration ( ) starting from rest, | p ivot| | p ivot| = 1.8 N  m    vot Ivot  pivot = Ipivot.  = 0.20 radians/s2    = 0   = 1  , 2 t2 x a . ANSWER: Correct Part C Now consider the situation in which and , but now a force with nonzero magnitude is acting on the rod. What does have to be to obtain equilibrium? Give a numerical answer, without trigonometric functions, in newtons, to two significant figures. Hint 1. Find the required component of Only the tangential (perpendicular) component of (call it ) provides a torque. What is ? Answer in terms of . You will need to evaluate any trigonometric functions. ANSWER: ANSWER: Correct x = 1 a 2 t2 t = 2.8 s F1 = 12 N F2 = 0 F3 F3 F 3 F  3 F3t F3t F3 F3t = 1 2 F3 F3 = 9.0 N Problem 12.32 A car tire is 55.0 in diameter. The car is traveling at a speed of 24.0 . Part A What is the tire’s rotation frequency, in rpm? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part C What is the speed of a point at the bottom edge of the tire? Express your answer as an integer and include the appropriate units. ANSWER: cm m/s 833 rpm 48.0 ms 0 ms Correct Problem 12.33 A 460 , 8.00-cm-diameter solid cylinder rolls across the floor at 1.30 . Part A What is the can’s kinetic energy? Express your answer with the appropriate units. ANSWER: Correct Problem 12.45 Part A What is the magnitude of the angular momentum of the 780 rotating bar in the figure ? g m/s 0.583 J g ANSWER: Correct Part B What is the direction of the angular momentum of the bar ? ANSWER: Correct Problem 12.46 Part A What is the magnitude of the angular momentum of the 2.20 , 4.60-cm-diameter rotating disk in the figure ? 3.27 kgm2/s into the page out of the page kg ANSWER: Correct Part B What is its direction? ANSWER: Correct Problem 12.60 A 3.0- -long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.46. 3.66×10−2 kgm /s 2 x direction -x direction y direction -y direction z direction -z direction m Part A What is the minimum angle the ladder can make with the floor without slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.61 The 3.0- -long, 90 rigid beam in the following figure is supported at each end. An 70 student stands 2.0 from support 1.  = 47 m kg kg m Part A How much upward force does the support 1 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much upward force does the support 2 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 12.63 A 44 , 5.5- -long beam is supported, but not attached to, the two posts in the figure . A 22 boy starts walking along the beam. You may want to review ( pages 330 – 334) . For help with math skills, you may want to review: F1 = 670 N F2 = 900 N kg m kg The Vector Cross Product Part A How close can he get to the right end of the beam without it falling over? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture of the four forces acting on the beam, indicating both their direction and the place on the beam that the forces are acting. Choose a coordinate system with a direction for the axis along the beam, and indicate the position of the boy. What is the net force on the beam if it is stationary? Just before the beam tips, the force of the left support on the beam is zero. Using the zero net force condition, what is the force due to the right support just before the beam tips? For the beam to remain stationary, what must be zero besides the net force on the beam? Choose a point on the beam, and compute the net torque on the beam about that point. Be sure to choose a positive direction for the rotation axis and therefore the torques. Using the zero torque condition, what is the position of the boy on the beam just prior to tipping? How far is this position from the right edge of the beam? ANSWER: Correct d = 2.0 m Problem 12.68 Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel’s energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.6 diameter and a mass of 270 . Its maximum angular velocity is 1500 . Part A A motor spins up the flywheel with a constant torque of 54 . How long does it take the flywheel to reach top speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much energy is stored in the flywheel? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.2 . What is the average power delivered to the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: m kg rpm N  m t = 250 s = 1.1×106 E J s Correct Part D How much torque does the flywheel exert on the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.71 The 3.30 , 40.0-cm-diameter disk in the figure is spinning at 350 . Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.10 ? P = 2.4×105 W  = 1800 Nm kg rpm s Express your answer with the appropriate units. ANSWER: Correct Problem 12.74 A 5.0 , 60- -diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. Part A What is the magnitude of the cylinder’s initial angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: 5.76 N kg cm  = 22 rad s2 Correct Part B What is the magnitude of the cylinder’s angular velocity when it is directly below the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.82 A 45 figure skater is spinning on the toes of her skates at 0.90 . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 , 20 average diameter, 160 tall) plus two rod-like arms (2.5 each, 67 long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 , 20- -diameter, 200- -tall cylinder. Part A What is her new rotation frequency, in revolutions per second? Express your answer to two significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Score Summary:  = 6.6 rad s kg rev/s kg cm cm kg cm kg cm cm 2 = Your score on this assignment is 95.7%. You received 189.42 out of a possible total of 198 points.

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Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g               , (Eq.2) M slope m g       (Eq.3) M b f        Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g                Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060  m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm      To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s   0.4% 100 9.80 9.76 9.80 100 . . . %        Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2           Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol  (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30  Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T   L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s    Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the  is dropped for simplicity), Vo is V at t = 0 (try it), and  (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating  and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o    Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g               , (Eq.2) M slope m g       (Eq.3) M b f        Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g                Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060  m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm      To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s   0.4% 100 9.80 9.76 9.80 100 . . . %        Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2           Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol  (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30  Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T   L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s    Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the  is dropped for simplicity), Vo is V at t = 0 (try it), and  (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating  and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o    Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

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Assignment 9 Due: 11:59pm on Friday, April 11, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 11.2 Part A Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Part B Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Problem 11.4  A B = 5 − 6 A i ^ j ^ = −9 − 5 B i ^ j ^ A  B  = -15  A B = −5 + 9 A i ^ j ^ = 5 + 6 B i ^ j ^ A  B  = 29 Part A What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct ± All Work and No Play Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as , where is the work done by force on the object that undergoes displacement directed at angle relative to .  A B A = 2 + 5 ı ^  ^ B = −2 − 4 ı ^  ^  = 175  A B A = −6 + 2 ı ^  ^ B = − − 3 ı ^  ^  = 90 W =  = cos  F  s  F   s  W F  s  F  Note that depending on the value of , the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force ? ANSWER: Correct When , the cosine of is zero, and therefore the work done is zero. Part B cos  F  1 It is positive. It is negative. It is zero. There is not enough information to answer the question.  = 90  What can be said about the work done by force ? ANSWER: Correct When , is positive, and so the work done is positive. Part C The work done by force is ANSWER: Correct When , is negative, and so the work done is negative. Part D The work done by force is ANSWER: F  2 It is positive. It is negative. It is zero. 0 <  < 90 cos  F  3 positive negative zero 90 <  < 180 cos  F  4 Correct Part E The work done by force is ANSWER: Correct positive negative zero F  5 positive negative zero Part F The work done by force is ANSWER: Correct Part G The work done by force is ANSWER: Correct In the next series of questions, you will use the formula to calculate the work done by various forces on an object that moves 160 meters to the right. F  6 positive negative zero F  7 positive negative zero W =  = cos  F  s  F   s  Part H Find the work done by the 18-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part I Find the work done by the 30-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part J Find the work done by the 12-newton force. Use two significant figures in your answer. Express your answer in joules. W W = 2900 J W W = 4200 J W ANSWER: Correct Part K Find the work done by the 15-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states , where is the work done by all forces that act on the object, and and are the initial and final kinetic energies, respectively. Part A The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. W = -1900 J W W = -1800 J Kf = Ki + Wall Wall Ki Kf Choose the best answer to fill in the blanks above: ANSWER: Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: Correct Part C To illustrate the work-energy concept, consider the case of a stone falling from to under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: distance / potential distance / kinetic vertical displacement / potential none of the above acceleration work distance potential energy xi xf ANSWER: Correct Potential Energy You should read about potential energy in your text before answering the following questions. Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: , where and are the final and initial potential energies, and is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy. Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic Kf + Uf = Ef = Wnc + Ei = Wnc + Ki + Ui Uf Ui Wnc Correct Part E This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: Correct Problem 11.7 Part A How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: work / potential force / kinetic change / potential sum / conserved sum / zero sum / not conserved difference / conserved F  = (− + i ^ ) N j ^ ! = r m i ^ Correct Part B How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.10 A 1.8 book is lying on a 0.80- -high table. You pick it up and place it on a bookshelf 2.27 above the floor. Part A How much work does gravity do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B W = -8.6 J F  = (− + i ^ ) N j ^ ! = r m? j ^ W = 26 J kg m m Wg = -26 J How much work does your hand do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.12 The three ropes shown in the bird's-eye view of the figure are used to drag a crate 3.3 across the floor. Part A How much work is done by each of the three forces? Express your answers using two significant figures. Enter your answers numerically separated by commas. ANSWER: WH = 26 J m W1 , W2 , W3 = 1.9,1.2,-2.1 kJ Correct Enhanced EOC: Problem 11.16 A 1.2 particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 4.6 at . You may want to review ( pages 286 - 287) . For help with math skills, you may want to review: The Definite Integral Part A What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: kg m/s x = 0m x = 2m x = 0 m x = 0 m x = 2 m x = 2 m x = 2 m Correct Part B What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Can the work be negative? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: Correct Work on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. A force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed. v = 6.2 ms x = 4m x = 0 m x = 0 m x = 4 m x = 4 m x = 4 m v = 4.6 ms w  F Part A The block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. Hint 1. What physical principle to use To find the total work done on the block, use the work-energy theorem: . Hint 2. Find the change in kinetic energy What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a distance ? Remember that the block is pulled at constant speed. Hint 1. Consider kinetic energy If the block's speed does not change, its kinetic energy cannot change. ANSWER: ANSWER: L Wtot Wtot = Kf − Ki L Kf − Ki = 0 Wtot = 0 Correct Part B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. Hint 1. Force diagram Hint 2. Force of gravity component What is the component of the force of gravity in the direction of the block's displacement (along the inclined plane)? Express your answer in terms of and . Hint 1. Relative direction of the force and the motion Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up the plane. ANSWER: Wg L w w  ANSWER: Correct Part C What is , the work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of and other given quantities. Hint 1. How to find the work done by a constant force Remember that the work done on an object by a particular force is the integral of the dot product of the force and the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is constant and the path is a straight segment of length up the inclined plane, the dot product becomes simple multiplication. ANSWER: Correct Part D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. Hint 1. First step in computing the work Fg|| = −wsin() Wg = −wLsin() WF F L F L WF = FL Wnormal L The work done by the normal force is equal to the dot product of the force vector and the block's displacement vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot product? ANSWER: ANSWER: Correct Problem 11.20 A particle moving along the -axis has the potential energy , where is in . Part A What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. N  L = 0 Wnormal = 0 y U = 3.2y3 J y m y y = 0 m Fy = 0 N y y = 1 m ANSWER: Correct Part C What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.28 A cable with 25.0 of tension pulls straight up on a 1.08 block that is initially at rest. Part A What is the block's speed after being lifted 2.40 ? Solve this problem using work and energy. Express your answer with the appropriate units. ANSWER: Correct Fy = -9.6 N y y = 2 m Fy = -38 N N kg m vf = 8.00 ms Problem 11.29 Part A How much work does an elevator motor do to lift a 1500 elevator a height of 110 ? Express your answer with the appropriate units. ANSWER: Correct Part B How much power must the motor supply to do this in 50 at constant speed? Express your answer with the appropriate units. ANSWER: Correct Problem 11.32 How many energy is consumed by a 1.20 hair dryer used for 10.0 and a 11.0 night light left on for 16.0 ? Part A Hair dryer: Express your answer with the appropriate units. kg m Wext = 1.62×106 J s = 3.23×104 P W kW min W hr ANSWER: Correct Part B Night light: Express your answer with the appropriate units. ANSWER: Correct Problem 11.42 A 2500 elevator accelerates upward at 1.20 for 10.0 , starting from rest. Part A How much work does gravity do on the elevator? Express your answer with the appropriate units. ANSWER: Correct W = 7.20×105 J = 6.34×105 W J kg m/s2 m −2.45×105 J Part B How much work does the tension in the elevator cable do on the elevator? Express your answer with the appropriate units. ANSWER: Correct Part C Use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10.0 . Express your answer with the appropriate units. ANSWER: Correct Part D What is the speed of the elevator as it reaches 10.0 ? Express your answer with the appropriate units. ANSWER: Correct 2.75×105 J m 3.00×104 J m 4.90 ms Problem 11.47 A horizontal spring with spring constant 130 is compressed 17 and used to launch a 2.4 box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Part A Use work and energy to find how far the box slides across the rough surface before stopping. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.49 Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 and the coefficient of rolling friction is 0.45. Part A Use work and energy to find the length of a ramp that will stop a 15,000 truck that enters the ramp at 30 . Express your answer to two significant figures and include the appropriate units. ANSWER: Correct N/m cm kg l = 53 cm kg m/s l = 83 m Problem 11.51 Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor. Part A Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is . Express your answer in terms of the variables , , , , and free fall acceleration . ANSWER: Part B Find an expression for the speed of the block if the table is frictionless. Express your answer in terms of the variables , , , and free fall acceleration . ANSWER: μk M m h μk g v = M m h g Problem 11.57 The spring shown in the figure is compressed 60 and used to launch a 100 physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the incline is 0.12 . Part A What is the student's speed just after losing contact with the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How far up the incline does the student go? Express your answer to two significant figures and include the appropriate units. ANSWER: v = cm kg 30 v = 17 ms Correct Score Summary: Your score on this assignment is 93.6%. You received 112.37 out of a possible total of 120 points. !s = 41 m

Assignment 9 Due: 11:59pm on Friday, April 11, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 11.2 Part A Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Part B Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Problem 11.4  A B = 5 − 6 A i ^ j ^ = −9 − 5 B i ^ j ^ A  B  = -15  A B = −5 + 9 A i ^ j ^ = 5 + 6 B i ^ j ^ A  B  = 29 Part A What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct ± All Work and No Play Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as , where is the work done by force on the object that undergoes displacement directed at angle relative to .  A B A = 2 + 5 ı ^  ^ B = −2 − 4 ı ^  ^  = 175  A B A = −6 + 2 ı ^  ^ B = − − 3 ı ^  ^  = 90 W =  = cos  F  s  F   s  W F  s  F  Note that depending on the value of , the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force ? ANSWER: Correct When , the cosine of is zero, and therefore the work done is zero. Part B cos  F  1 It is positive. It is negative. It is zero. There is not enough information to answer the question.  = 90  What can be said about the work done by force ? ANSWER: Correct When , is positive, and so the work done is positive. Part C The work done by force is ANSWER: Correct When , is negative, and so the work done is negative. Part D The work done by force is ANSWER: F  2 It is positive. It is negative. It is zero. 0 <  < 90 cos  F  3 positive negative zero 90 <  < 180 cos  F  4 Correct Part E The work done by force is ANSWER: Correct positive negative zero F  5 positive negative zero Part F The work done by force is ANSWER: Correct Part G The work done by force is ANSWER: Correct In the next series of questions, you will use the formula to calculate the work done by various forces on an object that moves 160 meters to the right. F  6 positive negative zero F  7 positive negative zero W =  = cos  F  s  F   s  Part H Find the work done by the 18-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part I Find the work done by the 30-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part J Find the work done by the 12-newton force. Use two significant figures in your answer. Express your answer in joules. W W = 2900 J W W = 4200 J W ANSWER: Correct Part K Find the work done by the 15-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states , where is the work done by all forces that act on the object, and and are the initial and final kinetic energies, respectively. Part A The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. W = -1900 J W W = -1800 J Kf = Ki + Wall Wall Ki Kf Choose the best answer to fill in the blanks above: ANSWER: Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: Correct Part C To illustrate the work-energy concept, consider the case of a stone falling from to under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: distance / potential distance / kinetic vertical displacement / potential none of the above acceleration work distance potential energy xi xf ANSWER: Correct Potential Energy You should read about potential energy in your text before answering the following questions. Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: , where and are the final and initial potential energies, and is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy. Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic Kf + Uf = Ef = Wnc + Ei = Wnc + Ki + Ui Uf Ui Wnc Correct Part E This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: Correct Problem 11.7 Part A How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: work / potential force / kinetic change / potential sum / conserved sum / zero sum / not conserved difference / conserved F  = (− + i ^ ) N j ^ ! = r m i ^ Correct Part B How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.10 A 1.8 book is lying on a 0.80- -high table. You pick it up and place it on a bookshelf 2.27 above the floor. Part A How much work does gravity do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B W = -8.6 J F  = (− + i ^ ) N j ^ ! = r m? j ^ W = 26 J kg m m Wg = -26 J How much work does your hand do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.12 The three ropes shown in the bird's-eye view of the figure are used to drag a crate 3.3 across the floor. Part A How much work is done by each of the three forces? Express your answers using two significant figures. Enter your answers numerically separated by commas. ANSWER: WH = 26 J m W1 , W2 , W3 = 1.9,1.2,-2.1 kJ Correct Enhanced EOC: Problem 11.16 A 1.2 particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 4.6 at . You may want to review ( pages 286 - 287) . For help with math skills, you may want to review: The Definite Integral Part A What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: kg m/s x = 0m x = 2m x = 0 m x = 0 m x = 2 m x = 2 m x = 2 m Correct Part B What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Can the work be negative? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: Correct Work on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. A force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed. v = 6.2 ms x = 4m x = 0 m x = 0 m x = 4 m x = 4 m x = 4 m v = 4.6 ms w  F Part A The block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. Hint 1. What physical principle to use To find the total work done on the block, use the work-energy theorem: . Hint 2. Find the change in kinetic energy What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a distance ? Remember that the block is pulled at constant speed. Hint 1. Consider kinetic energy If the block's speed does not change, its kinetic energy cannot change. ANSWER: ANSWER: L Wtot Wtot = Kf − Ki L Kf − Ki = 0 Wtot = 0 Correct Part B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. Hint 1. Force diagram Hint 2. Force of gravity component What is the component of the force of gravity in the direction of the block's displacement (along the inclined plane)? Express your answer in terms of and . Hint 1. Relative direction of the force and the motion Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up the plane. ANSWER: Wg L w w  ANSWER: Correct Part C What is , the work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of and other given quantities. Hint 1. How to find the work done by a constant force Remember that the work done on an object by a particular force is the integral of the dot product of the force and the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is constant and the path is a straight segment of length up the inclined plane, the dot product becomes simple multiplication. ANSWER: Correct Part D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. Hint 1. First step in computing the work Fg|| = −wsin() Wg = −wLsin() WF F L F L WF = FL Wnormal L The work done by the normal force is equal to the dot product of the force vector and the block's displacement vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot product? ANSWER: ANSWER: Correct Problem 11.20 A particle moving along the -axis has the potential energy , where is in . Part A What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. N  L = 0 Wnormal = 0 y U = 3.2y3 J y m y y = 0 m Fy = 0 N y y = 1 m ANSWER: Correct Part C What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.28 A cable with 25.0 of tension pulls straight up on a 1.08 block that is initially at rest. Part A What is the block's speed after being lifted 2.40 ? Solve this problem using work and energy. Express your answer with the appropriate units. ANSWER: Correct Fy = -9.6 N y y = 2 m Fy = -38 N N kg m vf = 8.00 ms Problem 11.29 Part A How much work does an elevator motor do to lift a 1500 elevator a height of 110 ? Express your answer with the appropriate units. ANSWER: Correct Part B How much power must the motor supply to do this in 50 at constant speed? Express your answer with the appropriate units. ANSWER: Correct Problem 11.32 How many energy is consumed by a 1.20 hair dryer used for 10.0 and a 11.0 night light left on for 16.0 ? Part A Hair dryer: Express your answer with the appropriate units. kg m Wext = 1.62×106 J s = 3.23×104 P W kW min W hr ANSWER: Correct Part B Night light: Express your answer with the appropriate units. ANSWER: Correct Problem 11.42 A 2500 elevator accelerates upward at 1.20 for 10.0 , starting from rest. Part A How much work does gravity do on the elevator? Express your answer with the appropriate units. ANSWER: Correct W = 7.20×105 J = 6.34×105 W J kg m/s2 m −2.45×105 J Part B How much work does the tension in the elevator cable do on the elevator? Express your answer with the appropriate units. ANSWER: Correct Part C Use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10.0 . Express your answer with the appropriate units. ANSWER: Correct Part D What is the speed of the elevator as it reaches 10.0 ? Express your answer with the appropriate units. ANSWER: Correct 2.75×105 J m 3.00×104 J m 4.90 ms Problem 11.47 A horizontal spring with spring constant 130 is compressed 17 and used to launch a 2.4 box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Part A Use work and energy to find how far the box slides across the rough surface before stopping. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.49 Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 and the coefficient of rolling friction is 0.45. Part A Use work and energy to find the length of a ramp that will stop a 15,000 truck that enters the ramp at 30 . Express your answer to two significant figures and include the appropriate units. ANSWER: Correct N/m cm kg l = 53 cm kg m/s l = 83 m Problem 11.51 Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor. Part A Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is . Express your answer in terms of the variables , , , , and free fall acceleration . ANSWER: Part B Find an expression for the speed of the block if the table is frictionless. Express your answer in terms of the variables , , , and free fall acceleration . ANSWER: μk M m h μk g v = M m h g Problem 11.57 The spring shown in the figure is compressed 60 and used to launch a 100 physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the incline is 0.12 . Part A What is the student's speed just after losing contact with the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How far up the incline does the student go? Express your answer to two significant figures and include the appropriate units. ANSWER: v = cm kg 30 v = 17 ms Correct Score Summary: Your score on this assignment is 93.6%. You received 112.37 out of a possible total of 120 points. !s = 41 m

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You are to choose 2 websites, with different purposes, and review the websites based on the criteria listed below. 1. Starting Point a. Composition Matches Site Purpose b. Target Audience Apparent c. Composition Appropriate for Target Audience 2. Site design a. Consistency within site b. Consistency among pages 3. Visually Pleasing Composition 4. Visual Style in Web Design a. Consistency b. Distinctiveness 5. Focus and Emphasis a. What is emphasized? b. How is emphasis achieved? 6. Consistency a. Real World b. Internal 7. Navigation and Flow a. Home page identifiable throughout b. Location within site apparent c. Navigation consistent; rule-based; appropriate 8. Grouping a. Grouping with White Space b. Grouping with Borders c. Grouping with Backgrounds 9. Response time 10. Links a. Titled b. Incoming c. Outgoing d. Color 11. Detailed content a. Meaningful headings b. Plain language c. Page chunking d. Long blocks of text e. Scrolling f. Use of “within” page links 12. Articles a. Clear headings b. Plain language 13. Presenting Information Simply and Meaningfully a. Legibility b. Readability c. Information in Usable Form d. Visual Lines Clear 14. Legibility of content a. Font color b. Font size c. Font style d. Background color e. Background graphic 15. Documentation

You are to choose 2 websites, with different purposes, and review the websites based on the criteria listed below. 1. Starting Point a. Composition Matches Site Purpose b. Target Audience Apparent c. Composition Appropriate for Target Audience 2. Site design a. Consistency within site b. Consistency among pages 3. Visually Pleasing Composition 4. Visual Style in Web Design a. Consistency b. Distinctiveness 5. Focus and Emphasis a. What is emphasized? b. How is emphasis achieved? 6. Consistency a. Real World b. Internal 7. Navigation and Flow a. Home page identifiable throughout b. Location within site apparent c. Navigation consistent; rule-based; appropriate 8. Grouping a. Grouping with White Space b. Grouping with Borders c. Grouping with Backgrounds 9. Response time 10. Links a. Titled b. Incoming c. Outgoing d. Color 11. Detailed content a. Meaningful headings b. Plain language c. Page chunking d. Long blocks of text e. Scrolling f. Use of “within” page links 12. Articles a. Clear headings b. Plain language 13. Presenting Information Simply and Meaningfully a. Legibility b. Readability c. Information in Usable Form d. Visual Lines Clear 14. Legibility of content a. Font color b. Font size c. Font style d. Background color e. Background graphic 15. Documentation

http://www.physicsclassroom.com/ http://www.usa.gov/ 1.                  Starting Point a.       Composition Matches Site Purpose … Read More...
Human Computer Interaction You are to choose 2 websites, with different purposes, and review the websites based on the criteria listed below. 1. Starting Point a. Composition Matches Site Purpose b. Target Audience Apparent c. Composition Appropriate for Target Audience 2. Site design a. Consistency within site b. Consistency among pages 3. Visually Pleasing Composition 4. Visual Style in Web Design a. Consistency b. Distinctiveness 5. Focus and Emphasis a. What is emphasized? b. How is emphasis achieved? 6. Consistency a. Real World b. Internal 7. Navigation and Flow a. Home page identifiable throughout b. Location within site apparent c. Navigation consistent; rule-based; appropriate 8. Grouping a. Grouping with White Space b. Grouping with Borders c. Grouping with Backgrounds 9. Response time 10. Links a. Titled b. Incoming c. Outgoing d. Color 11. Detailed content a. Meaningful headings b. Plain language c. Page chunking d. Long blocks of text e. Scrolling f. Use of “within” page links 12. Articles a. Clear headings b. Plain language 13. Presenting Information Simply and Meaningfully a. Legibility b. Readability c. Information in Usable Form d. Visual Lines Clear 14. Legibility of content a. Font color b. Font size c. Font style d. Background color e. Background graphic 15. Documentation a. Included b. Searchable c. Links to difficult concepts/words 16. Multimedia a. Animation/Audio/Video/Still images b. Load time given c. Add-in required d. Quality e. Appropriateness of use 17. Scrolling and Paging a. Usage b. Appropriate? 18. Amount of Information Presented Appropriate 19. Other factors to note?

Human Computer Interaction You are to choose 2 websites, with different purposes, and review the websites based on the criteria listed below. 1. Starting Point a. Composition Matches Site Purpose b. Target Audience Apparent c. Composition Appropriate for Target Audience 2. Site design a. Consistency within site b. Consistency among pages 3. Visually Pleasing Composition 4. Visual Style in Web Design a. Consistency b. Distinctiveness 5. Focus and Emphasis a. What is emphasized? b. How is emphasis achieved? 6. Consistency a. Real World b. Internal 7. Navigation and Flow a. Home page identifiable throughout b. Location within site apparent c. Navigation consistent; rule-based; appropriate 8. Grouping a. Grouping with White Space b. Grouping with Borders c. Grouping with Backgrounds 9. Response time 10. Links a. Titled b. Incoming c. Outgoing d. Color 11. Detailed content a. Meaningful headings b. Plain language c. Page chunking d. Long blocks of text e. Scrolling f. Use of “within” page links 12. Articles a. Clear headings b. Plain language 13. Presenting Information Simply and Meaningfully a. Legibility b. Readability c. Information in Usable Form d. Visual Lines Clear 14. Legibility of content a. Font color b. Font size c. Font style d. Background color e. Background graphic 15. Documentation a. Included b. Searchable c. Links to difficult concepts/words 16. Multimedia a. Animation/Audio/Video/Still images b. Load time given c. Add-in required d. Quality e. Appropriateness of use 17. Scrolling and Paging a. Usage b. Appropriate? 18. Amount of Information Presented Appropriate 19. Other factors to note?

Human Computer Interaction You are to choose 2 websites, with … Read More...
Assignment 4 Due: 11:59pm on Wednesday, February 26, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy ± Two Forces Acting at a Point Two forces, and , act at a point. has a magnitude of 9.80 and is directed at an angle of 56.0 above the negative x axis in the second quadrant. has a magnitude of 5.20 and is directed at an angle of 54.1 below the negative x axis in the third quadrant. Part A What is the x component of the resultant force? Express your answer in newtons. Hint 1. How to approach the problem The resultant force is defined as the vector sum of all forces. Thus, its x component is the sum of the x components of the forces, and its y component is the sum of the y components of the forces. Hint 2. Find the x component of Find the x component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . F 1 F  2 F  1 N  F  2 N  F 1 F  1 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 Hint 2. Find the direction of is directed at an angle of 56.0 above the x axis in the second quadrant. When you calculate the components of , however, the direction of the force is commonly expressed in terms of the angle that the vector representing the force forms with the positive x axis. What is the angle that forms with the positive x axis? Select an answer from the following list, where 56.0 . ANSWER: ANSWER: Hint 3. Find the x component of Find the x component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , F 1 F 1  F  1 F  1  =   180 −  180 +  90 +  -5.48 N F 2 F  2 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Typesetting math: 100% and if . Hint 2. Find the direction of is directed at an angle of 54.1 below the x axis in the third quadrant. When you calculate the components of , however, the direction of the force is commonly expressed in terms of the angle that the vector representing the force forms with the positive x axis. What is the angle that forms with the positive x axis? Select an answer from the following list, where 54.1 . ANSWER: ANSWER: ANSWER: Correct Part B What is the y component of the resultant force? Express your answer in newtons. Ax < 0 Ay <  <  < 3 2 F 2 F 2  F 2 F  2  =   180 −   − 180 −90 −  -3.05 N -8.53 N Typesetting math: 100% Hint 1. How to approach the problem Follow the same procedure that you used in Part A to find the x component of the resultant force, though now calculate the y components of the two forces. Hint 2. Find the y component of Find the y component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . ANSWER: Hint 3. Find the y component of Find the y component of . Express your answer in newtons. Hint 1. Components of a vector F 1 F  1 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 8.12 N F 2 F  2 Typesetting math: 100% Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . ANSWER: ANSWER: Correct Part C What is the magnitude of the resultant force? Express your answer in newtons. Hint 1. Magnitude of a vector Consider a vector , whose components are and . The magnitude of is . A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 -4.21 N 3.91 N A Ax Ay A A = A + 2 x A2 y −−−−−−−  Typesetting math: 100% ANSWER: Correct Enhanced EOC: Problem 5.9 The figure shows acceleration-versus-force graphs for two objects pulled by rubber bands. You may want to review ( pages 127 - 130) . For help with math skills, you may want to review: Finding the Slope of a Line from a Graph Part A What is the mass ratio ? Express your answer using two significant figures. 9.38 N m1 m2 Typesetting math: 100% Hint 1. How to approach the problem How are the acceleration and the force on an object related to its mass? How is the slope of each line in the figure related to each object's mass? For each line, what two points are easy to measure accurately to determine the slope of line? How is the slope determined from the x and y coordinates of the two points you chose for each line? ANSWER: Correct A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude . Part A How much horizontal force must a sprinter of mass 54 exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Hint 1. Newton's 2nd law of motion According to Newton's 2nd law of motion, if a net external force acts on a body, the body accelerates, and the net force is equal to the mass of the body times the acceleration of the body: . ANSWER: = 0.36 m1 m2 15 m/s2 F kg Fnet m a Fnet = ma F = 810 N Typesetting math: 100% Correct Part B Which body exerts the force that propels the sprinter, the blocks or the sprinter? Hint 1. How to approach the question To start moving forward, sprinters push backward on the starting blocks with their feet. Newton's 3rd law tells you that the blocks exert a force on the sprinter of the same magnitude, but opposite in direction. ANSWER: Correct To start moving forward, sprinters push backward on the starting blocks with their feet. As a reaction, the blocks push forward on their feet with a force of the same magnitude. This external force accelerates the sprinter forward. Problem 5.12 The figure shows an acceleration-versus-force graph for a 600 object. the blocks the sprinter g Typesetting math: 100% Part A What must equal in order for the graph to be correct? Express your answer with the appropriate units. ANSWER: Correct Part B What must equal in order for the graph to be correct? Express your answer with the appropriate units. ANSWER: Correct Free-Body Diagrams Learning Goal: To gain practice drawing free-body diagrams Whenever you face a problem involving forces, always start with a free-body diagram. a1 a1 = 1.67 m s2 a2 a2 = 3.33 m s2 Typesetting math: 100% To draw a free-body diagram use the following steps: Isolate the object of interest. It is customary to represent the object of interest as a point 1. in your diagram. Identify all the forces acting on the object and their directions. Do not include forces acting on other objects in the problem. Also, do not include quantities, such as velocities and accelerations, that are not forces. 2. Draw the vectors for each force acting on your object of interest. When possible, the length of the force vectors you draw should represent the relative magnitudes of the forces acting on the object. 3. In most problems, after you have drawn the free-body diagrams, you will explicitly label your coordinate axes and directions. Always make the object of interest the origin of your coordinate system. Then you will need to divide the forces into x and y components, sum the x and y forces, and apply Newton's first or second law. In this problem you will only draw the free-body diagram. Suppose that you are asked to solve the following problem: Chadwick is pushing a piano across a level floor (see the figure). The piano can slide across the floor without friction. If Chadwick applies a horizontal force to the piano, what is the piano's acceleration? To solve this problem you should start by drawing a free-body diagram. Part A Determine the object of interest for the situation described in the problem introduction. Hint 1. How to approach the problem You should first think about the question you are trying to answer: What is the acceleration of the piano? The object of interest in this situation will be the object whose acceleration you are asked to find. ANSWER: Typesetting math: 100% Correct Part B Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano. Check all that apply. ANSWER: Correct Now that you have identified the forces acting on the piano, you should draw the free-body diagram. Draw the length of your vectors to represent the relative magnitudes of the forces, but you don't need to worry about the exact scale. You won't have the exact value of all of the forces until you finish solving the problem. To maximize your learning, you should draw the diagram yourself before looking at the choices in the next part. You are on your honor to do so. Part C For this situation you should draw a free-body diagram for the floor. Chadwick. the piano. acceleration of the piano gravitational force acting on the piano (piano's weight) speed of the piano gravitational force acting on Chadwick (Chadwick's weight) force of the floor on the piano (normal force) force of the piano on the floor force of Chadwick on the piano force of the piano pushing on Chadwick Typesetting math: 100% Select the choice that best matches the free-body diagram you have drawn for the piano. Hint 1. Determine the directions and relative magnitudes of the forces Which of the following statements best describes the correct directions and relative magnitudes of the forces involved? ANSWER: ANSWER: The normal force and weight are both upward and the pushing force is horizontal. The normal force and weight are both downward and the pushing force is horizontal. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force has a greater magnitude than the weight. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force and weight have the same magnitude. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force has a smaller magnitude than the weight. Typesetting math: 100% Typesetting math: 100% Correct If you were actually going to solve this problem rather than just draw the free-body diagram, you would need to define the coordinate system. Choose the position of the piano as the origin. In this case it is simplest to let the y axis point vertically upward and the x axis point horizontally to the right, in the direction of the acceleration. Chadwick now needs to push the piano up a ramp and into a moving van. at left. The ramp is frictionless. Is Chadwick strong enough to push the piano up the ramp alone or must he get help? To solve this problem you should start by drawing a free-body diagram. Part D Determine the object of interest for this situation. ANSWER: Correct Now draw the free-body diagram of the piano in this new situation. Follow the same sequence of steps that you followed for the first situation. Again draw your diagram before you look at the choices For this situation, you should draw a free-body diagram for the ramp. Chadwick. the piano. Typesetting math: 100% below. Part E Which diagram accurately represents the free-body diagram for the piano? ANSWER: Typesetting math: 100% Typesetting math: 100% Correct In working problems like this one that involve an incline, it is most often easiest to select a coordinate system that is not vertical and horizontal. Instead, choose the x axis so that it is parallel to the incline and choose the y axis so that it is perpendicular to the incline. Problem 5.18 The figure shows two of the three forces acting on an object in equilibrium. Part A Redraw the diagram, showing all three forces. Label the third force . Draw the force vector starting at the black dot. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: F  3 Typesetting math: 100% Correct Problem 5.25 An ice hockey puck glides across frictionless ice. Part A Identify all forces acting on the object. ANSWER: Typesetting math: 100% Correct Part B Draw a free-body diagram of the ice hockey puck. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Normal force ; Gravity Normal force ; Gravity ; Kinetic friction Tension ; Weight Thrust ; Gravity n F  G n F  G fk  T  w Fthrust  F  G Typesetting math: 100% Correct Problem 5.26 Your physics textbook is sliding to the right across the table. Part A Identify all forces acting on the object. ANSWER: Typesetting math: 100% Correct Part B Draw a free-body diagram of the object. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Weight ; Kinetic friction Thrust ; Kinetic friction Normal force ; Weight ; Kinetic friction Normal force ; Weight ; Static friction w fk  Fthrust  fk  n w fk  n w fs  Typesetting math: 100% Correct Enhanced EOC: Problem 5.35 A constant force is applied to an object, causing the object to accelerate at 13 . You may want to review ( pages 127 - 130) . For help with math skills, you may want to review: Proportions I Proportions II Part A m/s2 Typesetting math: 100% What will the acceleration be if the force is halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the force is halved but the mass remains the same? ANSWER: Correct Part B What will the acceleration be if the object's mass is halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the mass is halved but the force remains the same? ANSWER: Correct Part C a = 6.50 m s2 a = 26.0 m s2 Typesetting math: 100% What will the acceleration be if the force and the object's mass are both halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if both the force and mass are reduced by a factor of two? ANSWER: Correct Part D What will the acceleration be if the force is halved and the object's mass is doubled? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the force is decreased by a factor of two and the mass is increased by a factor of two? Check your answer by choosing numerical values of the force and mass, and then halve the force and double the mass. ANSWER: Correct a = 13.0 m s2 a = 3.25 m s2 Typesetting math: 100% Problem 5.44 A rocket is being launched straight up. Air resistance is not negligible. Part A Which of the following is the correct motion diagram for the situation described above? Enter the letter that corresponds with the best answer. ANSWER: Correct Part B Draw a free-body diagram. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Typesetting math: 100% Correct Score Summary: Your score on this assignment is 99.7%. You received 63.82 out of a possible total of 64 points. Typesetting math: 100%

Assignment 4 Due: 11:59pm on Wednesday, February 26, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy ± Two Forces Acting at a Point Two forces, and , act at a point. has a magnitude of 9.80 and is directed at an angle of 56.0 above the negative x axis in the second quadrant. has a magnitude of 5.20 and is directed at an angle of 54.1 below the negative x axis in the third quadrant. Part A What is the x component of the resultant force? Express your answer in newtons. Hint 1. How to approach the problem The resultant force is defined as the vector sum of all forces. Thus, its x component is the sum of the x components of the forces, and its y component is the sum of the y components of the forces. Hint 2. Find the x component of Find the x component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . F 1 F  2 F  1 N  F  2 N  F 1 F  1 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 Hint 2. Find the direction of is directed at an angle of 56.0 above the x axis in the second quadrant. When you calculate the components of , however, the direction of the force is commonly expressed in terms of the angle that the vector representing the force forms with the positive x axis. What is the angle that forms with the positive x axis? Select an answer from the following list, where 56.0 . ANSWER: ANSWER: Hint 3. Find the x component of Find the x component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , F 1 F 1  F  1 F  1  =   180 −  180 +  90 +  -5.48 N F 2 F  2 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Typesetting math: 100% and if . Hint 2. Find the direction of is directed at an angle of 54.1 below the x axis in the third quadrant. When you calculate the components of , however, the direction of the force is commonly expressed in terms of the angle that the vector representing the force forms with the positive x axis. What is the angle that forms with the positive x axis? Select an answer from the following list, where 54.1 . ANSWER: ANSWER: ANSWER: Correct Part B What is the y component of the resultant force? Express your answer in newtons. Ax < 0 Ay <  <  < 3 2 F 2 F 2  F 2 F  2  =   180 −   − 180 −90 −  -3.05 N -8.53 N Typesetting math: 100% Hint 1. How to approach the problem Follow the same procedure that you used in Part A to find the x component of the resultant force, though now calculate the y components of the two forces. Hint 2. Find the y component of Find the y component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . ANSWER: Hint 3. Find the y component of Find the y component of . Express your answer in newtons. Hint 1. Components of a vector F 1 F  1 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 8.12 N F 2 F  2 Typesetting math: 100% Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . ANSWER: ANSWER: Correct Part C What is the magnitude of the resultant force? Express your answer in newtons. Hint 1. Magnitude of a vector Consider a vector , whose components are and . The magnitude of is . A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 -4.21 N 3.91 N A Ax Ay A A = A + 2 x A2 y −−−−−−−  Typesetting math: 100% ANSWER: Correct Enhanced EOC: Problem 5.9 The figure shows acceleration-versus-force graphs for two objects pulled by rubber bands. You may want to review ( pages 127 - 130) . For help with math skills, you may want to review: Finding the Slope of a Line from a Graph Part A What is the mass ratio ? Express your answer using two significant figures. 9.38 N m1 m2 Typesetting math: 100% Hint 1. How to approach the problem How are the acceleration and the force on an object related to its mass? How is the slope of each line in the figure related to each object's mass? For each line, what two points are easy to measure accurately to determine the slope of line? How is the slope determined from the x and y coordinates of the two points you chose for each line? ANSWER: Correct A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude . Part A How much horizontal force must a sprinter of mass 54 exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Hint 1. Newton's 2nd law of motion According to Newton's 2nd law of motion, if a net external force acts on a body, the body accelerates, and the net force is equal to the mass of the body times the acceleration of the body: . ANSWER: = 0.36 m1 m2 15 m/s2 F kg Fnet m a Fnet = ma F = 810 N Typesetting math: 100% Correct Part B Which body exerts the force that propels the sprinter, the blocks or the sprinter? Hint 1. How to approach the question To start moving forward, sprinters push backward on the starting blocks with their feet. Newton's 3rd law tells you that the blocks exert a force on the sprinter of the same magnitude, but opposite in direction. ANSWER: Correct To start moving forward, sprinters push backward on the starting blocks with their feet. As a reaction, the blocks push forward on their feet with a force of the same magnitude. This external force accelerates the sprinter forward. Problem 5.12 The figure shows an acceleration-versus-force graph for a 600 object. the blocks the sprinter g Typesetting math: 100% Part A What must equal in order for the graph to be correct? Express your answer with the appropriate units. ANSWER: Correct Part B What must equal in order for the graph to be correct? Express your answer with the appropriate units. ANSWER: Correct Free-Body Diagrams Learning Goal: To gain practice drawing free-body diagrams Whenever you face a problem involving forces, always start with a free-body diagram. a1 a1 = 1.67 m s2 a2 a2 = 3.33 m s2 Typesetting math: 100% To draw a free-body diagram use the following steps: Isolate the object of interest. It is customary to represent the object of interest as a point 1. in your diagram. Identify all the forces acting on the object and their directions. Do not include forces acting on other objects in the problem. Also, do not include quantities, such as velocities and accelerations, that are not forces. 2. Draw the vectors for each force acting on your object of interest. When possible, the length of the force vectors you draw should represent the relative magnitudes of the forces acting on the object. 3. In most problems, after you have drawn the free-body diagrams, you will explicitly label your coordinate axes and directions. Always make the object of interest the origin of your coordinate system. Then you will need to divide the forces into x and y components, sum the x and y forces, and apply Newton's first or second law. In this problem you will only draw the free-body diagram. Suppose that you are asked to solve the following problem: Chadwick is pushing a piano across a level floor (see the figure). The piano can slide across the floor without friction. If Chadwick applies a horizontal force to the piano, what is the piano's acceleration? To solve this problem you should start by drawing a free-body diagram. Part A Determine the object of interest for the situation described in the problem introduction. Hint 1. How to approach the problem You should first think about the question you are trying to answer: What is the acceleration of the piano? The object of interest in this situation will be the object whose acceleration you are asked to find. ANSWER: Typesetting math: 100% Correct Part B Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano. Check all that apply. ANSWER: Correct Now that you have identified the forces acting on the piano, you should draw the free-body diagram. Draw the length of your vectors to represent the relative magnitudes of the forces, but you don't need to worry about the exact scale. You won't have the exact value of all of the forces until you finish solving the problem. To maximize your learning, you should draw the diagram yourself before looking at the choices in the next part. You are on your honor to do so. Part C For this situation you should draw a free-body diagram for the floor. Chadwick. the piano. acceleration of the piano gravitational force acting on the piano (piano's weight) speed of the piano gravitational force acting on Chadwick (Chadwick's weight) force of the floor on the piano (normal force) force of the piano on the floor force of Chadwick on the piano force of the piano pushing on Chadwick Typesetting math: 100% Select the choice that best matches the free-body diagram you have drawn for the piano. Hint 1. Determine the directions and relative magnitudes of the forces Which of the following statements best describes the correct directions and relative magnitudes of the forces involved? ANSWER: ANSWER: The normal force and weight are both upward and the pushing force is horizontal. The normal force and weight are both downward and the pushing force is horizontal. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force has a greater magnitude than the weight. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force and weight have the same magnitude. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force has a smaller magnitude than the weight. Typesetting math: 100% Typesetting math: 100% Correct If you were actually going to solve this problem rather than just draw the free-body diagram, you would need to define the coordinate system. Choose the position of the piano as the origin. In this case it is simplest to let the y axis point vertically upward and the x axis point horizontally to the right, in the direction of the acceleration. Chadwick now needs to push the piano up a ramp and into a moving van. at left. The ramp is frictionless. Is Chadwick strong enough to push the piano up the ramp alone or must he get help? To solve this problem you should start by drawing a free-body diagram. Part D Determine the object of interest for this situation. ANSWER: Correct Now draw the free-body diagram of the piano in this new situation. Follow the same sequence of steps that you followed for the first situation. Again draw your diagram before you look at the choices For this situation, you should draw a free-body diagram for the ramp. Chadwick. the piano. Typesetting math: 100% below. Part E Which diagram accurately represents the free-body diagram for the piano? ANSWER: Typesetting math: 100% Typesetting math: 100% Correct In working problems like this one that involve an incline, it is most often easiest to select a coordinate system that is not vertical and horizontal. Instead, choose the x axis so that it is parallel to the incline and choose the y axis so that it is perpendicular to the incline. Problem 5.18 The figure shows two of the three forces acting on an object in equilibrium. Part A Redraw the diagram, showing all three forces. Label the third force . Draw the force vector starting at the black dot. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: F  3 Typesetting math: 100% Correct Problem 5.25 An ice hockey puck glides across frictionless ice. Part A Identify all forces acting on the object. ANSWER: Typesetting math: 100% Correct Part B Draw a free-body diagram of the ice hockey puck. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Normal force ; Gravity Normal force ; Gravity ; Kinetic friction Tension ; Weight Thrust ; Gravity n F  G n F  G fk  T  w Fthrust  F  G Typesetting math: 100% Correct Problem 5.26 Your physics textbook is sliding to the right across the table. Part A Identify all forces acting on the object. ANSWER: Typesetting math: 100% Correct Part B Draw a free-body diagram of the object. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Weight ; Kinetic friction Thrust ; Kinetic friction Normal force ; Weight ; Kinetic friction Normal force ; Weight ; Static friction w fk  Fthrust  fk  n w fk  n w fs  Typesetting math: 100% Correct Enhanced EOC: Problem 5.35 A constant force is applied to an object, causing the object to accelerate at 13 . You may want to review ( pages 127 - 130) . For help with math skills, you may want to review: Proportions I Proportions II Part A m/s2 Typesetting math: 100% What will the acceleration be if the force is halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the force is halved but the mass remains the same? ANSWER: Correct Part B What will the acceleration be if the object's mass is halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the mass is halved but the force remains the same? ANSWER: Correct Part C a = 6.50 m s2 a = 26.0 m s2 Typesetting math: 100% What will the acceleration be if the force and the object's mass are both halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if both the force and mass are reduced by a factor of two? ANSWER: Correct Part D What will the acceleration be if the force is halved and the object's mass is doubled? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the force is decreased by a factor of two and the mass is increased by a factor of two? Check your answer by choosing numerical values of the force and mass, and then halve the force and double the mass. ANSWER: Correct a = 13.0 m s2 a = 3.25 m s2 Typesetting math: 100% Problem 5.44 A rocket is being launched straight up. Air resistance is not negligible. Part A Which of the following is the correct motion diagram for the situation described above? Enter the letter that corresponds with the best answer. ANSWER: Correct Part B Draw a free-body diagram. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Typesetting math: 100% Correct Score Summary: Your score on this assignment is 99.7%. You received 63.82 out of a possible total of 64 points. Typesetting math: 100%

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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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ISTC3015 Human Computer Interaction Spring 2014 Assignment You are to choose 2 websites, with different purposes, and review the websites based on the criteria listed below. This assignment is due Thursday, March 20th and is worth 70 points. 1. Starting Point a. Composition Matches Site Purpose b. Target Audience Apparent c. Composition Appropriate for Target Audience 2. Site design a. Consistency within site b. Consistency among pages 3. Visually Pleasing Composition 4. Visual Style in Web Design a. Consistency b. Distinctiveness 5. Focus and Emphasis a. What is emphasized? b. How is emphasis achieved? 6. Consistency a. Real World b. Internal 7. Navigation and Flow a. Home page identifiable throughout b. Location within site apparent c. Navigation consistent; rule-based; appropriate 8. Grouping a. Grouping with White Space b. Grouping with Borders c. Grouping with Backgrounds 9. Response time 10. Links a. Titled b. Incoming c. Outgoing d. Color 11. Detailed content a. Meaningful headings b. Plain language c. Page chunking d. Long blocks of text e. Scrolling f. Use of “within” page links 12. Articles a. Clear headings b. Plain language 13. Presenting Information Simply and Meaningfully a. Legibility b. Readability c. Information in Usable Form d. Visual Lines Clear 14. Legibility of content a. Font color b. Font size c. Font style d. Background color e. Background graphic 15. Documentation a. Included b. Searchable c. Links to difficult concepts/words 16. Multimedia a. Animation/Audio/Video/Still images b. Load time given c. Add-in required d. Quality e. Appropriateness of use 17. Scrolling and Paging a. Usage b. Appropriate? 18. Amount of Information Presented Appropriate 19. Other factors to note?

ISTC3015 Human Computer Interaction Spring 2014 Assignment You are to choose 2 websites, with different purposes, and review the websites based on the criteria listed below. This assignment is due Thursday, March 20th and is worth 70 points. 1. Starting Point a. Composition Matches Site Purpose b. Target Audience Apparent c. Composition Appropriate for Target Audience 2. Site design a. Consistency within site b. Consistency among pages 3. Visually Pleasing Composition 4. Visual Style in Web Design a. Consistency b. Distinctiveness 5. Focus and Emphasis a. What is emphasized? b. How is emphasis achieved? 6. Consistency a. Real World b. Internal 7. Navigation and Flow a. Home page identifiable throughout b. Location within site apparent c. Navigation consistent; rule-based; appropriate 8. Grouping a. Grouping with White Space b. Grouping with Borders c. Grouping with Backgrounds 9. Response time 10. Links a. Titled b. Incoming c. Outgoing d. Color 11. Detailed content a. Meaningful headings b. Plain language c. Page chunking d. Long blocks of text e. Scrolling f. Use of “within” page links 12. Articles a. Clear headings b. Plain language 13. Presenting Information Simply and Meaningfully a. Legibility b. Readability c. Information in Usable Form d. Visual Lines Clear 14. Legibility of content a. Font color b. Font size c. Font style d. Background color e. Background graphic 15. Documentation a. Included b. Searchable c. Links to difficult concepts/words 16. Multimedia a. Animation/Audio/Video/Still images b. Load time given c. Add-in required d. Quality e. Appropriateness of use 17. Scrolling and Paging a. Usage b. Appropriate? 18. Amount of Information Presented Appropriate 19. Other factors to note?

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