1. (2 marks total) a. Multiply 109 x 309 b. Divide 1988 by 16 exactly 2. (4 marks total) a. Write 2/11 as a decimal to 2 decimal places b. Calculate 35% of 62 c. Add 103/4 to 92/3 d. Subtract 79.04 from 115.225 giving your answer correct to 2 decimal places 3. Circle the fractions in the list which are equivalent to 0.80 (2 marks) 2/7 32/40 8/10 8/20 8/25 9/24 36/45 40/50 4. Write the numerical value of: 3-3 (2 marks total) 5. Simplify z + 67 = 3z + 33 (1 mark total) 6. Solve to 1 decimal place 3y – 34 = 2y + 89 (1 mark total) 7. Solve the following equations to 2 decimal places (3 marks total) a. 37x + 1 = 35 b. 27 – a = 7.45 c. 3(y + 2) = 14 8. A 7-sided polygon is called a Heptagon. (3 marks total) a. What is the total of a Heptagon’s interior angles? b. If the Heptagon is regular (all angles the same), calculate the size of each interior angle to 2 decimal places. 9. Calculate the size of angle a and angle b. (2 mark total) 10. How many centilitres are there in 1.25 litres? (1 mark total) 11. The diagram below shows a stone carving with a hole on it; determine its volume (not including hole), if its thickness is 8 cm. Give your answer in cm3 to 2 decimal points. Assume π = 3.14 (6 marks total) 12. The diagram below shows a piece of alloy plate with a hole in it made from aluminium, copper and magnesium with a mass ratio of 35:3:2. Calculate the following to 2 decimal places. All measurements are in cm. (7 marks total) a. Using the formula A = 1/2(a+b)h calculate the height of the shape below. b. The volume of the solid part (not including the hole) of the shape below to 3 decimal places if it was 0.25cm thick. c. The mass of each material if the total mass of the plate is 62 kg. 10 cm Hole dia = 3 cm Cross sectional area of solid (not including hole) = 28.935 cm2 8 cm 13. A 66kg boy is running at 3 m/s. Calculate his Kinetic Energy using the formula KE = 1/2mv2 (2 marks total) 14. A rocket has a mass of 2,000 kg. What is its acceleration if the forces of its engines are 50kN? Show working out to receive full marks. (1 marks total) 250,000,000 m/s² 25 m/s² 25,000 m/s² 15. In the diagram below a force of 125N (F1) is applied to a lever 20cm (D1) away from the fulcrum, (4 marks total) Fulcrum (a) How far away in metres would a force of 5N (F2) need to be to balance the load? (b) How much force (F2) would need to be applied 0.7m away to balance the same load (F1)? 16. For the circuit shown in the diagram below, calculate: (3 mark total) a. The total circuit resistance. b. The value of the current I. c. Calculate the voltage of the battery cell if the current was 3amp and the resistors stayed the same. 17. In the diagram of a hydraulic system, the area of piston A is 8cm2 and the area of piston B is 48cm2. (2 mark total) If the Force IN is 16 N, calculate the force OUT. 18. Plot the graph 2y = x3 – 4 using a value range for x from 0 to 3 (3 marks total) 14 12 10 8 6 4 2 0 -2 Choosing appropriate scale (1 mark) Accurately plotting y values (1 mark) X 0 1 2 3 Y Accurately plotting line of best fit. (1 mark) SPARE PAPER

1. (2 marks total) a. Multiply 109 x 309 b. Divide 1988 by 16 exactly 2. (4 marks total) a. Write 2/11 as a decimal to 2 decimal places b. Calculate 35% of 62 c. Add 103/4 to 92/3 d. Subtract 79.04 from 115.225 giving your answer correct to 2 decimal places 3. Circle the fractions in the list which are equivalent to 0.80 (2 marks) 2/7 32/40 8/10 8/20 8/25 9/24 36/45 40/50 4. Write the numerical value of: 3-3 (2 marks total) 5. Simplify z + 67 = 3z + 33 (1 mark total) 6. Solve to 1 decimal place 3y – 34 = 2y + 89 (1 mark total) 7. Solve the following equations to 2 decimal places (3 marks total) a. 37x + 1 = 35 b. 27 – a = 7.45 c. 3(y + 2) = 14 8. A 7-sided polygon is called a Heptagon. (3 marks total) a. What is the total of a Heptagon’s interior angles? b. If the Heptagon is regular (all angles the same), calculate the size of each interior angle to 2 decimal places. 9. Calculate the size of angle a and angle b. (2 mark total) 10. How many centilitres are there in 1.25 litres? (1 mark total) 11. The diagram below shows a stone carving with a hole on it; determine its volume (not including hole), if its thickness is 8 cm. Give your answer in cm3 to 2 decimal points. Assume π = 3.14 (6 marks total) 12. The diagram below shows a piece of alloy plate with a hole in it made from aluminium, copper and magnesium with a mass ratio of 35:3:2. Calculate the following to 2 decimal places. All measurements are in cm. (7 marks total) a. Using the formula A = 1/2(a+b)h calculate the height of the shape below. b. The volume of the solid part (not including the hole) of the shape below to 3 decimal places if it was 0.25cm thick. c. The mass of each material if the total mass of the plate is 62 kg. 10 cm Hole dia = 3 cm Cross sectional area of solid (not including hole) = 28.935 cm2 8 cm 13. A 66kg boy is running at 3 m/s. Calculate his Kinetic Energy using the formula KE = 1/2mv2 (2 marks total) 14. A rocket has a mass of 2,000 kg. What is its acceleration if the forces of its engines are 50kN? Show working out to receive full marks. (1 marks total) 250,000,000 m/s² 25 m/s² 25,000 m/s² 15. In the diagram below a force of 125N (F1) is applied to a lever 20cm (D1) away from the fulcrum, (4 marks total) Fulcrum (a) How far away in metres would a force of 5N (F2) need to be to balance the load? (b) How much force (F2) would need to be applied 0.7m away to balance the same load (F1)? 16. For the circuit shown in the diagram below, calculate: (3 mark total) a. The total circuit resistance. b. The value of the current I. c. Calculate the voltage of the battery cell if the current was 3amp and the resistors stayed the same. 17. In the diagram of a hydraulic system, the area of piston A is 8cm2 and the area of piston B is 48cm2. (2 mark total) If the Force IN is 16 N, calculate the force OUT. 18. Plot the graph 2y = x3 – 4 using a value range for x from 0 to 3 (3 marks total) 14 12 10 8 6 4 2 0 -2 Choosing appropriate scale (1 mark) Accurately plotting y values (1 mark) X 0 1 2 3 Y Accurately plotting line of best fit. (1 mark) SPARE PAPER

No expert has answered this question yet. You can browse … Read More...
Derive an expression for the capillary-height change h for a fluid of surface tension Y and contact angle between two paraller plates W apart Evaluate h for water at 20c if W=0.5mm

Derive an expression for the capillary-height change h for a fluid of surface tension Y and contact angle between two paraller plates W apart Evaluate h for water at 20c if W=0.5mm

ELEC153 Circuit Theory II M2A3 Lab: AC Series Circuits Introduction Previously you worked with two simple AC series circuits, R-C and R-L circuits. We continue that work in this experiment. Procedure 1. Setup the following circuit in MultiSim.The voltage source is 10 volts peak at 1000 Hz. Figure 1: Circuit for analysis using MultiSim 2. Change R1 to 1 k and C1 to 0.1 uF. Connect the oscilloscope to measure both the source voltage and the voltage across the resistor.You should have the following arrangement. Figure 2: Circuit of figure 1 connected to oscilloscope To color the wires, right click the desired wire and select “Color Segment…” and follow the instructions. Start the simulation and open the oscilloscope. You should get the following plot: Figure 3: Source voltage (red) and the voltage (blue) across the resistor The red signal is the voltage of the source and the blue is the voltage across the resistor. The colors correspond to the colors of the wires from the oscilloscope. 3. From the resulting analysis plotdetermine the peak current. To determine the peak current measure the peak voltage across the resistor and divide by the value of the resistor (1000 Ohms). Record it here. Measured Peak Current 4. Determine the peak current by calculation. Record it here. Does it match the measured peak current? Explain. Calculated Peak Current 5 Determine the phase shift between the current in the circuit and the source voltage. We look at the time between zero crossings to determine the phase shift between two waveforms. In our plot, the blue waveform (representing the circuit current or the voltage across the resistor) crosses zero before the red waveform (the circuit voltage). So, current is leading voltage in this circuit. This is exactly what should happen when we have a capacitive circuit. 6. To determine the phase shift, we first have to measure the time between zero crossings on the red and blue waveforms. This is done by moving the oscillator probes to the two zero crossing as is shown in the following figure Figure 4: Determining the phase shift between the two voltage waveforms We can see from the figure that the zero crossing difference (T2 – T1) is approximately 134 us. The ratio of the zero-crossing time difference to the period of the waveform determines the phase shift, as follows: Using our time values, we have: How do we know if this phase shift is correct? In step 4 when you did your manual calculations to find the peak current, you had to find the total impedance of the circuit, which was: Now, the current will be: Here, the positive angle on the current indicates it is leading the circuit voltage. 7. Change the frequency of the voltage source to 5000 Hz. Estimulate and perform a Transient Analysis to find the new circuit current and phase angle. Measure them and record them here: Measured Current Measured Phase Shift 8. Perform the manual calculations needed to find the circuit current and phase shift. Record the calculated values here. Do they match the measured values within reason? What has happened to the circuit with an increase in frequency? Calculated Current Calculated Phase Shift Writeup and Submission In general, for each lab you do, you will be asked to setup certain circuits, simulate them, record the results, verify the results are correct by hand, and then discuss the solution. Your lab write-up should contain a one page, single spaced discussion of the lab experiment, what went right for you, what you had difficulty with, what you learned from the experiment, how it applies to our coursework, and any other comment you can think of. In addition, you should include screen shots from the MultiSim software and any other figure, table, or diagram as necessary.

ELEC153 Circuit Theory II M2A3 Lab: AC Series Circuits Introduction Previously you worked with two simple AC series circuits, R-C and R-L circuits. We continue that work in this experiment. Procedure 1. Setup the following circuit in MultiSim.The voltage source is 10 volts peak at 1000 Hz. Figure 1: Circuit for analysis using MultiSim 2. Change R1 to 1 k and C1 to 0.1 uF. Connect the oscilloscope to measure both the source voltage and the voltage across the resistor.You should have the following arrangement. Figure 2: Circuit of figure 1 connected to oscilloscope To color the wires, right click the desired wire and select “Color Segment…” and follow the instructions. Start the simulation and open the oscilloscope. You should get the following plot: Figure 3: Source voltage (red) and the voltage (blue) across the resistor The red signal is the voltage of the source and the blue is the voltage across the resistor. The colors correspond to the colors of the wires from the oscilloscope. 3. From the resulting analysis plotdetermine the peak current. To determine the peak current measure the peak voltage across the resistor and divide by the value of the resistor (1000 Ohms). Record it here. Measured Peak Current 4. Determine the peak current by calculation. Record it here. Does it match the measured peak current? Explain. Calculated Peak Current 5 Determine the phase shift between the current in the circuit and the source voltage. We look at the time between zero crossings to determine the phase shift between two waveforms. In our plot, the blue waveform (representing the circuit current or the voltage across the resistor) crosses zero before the red waveform (the circuit voltage). So, current is leading voltage in this circuit. This is exactly what should happen when we have a capacitive circuit. 6. To determine the phase shift, we first have to measure the time between zero crossings on the red and blue waveforms. This is done by moving the oscillator probes to the two zero crossing as is shown in the following figure Figure 4: Determining the phase shift between the two voltage waveforms We can see from the figure that the zero crossing difference (T2 – T1) is approximately 134 us. The ratio of the zero-crossing time difference to the period of the waveform determines the phase shift, as follows: Using our time values, we have: How do we know if this phase shift is correct? In step 4 when you did your manual calculations to find the peak current, you had to find the total impedance of the circuit, which was: Now, the current will be: Here, the positive angle on the current indicates it is leading the circuit voltage. 7. Change the frequency of the voltage source to 5000 Hz. Estimulate and perform a Transient Analysis to find the new circuit current and phase angle. Measure them and record them here: Measured Current Measured Phase Shift 8. Perform the manual calculations needed to find the circuit current and phase shift. Record the calculated values here. Do they match the measured values within reason? What has happened to the circuit with an increase in frequency? Calculated Current Calculated Phase Shift Writeup and Submission In general, for each lab you do, you will be asked to setup certain circuits, simulate them, record the results, verify the results are correct by hand, and then discuss the solution. Your lab write-up should contain a one page, single spaced discussion of the lab experiment, what went right for you, what you had difficulty with, what you learned from the experiment, how it applies to our coursework, and any other comment you can think of. In addition, you should include screen shots from the MultiSim software and any other figure, table, or diagram as necessary.

No expert has answered this question yet. You can browse … Read More...
Chapter 9 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 Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t) Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Part B Find the time derivative of the x component of the system’s total momentum. Express your answer in terms of , , , and . You did not open hints for this part. ANSWER: Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be useful in reaching our desired conclusion, if only for this very special case. Part C The quantity (mass times acceleration) is dimensionally equivalent to which of the following? ANSWER: p(t) t m1 m2 v1 (t) v2 (t) p(t) = dp(t)/dt a1 (t) a2 (t) m1 m2 dp(t)/dt = ma Part D Acceleration is due to which of the following physical quantities? ANSWER: Part E Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1? You did not open hints for this part. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force Part F This question will be shown after you complete previous question(s). Part G Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton’s second law? ANSWER: Part H Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true? You did not open hints for this part. the large mass of block 1 air resistance Earth’s gravitational attraction a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 F12 F21 and and and and F12 = m2a2 F21 = m1a1 F12 = m1a1 F21 = m2a2 F12 = m1a2 F21 = m2a1 F12 = m2a1 F21 = m1a2 F12 F21 m1 m2 ANSWER: Part I Now recall the expression for the time derivative of the x component of the system’s total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to . You did not open hints for this part. ANSWER: Impulse and Momentum Ranking Task Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest. Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. ANSWER: Both forces have equal magnitudes. |F12 | > |F21| |F21 | > |F12| dpx(t)/dt = Fx dpx(t)/dt 0 nonzero constant kt kt2 Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: A Game of Frictionless Catch Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is . Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie’s speed relative to the ground after she catches the ball is . When answering the questions in this problem, keep the following in mind: The original mass of Chuck and his cart does not include the 1. mass of the ball. 2. The speed of an object is the magnitude of its velocity. An object’s speed will always be a nonnegative quantity. mcart mball vc vb vj mcart Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . You did not open hints for this part. ANSWER: Part B What is the speed of the ball (relative to the ground) while it is in the air? Express your answer in terms of , , and . You did not open hints for this part. ANSWER: Part C What is Chuck’s speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . u vc vb u = vb mball mcart u vb = vc mball mcart u You did not open hints for this part. ANSWER: Part D Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: Part E Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: vc = vj vb vj mball mcart vb vj = vj u vj mball mcart u Momentum in an Explosion A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions. Part A What is the momentum of piece A before the explosion? Express your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: vj = pA,i Part B During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A? You did not open hints for this part. ANSWER: Part C The momentum of piece B is measured to be 500 after the explosion. Find the momentum of piece A after the explosion. Enter your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: pA,i = kg  m/s greater than less than equal to cannot be determined kg  m/s pA,f pA,f = kg  m/s ± PSS 9.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 9.1 for conservation of momentum problems. An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? PROBLEM-SOLVING STRATEGY 9.1 Conservation of momentum MODEL: Clearly define the system. If possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved. If it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or, as you will learn later, conservation of energy. VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. SOLVE: The mathematical representation is based on the law of conservation of momentum: . In component form, this is ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied. Part A Sort the following objects as part of the system or not. Drag the appropriate objects to their respective bins. ANSWER: kg kg m/s F = net 0 P = f P  i (pfx + ( + ( += ( + ( + ( + )1 pfx)2 pfx)3 pix)1 pix)2 pix)3 (pfy + ( + ( += ( + ( + ( + )1 pfy)2 pfy)3 piy)1 piy)2 piy)3 Part B This question will be shown after you complete previous question(s). Visualize Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that “momentum is conserved” in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement “momentum is conserved.” Part A What physical quantities are conserved in this collision? ANSWER: Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are and . What is the speed of the two-car system after the collision? the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually v1 v2 You did not open hints for this part. ANSWER: Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, what is the magnitude of their combined momentum? You did not open hints for this part. ANSWER: The answer depends on the directions in which the cars were moving before the collision. v1 + v2 v1 − v2 v2 − v1 v1v2 −−−− ” v1+v2 2 v1 + 2 v2 2 −−−−−−−  p1 p2 Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and . After the collision, their combined momentum is . Of what can one be certain? You did not open hints for this part. ANSWER: Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be certain? The answer depends on the directions in which the cars were moving before the collision. p1 + p2 p1 − p2 p2 − p1 p1p2 −−−− ” p1+p2 2 p1 + 2 p2 2 −−−−−−−  p 1 p 2 p p = p1 + # p2 # p = p1 − # p2 # p = p2 − # p1 # p1 p2 p You did not open hints for this part. ANSWER: Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown. Part A p1 + p2 $ p $ p1p2 −−−− ” p1 +p2 $ p $ p1+p2 2 p1 + p2 $ p $ |p1 − p2 | p1 + p2 $ p $ p1 + 2 p2 2 −−−−−−−  m1 m2 v1 v2 First, find the magnitude of , that is, the speed of the two-car unit after the collision. Express in terms of , , and the cars’ initial speeds and . You did not open hints for this part. ANSWER: Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Part C Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: v v v m1 m2 v1 v2 v = p1 p2 tan( ) = tan = 1 45′ The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. ± Catching a Ball on Ice Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.2 . Olaf’s mass is 67.1 . Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: Part B kg m/s kg vf vf = m/s If the ball hits Olaf and bounces off his chest horizontally at 8.00 in the opposite direction, what is his speed after the collision? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: A One-Dimensional Inelastic Collision Block 1, of mass = 2.90 , moves along a frictionless air track with speed = 25.0 . It collides with block 2, of mass = 17.0 , 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. m/s vf vf = m/s m1 kg v1 m/s m2 kg pi You did not open hints for this part. ANSWER: 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: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. pi = kg  m/s vf vf = m/s

Chapter 9 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 Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t) Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Part B Find the time derivative of the x component of the system’s total momentum. Express your answer in terms of , , , and . You did not open hints for this part. ANSWER: Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be useful in reaching our desired conclusion, if only for this very special case. Part C The quantity (mass times acceleration) is dimensionally equivalent to which of the following? ANSWER: p(t) t m1 m2 v1 (t) v2 (t) p(t) = dp(t)/dt a1 (t) a2 (t) m1 m2 dp(t)/dt = ma Part D Acceleration is due to which of the following physical quantities? ANSWER: Part E Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1? You did not open hints for this part. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force Part F This question will be shown after you complete previous question(s). Part G Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton’s second law? ANSWER: Part H Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true? You did not open hints for this part. the large mass of block 1 air resistance Earth’s gravitational attraction a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 F12 F21 and and and and F12 = m2a2 F21 = m1a1 F12 = m1a1 F21 = m2a2 F12 = m1a2 F21 = m2a1 F12 = m2a1 F21 = m1a2 F12 F21 m1 m2 ANSWER: Part I Now recall the expression for the time derivative of the x component of the system’s total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to . You did not open hints for this part. ANSWER: Impulse and Momentum Ranking Task Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest. Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. ANSWER: Both forces have equal magnitudes. |F12 | > |F21| |F21 | > |F12| dpx(t)/dt = Fx dpx(t)/dt 0 nonzero constant kt kt2 Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: A Game of Frictionless Catch Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is . Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie’s speed relative to the ground after she catches the ball is . When answering the questions in this problem, keep the following in mind: The original mass of Chuck and his cart does not include the 1. mass of the ball. 2. The speed of an object is the magnitude of its velocity. An object’s speed will always be a nonnegative quantity. mcart mball vc vb vj mcart Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . You did not open hints for this part. ANSWER: Part B What is the speed of the ball (relative to the ground) while it is in the air? Express your answer in terms of , , and . You did not open hints for this part. ANSWER: Part C What is Chuck’s speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . u vc vb u = vb mball mcart u vb = vc mball mcart u You did not open hints for this part. ANSWER: Part D Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: Part E Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: vc = vj vb vj mball mcart vb vj = vj u vj mball mcart u Momentum in an Explosion A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions. Part A What is the momentum of piece A before the explosion? Express your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: vj = pA,i Part B During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A? You did not open hints for this part. ANSWER: Part C The momentum of piece B is measured to be 500 after the explosion. Find the momentum of piece A after the explosion. Enter your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: pA,i = kg  m/s greater than less than equal to cannot be determined kg  m/s pA,f pA,f = kg  m/s ± PSS 9.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 9.1 for conservation of momentum problems. An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? PROBLEM-SOLVING STRATEGY 9.1 Conservation of momentum MODEL: Clearly define the system. If possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved. If it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or, as you will learn later, conservation of energy. VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. SOLVE: The mathematical representation is based on the law of conservation of momentum: . In component form, this is ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied. Part A Sort the following objects as part of the system or not. Drag the appropriate objects to their respective bins. ANSWER: kg kg m/s F = net 0 P = f P  i (pfx + ( + ( += ( + ( + ( + )1 pfx)2 pfx)3 pix)1 pix)2 pix)3 (pfy + ( + ( += ( + ( + ( + )1 pfy)2 pfy)3 piy)1 piy)2 piy)3 Part B This question will be shown after you complete previous question(s). Visualize Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that “momentum is conserved” in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement “momentum is conserved.” Part A What physical quantities are conserved in this collision? ANSWER: Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are and . What is the speed of the two-car system after the collision? the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually v1 v2 You did not open hints for this part. ANSWER: Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, what is the magnitude of their combined momentum? You did not open hints for this part. ANSWER: The answer depends on the directions in which the cars were moving before the collision. v1 + v2 v1 − v2 v2 − v1 v1v2 −−−− ” v1+v2 2 v1 + 2 v2 2 −−−−−−−  p1 p2 Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and . After the collision, their combined momentum is . Of what can one be certain? You did not open hints for this part. ANSWER: Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be certain? The answer depends on the directions in which the cars were moving before the collision. p1 + p2 p1 − p2 p2 − p1 p1p2 −−−− ” p1+p2 2 p1 + 2 p2 2 −−−−−−−  p 1 p 2 p p = p1 + # p2 # p = p1 − # p2 # p = p2 − # p1 # p1 p2 p You did not open hints for this part. ANSWER: Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown. Part A p1 + p2 $ p $ p1p2 −−−− ” p1 +p2 $ p $ p1+p2 2 p1 + p2 $ p $ |p1 − p2 | p1 + p2 $ p $ p1 + 2 p2 2 −−−−−−−  m1 m2 v1 v2 First, find the magnitude of , that is, the speed of the two-car unit after the collision. Express in terms of , , and the cars’ initial speeds and . You did not open hints for this part. ANSWER: Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Part C Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: v v v m1 m2 v1 v2 v = p1 p2 tan( ) = tan = 1 45′ The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. ± Catching a Ball on Ice Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.2 . Olaf’s mass is 67.1 . Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: Part B kg m/s kg vf vf = m/s If the ball hits Olaf and bounces off his chest horizontally at 8.00 in the opposite direction, what is his speed after the collision? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: A One-Dimensional Inelastic Collision Block 1, of mass = 2.90 , moves along a frictionless air track with speed = 25.0 . It collides with block 2, of mass = 17.0 , 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. m/s vf vf = m/s m1 kg v1 m/s m2 kg pi You did not open hints for this part. ANSWER: 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: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. pi = kg  m/s vf vf = m/s

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

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

please email info@checkyourstudy.com
1) Can two different forces, acting through the same point, produce the same torque on an object? Answer: Yes, as long as the component of the force perpendicular to the line joining the axis to the force is the same for both forces. 2) If you stand with your back towards a wall and your heels touching the wall, you cannot lean over to touch your toes. Why? Answer: As you bend over your center of gravity moves forward and eventually is beyond the area of the floor in touch with your feet. This does not happen when you do it away from the wall because part of your body moves back and the center of mass remains over your feet. 3) Two equal forces are applied to a door at the doorknob. The first force is applied perpendicular to the door; the second force is applied at 30° to the plane of the door. Which force exerts the greater torque? A) the first applied perpendicular to the door B) the second applied at an angle C) both exert equal non-zero torques D) both exert zero torques E) Additional information is needed. 4) A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw? A) It is impossible to say without knowing the masses. B) It is impossible to say without knowing the distances. C) The side the boy is sitting on will tilt downward. D) Nothing, the seesaw will still be balanced. E) The side the girl is sitting on will tilt downward. 5) A figure skater is spinning slowly with arms outstretched. She brings her arms in close to her body and her angular speed increases dramatically. The speed increase is a demonstration of A) conservation of energy: her moment of inertia is decreased, and so her angular speed must increase to conserve energy. B) conservation of angular momentum: her moment of inertia is decreased, and so her angular speed must increase to conserve angular momentum. C) Newton’s second law for rotational motion: she exerts a torque and so her angular speed increases. D) This has nothing to do with mechanics, it is simply a result of her natural ability to perform. 6) A girl weighing 450. N sits on one end of a seesaw that is 3.0 m long and is pivoted 1.3 m from the girl. If the seesaw is just balanced when a boy sits at the opposite end, what is his weight? Neglect the weight of the seesaw. 7) An 82.0 kg painter stands on a long horizontal board 1.55 m from one end. The 15.5 kg board is 5.50 m long. The board is supported at each end. (a) What is the total force provided by both supports? (b) With what force does the support, closest to the painter, push upward? FIGURE 11-4 8) The mobile shown in Figure 11-4 is perfectly balanced. What must be the masses of m1, m2, and m3?

1) Can two different forces, acting through the same point, produce the same torque on an object? Answer: Yes, as long as the component of the force perpendicular to the line joining the axis to the force is the same for both forces. 2) If you stand with your back towards a wall and your heels touching the wall, you cannot lean over to touch your toes. Why? Answer: As you bend over your center of gravity moves forward and eventually is beyond the area of the floor in touch with your feet. This does not happen when you do it away from the wall because part of your body moves back and the center of mass remains over your feet. 3) Two equal forces are applied to a door at the doorknob. The first force is applied perpendicular to the door; the second force is applied at 30° to the plane of the door. Which force exerts the greater torque? A) the first applied perpendicular to the door B) the second applied at an angle C) both exert equal non-zero torques D) both exert zero torques E) Additional information is needed. 4) A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw? A) It is impossible to say without knowing the masses. B) It is impossible to say without knowing the distances. C) The side the boy is sitting on will tilt downward. D) Nothing, the seesaw will still be balanced. E) The side the girl is sitting on will tilt downward. 5) A figure skater is spinning slowly with arms outstretched. She brings her arms in close to her body and her angular speed increases dramatically. The speed increase is a demonstration of A) conservation of energy: her moment of inertia is decreased, and so her angular speed must increase to conserve energy. B) conservation of angular momentum: her moment of inertia is decreased, and so her angular speed must increase to conserve angular momentum. C) Newton’s second law for rotational motion: she exerts a torque and so her angular speed increases. D) This has nothing to do with mechanics, it is simply a result of her natural ability to perform. 6) A girl weighing 450. N sits on one end of a seesaw that is 3.0 m long and is pivoted 1.3 m from the girl. If the seesaw is just balanced when a boy sits at the opposite end, what is his weight? Neglect the weight of the seesaw. 7) An 82.0 kg painter stands on a long horizontal board 1.55 m from one end. The 15.5 kg board is 5.50 m long. The board is supported at each end. (a) What is the total force provided by both supports? (b) With what force does the support, closest to the painter, push upward? FIGURE 11-4 8) The mobile shown in Figure 11-4 is perfectly balanced. What must be the masses of m1, m2, and m3?

info@checkyourstudy.com solution
/Two cables are tied togeather at B and are loaded , a. draw the free body diagram ((FBD) of the supporting book at B , This FBD must be a new figure , b) write the cable force the vector T in cartesian vector form, where the magnitude of the force vector is left as variable , c) write the cable force vector Tac in cartesian vector form, where the magnitude of the force vector is left as a variable, d) Determine the unit vector along the line from point A to C , e) Using the vector dot product calculate the angle between the two cable , f) write the two equations needed to calculate the tension in cable Tba and tbc

/Two cables are tied togeather at B and are loaded , a. draw the free body diagram ((FBD) of the supporting book at B , This FBD must be a new figure , b) write the cable force the vector T in cartesian vector form, where the magnitude of the force vector is left as variable , c) write the cable force vector Tac in cartesian vector form, where the magnitude of the force vector is left as a variable, d) Determine the unit vector along the line from point A to C , e) Using the vector dot product calculate the angle between the two cable , f) write the two equations needed to calculate the tension in cable Tba and tbc

Two cables are tied togeather at B and are loaded … Read More...
The diagram 3 forces F1, F2, F3 act on the body. The magnitude and direction of forces F1 and F2 are known and indicated . Also the resultant force R= F1+F2+F3 is know and indicated. Using force decomposition procedure in the Cartesian coordinate system find the magnitude of the force F3 and the angle between line of action of this force and the x direction.

The diagram 3 forces F1, F2, F3 act on the body. The magnitude and direction of forces F1 and F2 are known and indicated . Also the resultant force R= F1+F2+F3 is know and indicated. Using force decomposition procedure in the Cartesian coordinate system find the magnitude of the force F3 and the angle between line of action of this force and the x direction.

Laurentian University ENGR 1056: Applied Mechanics I 2015{2016 Assignment #2 Instructions: Complete all the questions. Show your work as marks are given for process. Submit your assignment as a single PDF le to the appropriate dropbox on D2L. You may use the photocopiers in the library as scanners if you do not have access to a scanner. Ensure that the scans are readable. Late assignments will NOT be accepted. Due: Tues. Sept. 29, 2015, 8:30am 2-1. Assuming that: A = 5i ? 3j + 2k B = ?2i + 2k then calculate the following and report your answers to 3 signi cant gures: (a) eA; (b) A  B; (c) B  A; and, (d) the component of A parallel to B. 2-2. Do question 2.159 from your text. Report your answers to four signi cant digits. 2-3. You are given the following directions: start at point A, walk north 5 ft to point B, turn 30 degrees to your right, walk forward for 8 ft to point C, turn 15 degrees to your right, walk forward x ft to point D. If the distance directly from A to D is 16 ft, what is the value of x? What is the angle  between AB and AD? Include a diagram of your route. Label your diagram with points A to D and the angle . Report you answers to 3 signi cant digits. 2-4. Do question 2.165 from your text. Report your answers in newtons to four signi - cant gures. W. Brent Lievers 2015-09-21 Page 1 of 1

Laurentian University ENGR 1056: Applied Mechanics I 2015{2016 Assignment #2 Instructions: Complete all the questions. Show your work as marks are given for process. Submit your assignment as a single PDF le to the appropriate dropbox on D2L. You may use the photocopiers in the library as scanners if you do not have access to a scanner. Ensure that the scans are readable. Late assignments will NOT be accepted. Due: Tues. Sept. 29, 2015, 8:30am 2-1. Assuming that: A = 5i ? 3j + 2k B = ?2i + 2k then calculate the following and report your answers to 3 signi cant gures: (a) eA; (b) A  B; (c) B  A; and, (d) the component of A parallel to B. 2-2. Do question 2.159 from your text. Report your answers to four signi cant digits. 2-3. You are given the following directions: start at point A, walk north 5 ft to point B, turn 30 degrees to your right, walk forward for 8 ft to point C, turn 15 degrees to your right, walk forward x ft to point D. If the distance directly from A to D is 16 ft, what is the value of x? What is the angle  between AB and AD? Include a diagram of your route. Label your diagram with points A to D and the angle . Report you answers to 3 signi cant digits. 2-4. Do question 2.165 from your text. Report your answers in newtons to four signi - cant gures. W. Brent Lievers 2015-09-21 Page 1 of 1

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

please email info@checkyourstudy.com