Workbook for Introductory Mechanics Problem-Solving. To the Student

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1 Workbook for Introductory Mechanics Problem-Solving Daniel M. Smith, Jr. South Carolina State University To the Student The Workbook should help you to learn how to think about physics problems before you try using equations. The format is designed to carefully build each level of your problem-solving ability before moving you to the next level. So instead of working one problem at a time, you will work several problems at once. Here is advice on how to use this Workbook. 1. After printing, place these materials in a binder before beginning work. (If possible, use double-sided or duplex printing.) 2. Attempt to answer every question by drawing or writing before looking at the answer. 3. Warning: Simply reading the questions, the answers, and the discussion without doing the work is a complete waste of time. Copyright by Daniel M. Smith, Jr., All rights reserved. Sponsored by FIPSE, U.S. Department of Education

2 Acknowledgments Students have provided invaluable assistance in word processing, and in creating graphics for this workbook. Thanks go to Andrick Anderson, Kyle Herbert, Tyesia Pompey, Tarryn Reeves, and Keilah Spann. Professors Theodore Hodapp and Dave Maloney are thanked for their criticism and suggestions. Clip art is taken from the Art Explosion collection published by the Nova Development Corporation, Calabasas, CA. I have had the pleasure of teaching and tutoring at South Carolina State University and Northeastern University, and in the process I have learned much from students about their difficulties in solving physics problems. Those who wish to further my enlightenment may send comments to dsmith@scsu.edu. Copyright by Daniel M. Smith, Jr. All rights reserved. Students using College Physics, 4th edition by Jerry Wilson and Anthony Buffa may print out one copy of the Workbook material for their own use but may not otherwise copy or distribute the material in part or in whole by any means whatsoever, electronic or otherwise, without express written permission from the author and Prentice- Hall, Inc.

3 5-93 Chapter 5 B ENERGY Wilson/Buffa Chapter 5: Work and Unit 1 Bar chart representation of energy conservation Unit 2 Restatement of the problem in the language of energy Unit 3 Translation of the energy statement into algebraic language; problem solutions

4 5-94 Copyright by Daniel M. Smith, Jr. All rights reserved. Students using College Physics, 4th edition by Jerry Wilson and Anthony Buffa may print out one copy of the Workbook material for their own use but may not otherwise copy or distribute the material in part or in whole by any means whatsoever, electronic or otherwise, without express written permission from the author and Prentice-Hall, Inc.

5 - Unit 5.1B - Problem 5.1B 5 95 m 2 Masses m 1 (0.3 kg) and m 2 (0.8 kg) are connected by a massless string which passes over a massless pulley. If the masses are released from rest, find the speed of the masses at the instant that m 2 has moved a distance of 0.4 m. m What physics principle or principles should be used to solve this problem? Workbook for Introductory Mechanics Problem-Solving Copyright by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

6 Unit 5.1B - Problem 5.1B m 2 Masses m 1 (0.3 kg) and m 2 (0.8 kg) are connected by a massless string which passes over a massless pulley. If the masses are released from rest, find the speed of the masses at the instant that m 2 has moved a distance of 0.4 m. m The problem can be solved in at least two ways, by using either Newton s 2nd Law, or by using energy conservation. This amounts to saying that the problem can be described either in the language of forces or in the language of energy.

7 - Unit 5.1B - Problem 5.1B 5 97 m 2 Masses m 1 (0.3 kg) and m 2 (0.8 kg) are connected by a massless string which passes over a massless pulley. If the masses are released from rest, find the speed of the masses at the instant that m 2 has moved a distance of 0.4 m. m Use the language of forces to describe the problem by drawing vectors on the diagram (free-body diagram), and writing force equations. Without solving the problem, explain in words how to continue to a problem solution. m 2 m 1

8 Unit 5.1B - Problem 5.1B m 2 Masses m 1 (0.3 kg) and m 2 (0.8 kg) are connected by a massless string which passes over a massless pulley. If the masses are released from rest, find the speed of the masses at the instant that m 2 has moved a distance of 0.4 m. m T y m 2 T m 1 m 2 g m 1 g For mass m 1, Newton s 2nd Law gives T m 1 g = m 1 a, (5.133) while for mass m 2, T m 2 g = m 2 a. (5.134) We have two equations in two unknowns, so both a and T can be determined. Using the value of a in the kinematics equation v 2 = v a(y y 0 ) gives us the desired result, because the masses have the same speed.

9 - Unit 5.1B - Problem 5.1B 5 99 m 2 Masses m 1 (0.3 kg) and m 2 (0.8 kg) are connected by a massless string which passes over a massless pulley. If the masses are released from rest, find the speed of the masses at the instant that m 2 has moved a distance of 0.4 m. m Sketch the positions of both masses after mass m 2 has moved 0.4 m. Choose a reference level where the gravitational potential energy is 0, and draw this on the diagram also. m 2 m 1

10 Unit 5.1B - Problem 5.1B m 2 Masses m 1 (0.3 kg) and m 2 (0.8 kg) are connected by a massless string which passes over a massless pulley. If the masses are released from rest, find the speed of the masses at the instant that m 2 has moved a distance of 0.4 m. m The reference level is chosen to be below both masses, but any choice is valid. m 2 m 1 GPE = 0

11 - Unit 5.1B - Problem 5.1B m 2 Masses m 1 (0.3 kg) and m 2 (0.8 kg) are connected by a massless string which passes over a massless pulley. If the masses are released from rest, find the speed of the masses at the instant that m 2 has moved a distance of 0.4 m. m Make bar graphs, indicating qualitatively the relative amounts of the different forms of energy, both initially and after mass m 2 has moved by 0.4 m. For example, mass m 1 has GPE initially, as indicated by the solid bar. Your bar graphs need not look like the ones in the next frame, but they must show energy conservation. (GPE = gravitational potential energy, KE = kinetic energy) Mass m 1 Mass m 2 initial energy initial energy Mass m 1 Mass m 2 final energy final energy GPE1 KE1 GPE2 KE2 GPE1 KE1 GPE2 KE2 (a) (b) (d)

12 Unit 5.1B - Problem 5.1B m 2 Masses m 1 (0.3 kg) and m 2 (0.8 kg) are connected by a massless string which passes over a massless pulley. If the masses are released from rest, find the speed of the masses at the instant that m 2 has moved a distance of 0.4 m. m Both m 1 and m 2 have GPE initially. Mass m 2 has a larger GPE than mass m 1 initially because m 2 is higher above the reference level, and because m 2 > m 1 (recall that E GP = mgh). Otherwise, your initial values for GPE are unimportant. But because the system starts from rest, both masses have KE = 0 initially. Mass m 1 initial energy Mass m 2 initial energy Mass m 1 final energy Mass m 2 final energy GPE1 KE1 GPE2 KE2 GPE1 KE1 GPE2 KE2 (a) (b) (d) Because mass m 2 has fallen by 0.4 m, it has lost GPE (compare bar charts (b) and (d)), while m 1 has gained GPE because it is raised by 0.4 m. Both masses are moving with the same speed after m 2 s 0.4 m fall, so they both have KE. But because m 2 > m 1, mass m 2 has the larger KE (recall that E k = 1 2 mv2 ). Notice that energy is conserved: the total number of shaded blocks in diagrams (a) and (b) is equal to the total number in diagrams and (d); this must be true in your diagrams also. Although we show GPE1>GPE2 in figures and (d), this is not always true because the relationship between GPE1 and GPE2 really depends upon the chosen reference level (GPE = 0) and initial positions of the masses. For example, if the masses are at the ends of a very long string, m 2 would remain higher than m 1 even after m 2 moved by 0.4 m ; then GPE2>GPE1.

13 - Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? What principle(s) should be used to solve this problem? Although you may see that energy conservation is the principle allowing this problem to be solved, can you tell why Newton s 2nd Law is not useful here? Try to answer this by considering the descending car at two places on the track. Workbook for Introductory Mechanics Problem-Solving Copyright by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

14 Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? Notice that the component of gravitational forceparallel to the track is different in each case, because the angle changes. This component changes continuously as the car moves down the track. For this reason, writing a single force equation using the 2nd Law is impossible.

15 - Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? Draw the coaster car on the diagram at the car s two important positions in part (a) of the problem. Then, after choosing the reference level for GPE=0, make bar graphs indicating the relative amounts of the different forms of energy at the two events. ground (a) at at GPE KE GPE KE (b)

16 Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? The two important positions of the car in part (a) are its initial height (at A), and its height at the lower level (at B). Your bar graphs may not be identical to the ones below, but at least they must show that energy is conserved. A B ground (a) GPE = 0 at A at B GPE KE GPE KE (b)

17 - Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? GPE = 0 can be assigned at any height on diagram (a) of the previous frame. If it is assigned at a height greater than that of the track at B, values of GPE below that height are negative. Because the roller coaster car s height above ground level is reduced as it moves from position A to position B, its GPE is reduced as well. What the car loses in GPE, however, it gains in KE, so the car is moving faster at position B. How is the principle of energy conservation reflected in the bar graphs?

18 Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? That the total number of shaded blocks in diagram (b) is equal to the total number in diagram indicates that energy is conserved.

19 - Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? Draw the roller-coaster car at two important positions for part (b) of the problem (including a GPE=0 reference level). Then, draw bar graphs indicating relative magnitudes of the energy at these two positions. Again, your bar graphs will not necessarily be identical to the ones in the following frame. ground (a) at at GPE KE GPE KE (b)

20 Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? You might have drawn either of the diagrams ((a) or (b)) below because there are two possibilities, positions A and C, or positions B and C. Again the GPE=0 level is arbitrary but we have chosen it at ground level. A C C B ground at A (a) at C (b) GPE = 0 GPE KE GPE KE (d) at B at C GPE KE GPE KE (e) (f)

21 - Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? For diagram (a) of the previous frame, the car has lost GPE and gained KE in moving from A to C, so the car moves faster at C than at A. Note that the numbers of shaded blocks in diagrams and (d) are equal, indicative of energy conservation. For diagram (b) of the previous frame, the roller coaster car gains GPE in moving from B to C, and loses KE, so the car slows in moving from B to C. conservation is represented by the equality in the numbers of shaded blocks in diagrams (e) and (f). For part of the problem, consider the diagram below, then complete the bar graphs to show the relative energy distribution for the car at positions A and C. A C ground GPE = 0 (a) at A at C GPE KE GPE KE (b)

22 Unit 5.1B - Problem 5.2B A roller-coaster car of mass 480 kg is raised to an initial height of 18.0 m above ground level. With a speed of 1.3 m s the car begins its descent. (a) Ignoring friction, what is the car s speed when it reaches a horizontal section of track of height 4.0 m above ground level? The car ascends its first hill to reach a level of 8.5 m above ground. (b) What is the car s speed at this level, again ignoring friction? If the car s speed after ascending the hill is actually 10.1 m s, how much energy has been lost from the start because of the friction between the track and the coaster car? At positon C in the diagram of the previous frame, the roller-coaster car has the same GPE as in the no-friction case. But the KE must be less than that in the frictionless case because its speed is less: some of the original GPE has been converted to heat energy (HE). Your bar graphs should be similar to those below, not necessarily identical, but they must show energy conservation. at A at C GPE KE GPE KE HE (a) (b)

23 - Unit 5.1B - Problem 5.3B A small child of mass 18.2 kg (weight of 40 lb) jumps onto a mattress and compresses it by 10.3 cm. If the mattress compresses with an effective spring constant of 8574 N, (a) what is the m maximum height above the mattress to which the child is propelled? (b) What is the child s speed as it leaves the mattress? What principle should be used to solve this problem? Again energy conservation is most convenient. Explain why this problem is less conveniently solved using Newton s 2nd Law by considering the forces exerted on the child. You may represent the child by a block, and the mattress by a spring. Workbook for Introductory Mechanics Problem-Solving Copyright by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

24 Unit 5.1B - Problem 5.3B A small child of mass 18.2 kg (weight of 40 lb) jumps onto a mattress and compresses it by 10.3 cm. If the mattress compresses with an effective spring constant of 8574 N, (a) what is the m maximum height above the mattress to which the child is propelled? (b) What is the child s speed as it leaves the mattress? The vectors in the diagram below represent the two forces exerted on the child, the gravitational force (weight), and the force exerted by the spring. Remember that the magnitude of the spring s force depends upon how much the spring is compressed. This means that the force equation is complicated, because the amount of the spring s compression must be taken into account. This complication renders the force equation impossible to solve algebraically What are the two important positions of the child in part (a) of the problem? What are they in part (b)? For part (a), the child is at important positions when he compresses the mattress by 10.3 cm, and when he reaches maximum height. For part (b), the child is at important positions when the mattress is compressed by 10.3 cm, and when he just leaves the mattress moving upward.

25 - Unit 5.1B - Problem 5.3B A small child of mass 18.2 kg (weight of 40 lb) jumps onto a mattress and compresses it by 10.3 cm. If the mattress compresses with an effective spring constant of 8574 N, (a) what is the m maximum height above the mattress to which the child is propelled? (b) What is the child s speed as it leaves the mattress? After the child jumps onto the mattress, it is compressed by a certain amount before recoil. Draw a diagram of the maximum compression for part (a), then draw a diagram of the child at maximum height above the mattress. Use a block and spring to represent the child and mattress in both cases. Choose a GPE=0 reference level, then complete the bar graphs indicating the relative energy distribution in the two cases. Spring energy for (a) Block energy for (a) (a) Spring at ma xi mum compressi on Spring energy for (b) Block energy for (b) (d) (b) Spring uncompressed, Block at maximum height (e) (f)

26 Unit 5.1B - Problem 5.3B A small child of mass 18.2 kg (weight of 40 lb) jumps onto a mattress and compresses it by 10.3 cm. If the mattress compresses with an effective spring constant of 8574 N, (a) what is the m maximum height above the mattress to which the child is propelled? (b) What is the child s speed as it leaves the mattress? For part (a) of the problem, all of the spring s elastic potential energy (diagrams (a) and ) has been converted to the block s gravitational potential energy (diagrams (b) and (f)). Recall that at maximum height, the block s velocity is zero, then so is its KE. Your bar charts may differ quantitatively from these, but they must show energy conservation. Spring energy for (a) EPE Block energy for (a) GPE GPE = 0 (a) Spring at ma xi mum compressi on Spring energy for (b) Block energy for (b) (d) EPE GPE GPE = 0 (b) Spring uncompressed, bl ock at maxi mum h ei gh t (e) (f)

27 - Unit 5.1B - Problem 5.3B A small child of mass 18.2 kg (weight of 40 lb) jumps onto a mattress and compresses it by 10.3 cm. If the mattress compresses with an effective spring constant of 8574 N, (a) what is the m maximum height above the mattress to which the child is propelled? (b) What is the child s speed as it leaves the mattress? For part (b) of the problem we first consider the child when the mattress is at maximum compression, and next when the child just leaves the fully uncompressed mattress. Draw a diagram of the latter event. Complete the bar graphs indicating the relative energy distributions in the two cases. Spring energy for (a) Block energy for (a) GPE = 0 (a) Spring at ma xi mum compressi on Spring energy for (b) Block energy for (b) (d) (b) Block in motion, just leaving fully uncompressed spri ng (e) (f)

28 Unit 5.1B - Problem 5.3B A small child of mass 18.2 kg (weight of 40 lb) jumps onto a mattress and compresses it by 10.3 cm. If the mattress compresses with an effective spring constant of 8574 N, (a) what is the m maximum height above the mattress to which the child is propelled? (b) What is the child s speed as it leaves the mattress? All of the spring s elastic potential energy (diagrams (a) and ) is converted partly to the block s gravitational potential energy, and partly to the block s kinetic energy (diagrams (b) and (f)). Therefore, the total number of blocks in diagrams and (d) equals the number of blocks in diagram (f). Spring energy for (a) Block energy for (a) EPE GPE KE GPE = 0 (a) Spring at ma xi mum compressi on Spring energy for (b) Block energy for (b) (d) EPE GPE KE GPE = 0 (b) Spring uncompressed, bl ock i n moti on, just leaving spring (e) (f)

29 - Unit 5.1B - Problem 5.4B A child of mass 31.5 kg (weight of 69.3 lb) sits in a playground swing whose seat, 0.6 m above the ground, is supported by ropes of length 2.4 m. The child is given a push, and the swing rises to a maximum angle of 36 with the vertical. (a) What is the child s speed as the swing passes back through its rest position? (b) The child releases her grasp on the ropes as the swing moves forward at the bottom of its trajectory, and the child slides out of the swing. How far away from the release point does she land? Ignore air resistance What principle should be used to solve part (a) of the problem? conservation, of course. But why is it difficult to apply Newton s 2nd Law in this circumstance? Draw a diagram for your explanation by representing the child in the swing by a mass at the end of a string, then draw vectors to indicate the forces exerted on the mass. Workbook for Introductory Mechanics Problem-Solving Copyright by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

30 Unit 5.1B - Problem 5.4B A child of mass 31.5 kg (weight of 69.3 lb) sits in a playground swing whose seat, 0.6 m above the ground, is supported by ropes of length 2.4 m. The child is given a push, and the swing rises to a maximum angle of 36 with the vertical. (a) What is the child s speed as the swing passes back through its rest position? (b) The child releases her grasp on the ropes as the swing moves forward at the bottom of its trajectory, and the child slides out of the swing. How far away from the release point does she land? Ignore air resistance θ The angle θ changes continuously, so there is no single 2nd Law equation in which the tension is constant. The force equation would need to take into account the dependence of the tension on the angle throughout the motion. Such an equation cannot be solved algebraically.

31 - Unit 5.1B - Problem 5.4B A child of mass 31.5 kg (weight of 69.3 lb) sits in a playground swing whose seat, 0.6 m above the ground, is supported by ropes of length 2.4 m. The child is given a push, and the swing rises to a maximum angle of 36 with the vertical. (a) What is the child s speed as the swing passes back through its rest position? (b) The child releases her grasp on the ropes as the swing moves forward at the bottom of its trajectory, and the child slides out of the swing. How far away from the release point does she land? Ignore air resistance Complete the bar graphs to show the relative energy distribution in part (a) of the problem as the mass moves from position A to position B. 36 o A B GPE = 0 (a) at A at B GPE KE GPE KE (b)

32 Unit 5.1B - Problem 5.4B A child of mass 31.5 kg (weight of 69.3 lb) sits in a playground swing whose seat, 0.6 m above the ground, is supported by ropes of length 2.4 m. The child is given a push, and the swing rises to a maximum angle of 36 with the vertical. (a) What is the child s speed as the swing passes back through its rest position? (b) The child releases her grasp on the ropes as the swing moves forward at the bottom of its trajectory, and the child slides out of the swing. How far away from the release point does she land? Ignore air resistance The child s velocity is 0 at A, then so is her kinetic energy. All of the GPE at A is converted to KE, the energy of the motion, at B. Your bar charts may differ quantitatively (if you choose GPE=0 at position B, then GPE=0 for chart ), but they must show that energy is conserved. 36 o A B GPE = 0 (a) at A at B GPE KE GPE KE (b)

33 - Unit 5.1B - Problem 5.5B A roller coaster car starts from rest on a track 21.3 m above the ground. It moves down an incline, then around a vertical loop. In the absence of friction, what is the speed of the car at the top of the loop, which is 15.4 m above the ground level? conservation is the principle used to solve the problem. Draw the car on the diagram at the two important events of the problem. Choose a GPE=0 level, then fill in the bar graphs to indicate the relative energy distribution for these important events. (a) at at GPE KE GPE KE (b) Workbook for Introductory Mechanics Problem-Solving Copyright by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

34 Unit 5.1B - Problem 5.5B A roller coaster car starts from rest on a track 21.3 m above the ground. It moves down an incline, then around a vertical loop. In the absence of friction, what is the speed of the car at the top of the loop, which is 15.4 m above the ground level? Although the total number of shaded blocks is unimportant, note the equality of the number of blocks in diagrams (b) and, again indicating energy conservation. A B GPE = 0 (a) at A at B GPE KE GPE KE (b)

35 - Unit 5.1B - Problem 5.6B A playground slide has a straight inclined section which ends at the bottom in a short, horizontal section. The inclined section is 2.7 m in length and forms an angle of 40 with the horizontal. A 45 kg child moves down the whole length of the slide. (a) If she starts from rest, what is her speed at the bottom of the incline for a coefficient of kinetic friction of 0.12 between the child and the slide? (b) How much energy has been lost to heat when she reaches the incline s bottom? What is the child s speed for the same heat energy loss if a curved section replaces the straight, inclined section? conservation will be used to solve this problem. The advantage of this method over Newton s 2nd Law will become clear when part of the problem is considered. Draw a block to represent the child at the two important events of the problem. Also fill-in bar graphs to represent the relative energy distribution at those two events, after choosing a GPE=0 level. (a) at at GPE KE HE GPE KE HE Loss (b) Workbook for Introductory Mechanics Problem-Solving Copyright by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

36 Unit 5.1B - Problem 5.6B A playground slide has a straight inclined section which ends at the bottom in a short, horizontal section. The inclined section is 2.7 m in length and forms an angle of 40 with the horizontal. A 45 kg child moves down the whole length of the slide. (a) If she starts from rest, what is her speed at the bottom of the incline for a coefficient of kinetic friction of 0.12 between the child and the slide? (b) How much energy has been lost to heat when she reaches the incline s bottom? What is the child s speed for the same heat energy loss if a curved section replaces the straight, inclined section? You might have chosen GPE = 0 at the ground level, in which case GPE 0 in your diagram. Rubbing between the child s bottom and the slide creates heat as the child s KE increases. Your bar charts must show energy conservation, and your bar chart (b) must be identical to the one below. A B GPE = 0 (a) at A at B GPE KE HE GPE KE HE Loss (b)

37 - Unit 5.1B - Problem 5.7B Starting from rest, a 1.7 kg block slides a distance of 2.1 m down a rough (µ k = 0.2), 20 incline before encountering a spring (k=23.5 N ). (a) What is the block s speed as it first touches the m spring? (b) By how much is the spring compressed? How far back up the incline does the block move after it leaves the spring? Draw the block in the diagram at the important events of part (a). Complete the bar graphs showing the relative energy distribution at these two events after choosing a GPE=0 reference level. (a) at at GPE KE GPE KE HE Loss (b) Workbook for Introductory Mechanics Problem-Solving Copyright by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

38 Unit 5.1B - Problem 5.7B Starting from rest, a 1.7 kg block slides a distance of 2.1 m down a rough (µ k = 0.2), 20 incline before encountering a spring (k=23.5 N ). (a) What is the block s speed as it first touches the m spring? (b) By how much is the spring compressed? How far back up the incline does the block move after it leaves the spring? Your bar charts need not be identical to the ones below, but they must show that energy is conserved. A B (a) GPE = 0 at A at B GPE KE GPE KE HE Loss (b) As the block slides down the plane, some of the initial GPE is lost as heat because of the rubbing between the block and the plane. At the same time, the block speeds up, thereby gaining KE. Some GPE remains when the block first touches the spring because of the choice of the GPE = 0 reference level.

39 - Unit 5.1B - Problem 5.7B Starting from rest, a 1.7 kg block slides a distance of 2.1 m down a rough (µ k = 0.2), 20 incline before encountering a spring (k=23.5 N ). (a) What is the block s speed as it first touches the m spring? (b) By how much is the spring compressed? How far back up the incline does the block move after it leaves the spring? Sketch in the important positions of the block for part (b) of the problem. Then complete the bar charts to show relative energy distributions at those positions. (a) GPE = 0 GPE = 0 (b) at at GPE KE GPE KE HE Loss (d)

40 Unit 5.1B - Problem 5.7B Starting from rest, a 1.7 kg block slides a distance of 2.1 m down a rough (µ k = 0.2), 20 incline before encountering a spring (k=23.5 N ). (a) What is the block s speed as it first touches the m spring? (b) By how much is the spring compressed? How far back up the incline does the block move after it leaves the spring? Notice that only the GPE and KE from diagram, frame are available for conversion into mostly EPE. The heat energy in that same diagram is lost forever. Note also that because of the friction between the block and the incline, more heat energy is lost as the spring is compressed. This is indicated in diagram (d). B C (a) GPE = 0 GPE = 0 (b) at B at C GPE KE EPE HE GPE KE EPE HE Loss Loss (d)

41 - Unit 5.1B - Problem 5.7B Starting from rest, a 1.7 kg block slides a distance of 2.1 m down a rough (µ k = 0.2), 20 incline before encountering a spring (k=23.5 N ). (a) What is the block s speed as it first touches the m spring? (b) By how much is the spring compressed? How far back up the incline does the block move after it leaves the spring? Complete the diagram and bar chart in the same manner as before for part of the problem. C (a) GPE = 0 (b) GPE = 0 at C at GPE KE EPE HE GPE KE EPE HE Loss Loss (d)

42 Unit 5.1B - Problem 5.7B Starting from rest, a 1.7 kg block slides a distance of 2.1 m down a rough (µ k = 0.2), 20 incline before encountering a spring (k=23.5 N ). (a) What is the block s speed as it first touches the m spring? (b) By how much is the spring compressed? How far back up the incline does the block move after it leaves the spring? Heat energy is lost as the spring decompresses, again because of the rubbing between the block and the plane. The block loses additional heat energy as it moves to its maximum height up the plane (position D). At position D, the block sspeed is 0, so the KE is 0 (diagram (d)). D C (a) GPE = 0 (b) GPE = 0 at C at D GPE KE EPE HE GPE KE EPE HE Loss Loss (d)