Workbook for Introductory Mechanics Problem-Solving - Physics...

Copyright 1996-99 by Daniel M. Smith, Jr., All rights reserved.Sponsored by FIPSE, U.S. Department of Education

Workbook for Introductory MechanicsProblem-Solving

Daniel M. Smith, Jr.South Carolina State University

To the Student

The Workbook should help you to learn how to think about physicsproblems before you try using equations. The format is designed to carefullybuild each level of your problem-solving ability before moving you to thenext level. So instead of working one problem at a time, you will workseveral problems at once. Here is advice on how to use this Workbook.

1. After printing, place these materials in a binder before beginning work. (Ifpossible, use double-sided or duplex printing.)

2. Attempt to answer every question by drawing or writing before looking atthe answer.

3. Warning: Simply reading the questions, the answers, and the discussionwithout doing the work is a complete waste of time.

Acknowledgments

Students have provided invaluable assistance in word processing, andin creating graphics for this workbook. Thanks go to Andrick Anderson, KyleHerbert, Tyesia Pompey, Tarryn Reeves, and Keilah Spann. ProfessorsTheodore Hodapp and Dave Maloney are thanked for their criticism andsuggestions. Clip art is taken from the Art Explosion collection published bythe Nova Development Corporation, Calabasas, CA.

I have had the pleasure of teaching and tutoring at South CarolinaState University and Northeastern University, and in the process I havelearned much from students about their difficulties in solving physicsproblems. Those who wish to further my enlightenment may sendcomments to [email protected].

Copyright 1996-99 by Daniel M. Smith, Jr.

All rights reserved. Students using College Physics, 4th edition by JerryWilson and Anthony Buffa may print out one copy of the Workbookmaterial for their own use but may not otherwise copy or distribute thematerial in part or in whole by any means whatsoever, electronic orotherwise, without express written permission from the author and Prentice-Hall, Inc.

5-93

Chapter 5 B

ENERGY

Wilson/Buffa Chapter 5: Work and Energy

Unit 1Bar chart representation of energy conservation

Unit 2Restatement of the problem in the language of energy

Unit 3Translation of the energy statement into algebraic

language; problem solutions

5-94

Copyright 1996-99 by Daniel M. Smith, Jr.

All rights reserved. Students using College Physics, 4thedition by Jerry Wilson and Anthony Buffa may print outone copy of the Workbook material for their own use butmay not otherwise copy or distribute the material in part orin whole by any means whatsoever, electronic or otherwise,without express written permission from the author andPrentice-Hall, Inc.

Energy - Unit 5.1B - Problem 5.1B 5–95

Masses m1 (0.3 kg) and m2 (0.8 kg) are connected by a masslessstring which passes over a massless pulley. If the masses arereleased from rest, find the speed of the masses at the instantthat m2 has moved a distance of 0.4 m.

Workbook for Introductory Mechanics Problem-SolvingCopyright 1996-99 by Daniel M. Smith, Jr.Sponsored by FIPSE (U.S. Department of Education)

m1

m2

5.103What physics principle or principles should be used to

solve this problem?

5–96 Energy - Unit 5.1B - Problem 5.1B


m1

m2

5.104The problem can be solved in at least two ways, by using

either Newton’s 2nd Law, or by using energy conservation. Thisamounts to saying that the problem can be described either inthe language of forces or in the language of energy.



m1

m2

5.105Use the language of forces to describe the problem by

drawing vectors on the diagram (free-body diagram), and writingforce equations. Without solving the problem, explain in wordshow to continue to a problem solution.

m1

m2



m1

m2

5.106

For mass m1, Newton’s 2nd Law givesT – m1g = m1a, (5.133)

while for mass m2,T – m2g = –m2a. (5.134)

We have two equations in two unknowns, so both a andT can be determined. Using the value of a in the kinematicsequation v2 = v0

2 + 2a(y–y0) gives us the desired result, becausethe masses have the same speed.

m1

m2

T

T

m1 g

m2 g

y



m1

m2

5.107Sketch the positions of both masses after mass m2 has

moved 0.4 m. Choose a reference level where the gravitationalpotential energy is 0, and draw this on the diagram also.

m1

m2



m1

m2

5.108The reference level is chosen to be below both masses, but

any choice is valid.

m1

m2

GPE = 0



m1

m2

5.109Make bar graphs, indicating qualitatively the relative

amounts of the different forms of energy, both initially and aftermass m2 has moved by 0.4 m. For example, mass m1 has GPEinitially, as indicated by the solid bar. Your bar graphs need notlook like the ones in the next frame, but they must show energyconservation. (GPE = gravitational potential energy, KE = kineticenergy)

Mass m ini tial energy

1 Mass m ini tial energy

2

GPE1 KE1 GPE2 KE2

(a) (b)

Mass m final energy

Mass m final energy

1 2

GPE1 KE1 GPE2 KE2

(c) (d)



m1

m2

5.110Both m1 and m2 have GPE initially. Mass m2 has a larger

GPE than mass m1 initially because m2 is higher above thereference level, and because m2 > m1 (recall that EGP = mgh).Otherwise, your initial values for GPE are unimportant. Butbecause the system starts from rest, both masses have KE = 0initially.

Because mass m2 has fallen by 0.4 m, it has lost GPE(compare bar charts (b) and (d)), while m1 has gained GPEbecause it is raised by 0.4 m. Both masses are moving with thesame speed after m2’s 0.4 m fall, so they both have KE. Butbecause m2 > m1 , mass m2 has the larger KE (recall thatEk = 1

2 mv2). Notice that energy is conserved: the total number ofshaded blocks in diagrams (a) and (b) is equal to the total numberin diagrams (c) and (d); this must be true in your diagrams also.

Although we show GPE1>GPE2 in figures (c) and (d), thisis not always true because the relationship between GPE1 andGPE2 really depends upon the chosen reference level (GPE = 0)and initial positions of the masses. For example, if the massesare at the ends of a very long string, m2 would remain higherthan m1 even after m2 moved by 0.4 m ; then GPE2>GPE1.


1

GPE1 KE1


2

GPE2 KE2

(a) (b)

Mass m final energy

1

GPE1 KE1

(c)

Mass m final energy

2

GPE2 KE2

(d)


A roller-coaster car of mass 480 kg is raised to an initial height of18.0 m above ground level. With a speed of 1.3

ms the car begins

its descent. (a) Ignoring friction, what is the car’s speed when itreaches a horizontal section of track of height 4.0 m aboveground level? The car ascends its first hill to reach a level of8.5 m above ground. (b) What is the car’s speed at this level,again ignoring friction? (c) If the car’s speed after ascending thehill is actually 10.1

ms , how much energy has been lost from the

start because of the friction between the track and the coaster car?


5.111What principle(s) should be used to solve this problem?

5.112Although you may see that energy conservation is the

principle allowing this problem to be solved, can you tell whyNewton’s 2nd Law is not useful here? Try to answer this byconsidering the descending car at two places on the track.



ms the car begins




5.113

Notice that the component of gravitational forceparallelto the track is different in each case, because the angle changes.This component changes continuously as the car moves downthe track. For this reason, writing a single force equation usingthe 2nd Law is impossible.



ms the car begins




5.114Draw the coaster car on the diagram at the car’s two

important positions in part (a) of the problem. Then, afterchoosing the reference level for GPE=0, make bar graphsindicating the relative amounts of the different forms of energyat the two events.

ground(a)

GPE KE GPE KE

(b) (c)

Energy at ___

Energy at ___



ms the car begins




5.115The two important positions of the car in part (a) are its

initial height (at A), and its height at the lower level (at B). Yourbar graphs may not be identical to the ones below, but at leastthey must show that energy is conserved.

ground(a)

GPE KE GPE KE

(b) (c)

Energy at ___

Energy at ___

A

B

BA

GPE = 0



ms the car begins




5.116GPE = 0 can be assigned at any height on diagram (a) of the

previous frame. If it is assigned at a height greater than that ofthe track at B, values of GPE below that height are negative.

Because the roller coaster car’s height above ground levelis reduced as it moves from position A to position B, its GPE isreduced as well. What the car loses in GPE, however, it gains inKE, so the car is moving faster at position B.

How is the principle of energy conservation reflected inthe bar graphs?



ms the car begins




5.117That the total number of shaded blocks in diagram (b) is

equal to the total number in diagram (c) indicates that energy isconserved.



ms the car begins




5.118Draw 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 theenergy at these two positions. Again, your bar graphs will notnecessarily be identical to the ones in the following frame.

ground(a)

GPE KE GPE KE

(b) (c)

Energy at ___

Energy at ___



ms the car begins




5.119You might have drawn either of the diagrams ((a) or (b))

below because there are two possibilities, positions A and C, orpositions B and C. Again the GPE=0 level is arbitrary but wehave chosen it at ground level.

(a)

GPE KE GPE KE

(c) (d)

Energy at ___

Energy at ___CA

ground

A

C

B

C

GPE = 0

(b)

GPE KE

(e)

Energy at ___B

GPE KE

(f)

Energy at ___C



ms the car begins




5.120For 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 atC than at A. Note that the numbers of shaded blocks in diagrams(c) and (d) are equal, indicative of energy conservation.

For diagram (b) of the previous frame, the roller coastercar gains GPE in moving from B to C, and loses KE, so the carslows in moving from B to C. Energy conservation isrepresented by the equality in the numbers of shaded blocks indiagrams (e) and (f).

For part (c) of the problem, consider the diagram below,then complete the bar graphs to show the relative energydistribution for the car at positions A and C.

ground GPE = 0

A

C

GPE KE GPE KE

(b) (c)

Energy at ___

Energy at ___CA

(a)



ms the car begins




5.121At positon C in the diagram of the previous frame, the

roller-coaster car has the same GPE as in the no-friction case. Butthe KE must be less than that in the frictionless case because itsspeed is less: some of the original GPE has been converted toheat energy (HE). Your bar graphs should be similar to thosebelow, not necessarily identical, but they must show energyconservation.

GPE KE GPE KE

(a) (b)

Energy at ___

Energy at ___CA

HE


A small child of mass 18.2 kg (weight of 40 lb) jumps onto amattress and compresses it by 10.3 cm. If the mattress compresseswith an effective spring constant of 8574

Nm , (a) what is the

maximum height above the mattress to which the child ispropelled? (b) What is the child’s speed as it leaves the mattress?


5.122What principle should be used to solve this problem?

Again energy conservation is most convenient. Explain why thisproblem is less conveniently solved using Newton’s 2nd Law byconsidering the forces exerted on the child. You may representthe child by a block, and the mattress by a spring.





5.123The vectors in the diagram below represent the two forces

exerted on the child, the gravitational force (weight), and theforce exerted by the spring. Remember that the magnitude of thespring’s force depends upon how much the spring iscompressed. This means that the force equation is complicated,because the amount of the spring’s compression must be takeninto account. This complication renders the force equationimpossible to solve algebraically.

5.124What are the two important positions of the child in part

(a) of the problem? What are they in part (b)?

5.125For part (a), the child is at important positions when he

compresses the mattress by 10.3 cm, and when he reachesmaximum height. For part (b), the child is at important positionswhen the mattress is compressed by 10.3 cm, and when he justleaves the mattress moving upward.





5.126After the child jumps onto the mattress, it is compressed

by a certain amount before recoil. Draw a diagram of themaximum compression for part (a), then draw a diagram of thechild at maximum height above the mattress. Use a block andspring to represent the child and mattress in both cases. Choose aGPE=0 reference level, then complete the bar graphs indicatingthe relative energy distribution in the two cases.

Spring energy for (a)

Block energy for (a)

Spring energy for (b)

Block energy for (b)

(c) (d)

(e) (f)

(a) Spring at maximum compression

(b) Spring uncompressed, Block at maximum height





5.127For part (a) of the problem, all of the spring’s elastic

potential energy (diagrams (a) and (c)) has been converted to theblock’s gravitational potential energy (diagrams (b) and (f)).Recall that at maximum height, the block’s velocity is zero, thenso is its KE. Your bar charts may differ quantitatively from these,but they must show energy conservation.





(c) (d)

(e) (f)


(b) Spring uncompressed, block at maximum height

GPE = 0

GPE = 0

EPE

GPE

GPE

EPE





5.128For part (b) of the problem we first consider the child

when the mattress is at maximum compression, and next whenthe child just leaves the fully uncompressed mattress. Draw adiagram of the latter event. Complete the bar graphs indicatingthe relative energy distributions in the two cases.





(c) (d)

(e) (f)


(b) Block in motion, just leaving ful ly uncompressed spring

GPE = 0





5.129All of the spring’s elastic potential energy (diagrams (a)

and (c)) is converted partly to the block’s gravitational potentialenergy, and partly to the block’s kinetic energy (diagrams (b) and(f)). Therefore, the total number of blocks in diagrams (c) and (d)equals the number of blocks in diagram (f).





(c) (d)

(e) (f)


(b) Spring uncompressed, block in motion, just leaving spring

GPE = 0

GPE = 0

EPE

GPE KEEPE

GPE KE


A child of mass 31.5 kg (weight of 69.3 lb) sits in a playgroundswing whose seat, 0.6 m above the ground, is supported by ropesof length 2.4 m. The child is given a push, and the swing rises toa maximum angle of 36° with the vertical. (a) What is the child’sspeed as the swing passes back through its rest position? (b) Thechild releases her grasp on the ropes as the swing moves forwardat the bottom of its trajectory, and the child slides out of theswing. How far away from the release point does she land?Ignore air resistance.


5.130What principle should be used to solve part (a) of the

problem? Energy conservation, of course. But why is it difficultto apply Newton’s 2nd Law in this circumstance? Draw adiagram for your explanation by representing the child in theswing by a mass at the end of a string, then draw vectors toindicate the forces exerted on the mass.



5.131

The angle θ changes continuously, so there is no single2nd Law equation in which the tension is constant. The forceequation would need to take into account the dependence of thetension on the angle throughout the motion. Such an equationcannot be solved algebraically.

θ



5.132Complete the bar graphs to show the relative energy

distribution in part (a) of the problem as the mass moves fromposition A to position B.

36o

GPE = 0(a)

Energyat A

(b)

GPE

Energyat B

(c)

GPEKE KE

B

A



5.133The 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 chooseGPE=0 at position B, then GPE=0 for chart (c)), but they mustshow that energy is conserved.

36o

GPE = 0(a)

Energyat A

(b)

GPE

Energyat B

(c)

GPEKE KE

B

A


A roller coaster car starts from rest on a track 21.3 m above theground. 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 ofthe loop, which is 15.4 m above the ground level?


5.134Energy conservation is the principle used to solve the

problem. Draw the car on the diagram at the two importantevents of the problem. Choose a GPE=0 level, then fill in the bargraphs to indicate the relative energy distribution for theseimportant events.

(a)

Energy at______

(b)

Energy at _______

(c)

GPE KE GPE KE


A roller coaster car starts from rest on a track 21.3 m above theground. 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 ofthe loop, which is 15.4 m above the ground level?

5.135Although the total number of shaded blocks is

unimportant, note the equality of the number of blocks indiagrams (b) and (c), again indicating energy conservation.

(a)

B

A

GPE = 0

Energy atA

(b)

Energy at B

(c)

GPE KE GPE KE


A playground slide has a straight inclined section which ends atthe bottom in a short, horizontal section. The inclined section is2.7 m in length and forms an angle of 40° with the horizontal. A45 kg child moves down the whole length of the slide. (a) If shestarts from rest, what is her speed at the bottom of the incline fora coefficient of kinetic friction of 0.12 between the child and theslide? (b) How much energy has been lost to heat when shereaches the incline’s bottom? (c) What is the child’s speed for thesame heat energy loss if a curved section replaces the straight,inclined section?


5.136Energy conservation will be used to solve this problem.

The advantage of this method over Newton’s 2nd Law willbecome clear when part (c) of the problem is considered. Draw ablock to represent the child at the two important events of theproblem. Also fill-in bar graphs to represent the relative energydistribution at those two events, after choosing a GPE=0 level.

Energy at

GPE KE

(b)

Energy at

GPE KE

(c)

(a)

HE HELoss

5–126 Energy- Unit 5.1B - Problem 5.6B

A playground slide has a straight inclined section which ends atthe bottom in a short, horizontal section. The inclined section is2.7 m in length and forms an angle of 40° with the horizontal. A45 kg child moves down the whole length of the slide. (a) If shestarts from rest, what is her speed at the bottom of the incline fora coefficient of kinetic friction of 0.12 between the child and theslide? (b) How much energy has been lost to heat when shereaches the incline’s bottom? (c) What is the child’s speed for thesame heat energy loss if a curved section replaces the straight,inclined section?

5.137You might have chosen GPE = 0 at the ground level, in

which case GPE ≠ 0 in your diagram (c). Rubbing between thechild’s bottom and the slide creates heat as the child’s KEincreases. Your bar charts must show energy conservation, andyour bar chart (b) must be identical to the one below.

Energy at

GPE KE

(b)

Energy at

GPE KE

(c)

(a)

HE HE

A

B

A B

GPE = 0

Loss


Starting from rest, a 1.7 kg block slides a distance of 2.1 m down arough (µk = 0.2), 20° incline before encountering a spring(k=23.5

Nm ). (a) What is the block’s speed as it first touches the

spring? (b) By how much is the spring compressed? (c) How farback up the incline does the block move after it leaves thespring?


5.138Draw the block in the diagram at the important events of

part (a). Complete the bar graphs showing the relative energydistribution at these two events after choosing a GPE=0 referencelevel.

Energy at

GPE KE

(b)

Energy at

GPE KE

(c)

(a)

HELoss





5.139Your bar charts need not be identical to the ones below,

but they must show that energy is conserved.

As the block slides down the plane, some of the initialGPE is lost as heat because of the rubbing between the block andthe plane. At the same time, the block speeds up, thereby gainingKE. Some GPE remains when the block first touches the springbecause of the choice of the GPE = 0 reference level.

GPE = 0

Energy at

GPE KE

(b)

Energy at

GPE KE

(c)

(a)

HELoss

B

A

A B





5.140Sketch in the important positions of the block for part (b)

of the problem. Then complete the bar charts to show relativeenergy distributions at those positions.

Energy at

GPE KE

(c)

Energy at

GPE KE

(d)

HELoss

(a) (b)GPE = 0 GPE = 0





5.141Notice that only the GPE and KE from diagram (c),

frame 5.139 are available for conversion into mostly EPE. Theheat energy in that same diagram is lost forever.

Note also that because of the friction between the blockand the incline, more heat energy is lost as the spring iscompressed. This is indicated in diagram (d).

Energyat B

(c)

Energy at C

(d)

(a) (b)GPE = 0 GPE = 0

GPE KE HELoss

EPE GPE KE HELoss

EPE

B C





5.142Complete the diagram and bar chart in the same manner

as before for part (c) of the problem.

Energyat C

(c)

Energy at

(d)

(a) (b)GPE = 0

GPE KE HELoss

EPE GPE KE HELoss

EPE

GPE = 0

C





5.143Heat energy is lost as the spring decompresses, again

because of the rubbing between the block and the plane. Theblock loses additional heat energy as it moves to its maximumheight up the plane (position D). At position D, the block’sspeedis 0, so the KE is 0 (diagram (d)).

Energyat C

(c)

Energy at D

(d)

(a) (b)GPE = 0

GPE KE HELoss

EPE GPE KE HELoss

EPE

GPE = 0

C

D

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