PHYS 1111 - Summer 2007 - Professor Caillault

Homework Solutions

Chapter 8

5. Picture the Problem: The physicalsituation is depicted at right.

Strategy: Use Wsp =

12

k xi2 − xf

2( ) for the

work done by the spring. That way thework will always be negative if you startout at

xi = 0 because the spring force will

always be in the opposite direction fromthe stretch or compression. The workdone by kinetic friction is

Wfr = −µkmgd ,

where d is the distance the box is pushedirregardless of direction, because thefriction force always acts in a directionopposite the motion.

Solution: 1. (a) Sum thework done by thespring for each segment ofpath 2:

Wsp =12

k x12 − x4

2( ) + x42 − x3

2( )

= 12

480 N/m( ) 02 − − 0.020 m( )2

+ − 0.020 m( )2 − 0.020 m2( )

Wsp = − 0.096 J( ) + 0 J( ) = − 0.096 J

2. Sum the work done byfriction foreach segment of path 2:

Wfr = −µkmg d1 + d2( )= − 0.16( ) 2.7 kg( ) 9.81 m/s2( ) 0.020 + 0.040 m( ) = − 0.25 J

3. (b) Sum the work done bythespring for the direct pathfrom A to B:

Wsp =12

k xA2 − xB

2( )= 1

2480 N/m( ) 02 − 0.020 m( )2

= − 0.096 J

4. Sum the work done byfriction forthe direct path from A to B:

Wfr = −µkmgd = − 0.16( ) 2.7 kg( ) 9.81 m/s2( ) 0.020 m( ) = − 0.085 J

Insight: The work done by friction is always negative, and increases in magnitude with the distancetraveled.

11. Picture the Problem: The spring in the soap dispenser is compressed by the applied force.

Strategy: Use equation 8-5 to find the spring constant using the given energy and compressiondistance data. Solve the same equation for x in order to answer part (b).

Solution: 1. (a) Solve equation 8-5 for k:

k =2U

x2=

2 0.0025 J( )0.0050 m( )2

= 200 N/m = 0.20 kN/m

2. (b) Solve equation 8-5 for x: x =

2Uk

=2 0.0084 J( )

200 N/m= 0.92 cm

Insight: To compress the spring of this dispenser 0.50 cm requires 1.0 N or about � lb of force.

13. Picture the Problem: A graph of the potentialenergy vs. stretch distance is depicted at right.

Strategy: The work that you must do to stretch aspring is equal to minus the work done by the springbecause the force you exert is in the oppositedirection from the force the spring exerts. Useequations 8-1 and 8-5 together to find the springconstant and the required work to stretch the springthe specified distance.

Solution: 1. (a) Because the stored potential energyin a spring is proportional to the stretch distancesquared, the work required to stretch the spring from5.00 cm to 6.00 cm will be greater than the workrequired to stretch it from 4.00 cm to 5.00 cm.

2. (b) Useequations 8-1and 8-5 tofind k:

Wreq = −Wspring = − −ΔU( ) =U5 −U4

= 12

kx52 − 1

2kx4

2 = 12

k x52 − x4

2( )k =

2Wreq

x52 − x4

2=

2 30.5 J( )0.0500 m( )2 − 0.0400 m( )2

= 6.78 ×104 N/m

3. Use k andequations 8-1and 8-5 to findthe new

Wreq :

Wreq =

12

k x22 − x1

2( ) = 12

6.78 ×104 N/m( ) 0.0600 m( )2

− 0.0500 m( )2 = 37.3 J

Insight: Using the same procedure we discover that it would take 44.1 J to stretch the spring from 6.00cm to 7.00 cm.

21. Picture the Problem: The trajectory of the rock is depicted at right.

Strategy: The rock starts at height h, rises to ymax , comes briefly to

rest, then falls down to the base of the cliff at y = 0. Set themechanical energy at the point of release equal to the mechanicalenergy at the base of the cliff and at the maximum height

ymax in order

to find vi and

ymax .

Solution: 1. (a)Set

Ei = Ef and

solve for vi :

Ki +U i = Kf +U f

12

mvi2 + mgh = 1

2mvf

2 + 0

vi = vf2 − 2gh

= 29 m/s( )2 − 2 9.81 m/s2( ) 32 m( )vi = 15 m/s

2. (b) Now set

Eymax = Ef

and solve for

ymax :

Kymax +Uymax = Kf +U f

0 + mgymax =12

mvf2 + 0 ⇒ ymax =

vf2

2g=

29 m/s( )2

2 9.81 m/s2( )= 43 m

Insight: In part (a) the initial energy is a combination of potential and kinetic, but becomes all kineticjust before impact with the ground. In part (b) the rock at the peak of its flight has zero kinetic energy;all of its energy is potential energy.

all of its energy is potential energy.

25. Picture the Problem: The pendulum bob swings from point B to point Aand gains altitude and thus gravitational potential energy. See the figure atright.

Strategy: Use equation 7-6 to find the kinetic energy of the bob atpoint B. Use the geometry of the problem to find the maximum change inaltitude

Δymax of the pendulum bob, and then use equation 8-3 to find its

maximum change in gravitational potential energy. Apply conservation ofenergy between points B and the endpoint of its travel to find themaximum angle

θmax the string makes with the vertical.

Solution: 1. (a) Useequation 7-6 to find

KB :

KB = 12

mvB2 = 1

20.33 kg( ) 2.4 m/s( )2 = 0.95 J

2. (b) Since there is no friction, mechanical energy is conserved. If wetake the potential energy at point B to be zero, we can say that all of thebob’s kinetic energy will become potential energy when the bob reachesits maximum height and comes momentarily to rest. Therefore the changein potential energy between point B and the point where the bob comes torest is 0.95 J.

3. (c) Find the heightchange

Δymax of the

pendulum bob:

Δymax = L − Lcosθmax = L 1− cosθmax( )

4. Use equation 8-3 andthe result of part (b) tosolve for

θmax :

ΔU = mgΔymax = mgL 1− cosθmax( )

θmax = cos−1 1−ΔUmgL

= cos−1 1−

0.95 J

0.33 kg( ) 9.81 m/s2( ) 1.2 m( )

= 41°

Insight: The pendulum bob cannot swing any further than 41° because there is not enough energyavailable to raise the mass to a higher elevation.

33.Picture the Problem: The initial and final statesof the system are depicted at right. The valuesgiven in the example are

m1 = 2.40 kg,

m2 = 1.80 kg, d = 0.500 m, and

µk = 0.450.

Strategy: The nonconservative work done byfriction changes the mechanical energy of thesystem. Use equation 8-9 to find ΔE and set itequal to the work done by friction. Solve theresulting expression for the final velocity of thesystem.Solution: 1.Write equation8-9 to find

Wnc :

Wnc = Ef − Ei = Kf +U f( ) − Ki +U i( )= 1

2m1 + m2( )vf

2 + m1gh + m2gy2f

−

12

m1 + m2( )vi2 + m1gh + m2gy2i

Wnc =12

m1 + m2( ) vf2 − vi

2( ) + m2g y2f − y2i( )2. Thenonconservativework is done byfriction:

Wnc = − fkd = −µkm1gd

3. Substitute theexpression fromstep 2 into step1 and solve for

vf

2 − vi2 :

−µkm1gd = 12

m1 + m2( ) vf2 − vi

2( ) + m2g 0 − d( )gd m2 − µkm1( ) = 1

2m1 + m2( ) vf

2 − vi2( )

2gd m2 − µkm1( )m1 + m2( )

= vf2 − vi

2 =2 9.81 m/s2( ) 0.500 m( ) 1.80 kg − 0.450( ) 2.40 kg( )

2.40 +1.80 kg( )= 1.68 m2 / s2

4. Finally, solvefor

vf :

vf = vi2 +1.68 m2 /s2 = 1.3 m/s( )2 +1.68 m2 /s2 = 1.8 m/s

Insight: If the blocks are not given the initial speed of 1.3 m/s, their final speed is 1.30 m/s, as inexample 8-10.

37. Picture the Problem: The skater travels up a hill (we know this for reasons given below), changing hiskinetic and gravitational potential energies, while both his muscles and friction do nonconservative workon him.Strategy: The total nonconservative work done on the skater changes his mechanical energy according toequation 8-9. This nonconservative work includes the positive work

Wnc1 done by his muscles and the

negative work Wnc2 done by the friction. Use this relationship and the known change in potential energy

to find Δy.

Solution: 1. (a) The skater has gone uphill because the work done by the skater is larger than that doneby friction, so the skater has gained mechanical energy. However, the final speed of the skater is less thanthe initial speed, so he has lost kinetic energy. Therefore he must have gained potential energy, and hasgone uphill.

2. (b) Set thenonconservativework equal tothe change inmechanicalenergy andsolvefor Δy :

Wnc =Wnc1 +Wnc2 = ΔE = Ef − Ei

Wnc1 +Wnc2 = Kf +U f( ) − Ki +U i( ) = 12

m vf2 − vi

2( ) + mgΔy

Δy = Wnc1 +Wnc2 −12

m vf2 − vi

2( )

mg

=3420 J( ) + −715 J( ) − 1

281.0 kg( ) 1.22 m/s( )2 − 2.50 m/s( )2

81.0 kg( ) 9.81 m/s2( )= 3.65 m

Insight: Verify for yourself that if the skates had been frictionless but the skater’s muscles did the sameamount of work, the skater’s final speed would have been 4.37 m/s. He would have sped up if it weren’tfor friction!

41. Picture the Problem: The U vs. x plot is depicted at right.

Strategy: Describe the motion of the object, keeping inmind that objects tend to move to the minimum potentialenergy, and when they do their kinetic energy is maximum.The turning points on the potential energy plot are points Aand E.

Solution: At point A, the object is at rest. As the objecttravels from point A to point B, some of its potential energyis converted into kinetic energy and the object’s speedincreases. As the object travels from point B to point C,some of its kinetic energy is converted back into potentialenergy and its speed decreases. From point C to point D, thespeed increases again, and from point D to point E, thespeed decreases.

Insight: The object momentarily comes to rest at point E, but then turns around and accelerates backtowards D and retraces its path all the way to point A, at which time the cycle begins again.

61. Picture the Problem: The child slides fromrest at point A and lands at point B as indicatedin the figure at right.

Strategy: Use the conservation of mechanicalenergy to find the horizontal speed of the childat the bottom of the slide in terms of h. Thenuse equation 4-9, the landing site of a projectilelaunched horizontally, to find the speed thechild should have in order to land 2.50 m downrange. Set the speeds equal to each other andsolve for h.

Solution: 1. Use equation 8-3 and let

EA = Ebottom to find

vbottom :

KA +UA = Kbottom +Ubottom

12

mvA2 + mgyA = 1

2mvbottom

2 + mgybottom

0 + mg h +1.50 m( ) = 12

mvbottom2 + mg 1.50 m( )

2gh = vbottom

2. Use equation 4-9 to find theappropriate

vbottom for the child to land

2.50 m down range:

x = vbottom 2ybottom g ⇒ vbottom = x g 2ybottom

3. Set the two velocities equal to each other andsolve for h:

2gh = x g 2ybottom

2gh = x2g 2ybottom

h =x2

4ybottom

=2.50 m( )2

4 1.50 m( )= 1.04 m

Insight: These are the sort of calculations an engineer might make to determine how high to build aslide so that a child will land in a certain place. However, an engineer should account for the nonzerofriction when making his design.

71. Picture the Problem: The mass oscillates back and forth on a frictionless horizontal surface.

Strategy: As the mass oscillates back and forth, its kinetic energy is converted into spring potentialenergy and back again. Use equation 8-6 and the conservation of mechanical energy to find theturning points of the motion.

Solution: 1. (a) Set Ei = Ef , where the

initial state corresponds to maximum speedand the final state isat the turning point.

U i + Ki =U f + Kf

0 + 12

mvi2 = 1

2kx2 + 0

2. Solve the expression from step 1 for x: x = ±vi

mk= ± 1.7 m/s( ) 2.2 kg

560 N/m= ±0.11 m

3. (b) Write a ratio to find the multiplicativefactorfor x when k is doubled:

xnew

xold

=±vi m 2k

vi m k=

1

2

Insight: The maximum compression distance is reduced because the stiffer spring stores more energywith less compression.