Download Chapter 8: Potential Energy and Conservation of

Chapter 8: Potential Energy and Conservation of Energy
Answers to Even-Numbered Conceptual Questions
6.
The object’s kinetic energy is a maximum when it is released, and a minimum when it reaches its greatest height. The gravitational potential of the system is a minimum when the object is released, and a maximum when the object reaches its greatest height.
12.
The total mechanical energy decreases with time if air resistance is present.
Answers to Even-Numbered Conceptual Exercises
4.
The two balls have the same change in gravitational potential energy. Thus, ignoring air resistance, they will also have the same change in kinetic energy.
6.
(a) Your answers will disagree. (b) Your answers will agree. (c) Your answers will agree.
12.
(a) The potential energy of the system—which is gravitational potential energy—decreases as you move down the hill. (b) Your kinetic energy remains the same, since your speed is constant. (c) In order for your speed and kinetic energy to remain constant as you pedal down the hill, a nonconservative force must have done negative work on you and your bicycle. For example, you may have applied the brakes to control your speed, or the ground may be soft or muddy. In any case, the mechanical energy of this system has decreased.
14.
The mechanical energy of the shuttle when it lands is considerably less than when it is in orbit. First, its speed on landing is low compared to its speed in orbit, giving a much smaller kinetic energy. Second, the source of energy responsible for heating the tiles is the gravitational potential energy that is released as the shuttle loses altitude. Therefore, both the kinetic and potential energy of the shuttle have much smaller values when the shuttle lands, leading to a smaller value for the total mechanical energy.
16.
Ignoring any type of frictional force, the speed of the object is the same at points A and G. Similarly, the speed of the object at point B is the same as its speed at points D and F. In addition, the lower a point on the curve, the greater its speed. Combining these observations, we arrive at the following ranking: A = G < B = D = F < E < C.
Solutions to Problems
6.
Picture the Problem: The cliff diver plunges straight downward due to the force of gravity.
Strategy: Solve equation 8-3 for the weight of the diver. Let y = 0 correspond to the surface of the water.
Solution: Solve equation 8-3 for mg:
U = mgy ⇒ mg =
U 25, 000 J
=
= 540 N = 0.54 kN
y
46 m
Insight: If you set U = 0 at the top of the cliff, then U = −25 kJ and y = − 46 m when the diver enters the water.
14. Picture the Problem: The pendulum bob swings from point A to point B and loses
altitude and thus gravitational potential energy. See the figure at right.
Strategy:
Use the geometry of the problem to find the change in altitude ∆y of
the pendulum bob, and then use equation 8-3 to find its change in gravitational potential
energy.
Solution: 1. Find the height change
∆y of the pendulum bob:
∆y = L cos θ − L = L ( cos θ − 1)
2. Use ∆y to find ∆U :
Insight: Note that ∆y is negative because the pendulum swings from A to B. Likewise, ∆y is positive and the
pendulum gains potential energy if it swings from B to A.
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 the mechanical energy at the point of release equal to the mechanical energy 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 :
vi
ymax
h
2. (b) Now set Eymax = Ef
and solve for ymax :
Insight: In part (a) the initial energy is a combination of potential and kinetic, but becomes all kinetic just 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.
24. Picture the Problem: The pendulum bob swings from point B to point A and gains
altitude and thus gravitational potential energy. See the figure at right.
Strategy:
Use the geometry of the problem to find the change in altitude ∆y of
the pendulum bob, and then use equation 8-3 to find its change in gravitational potential
energy. Apply conservation of energy between points B and A to find the speed at A.
Solution: 1. (a) Find the height
change ∆y of the pendulum bob:
2. Use ∆y to find ∆U :
∆y = L − L cos θ = L ( 1 − cos θ )
3. (b) Set EB = EA and solve for vA :
KB + U B = KA + U A
1
2
mvB2 = 12 mvA2 + ∆U
vA = vB2 −
2∆U
=
m
( 2.4 m/s )
2
−
2 ( 0.70 J )
0.33 kg
= 1.2 m/s
4. (c) If the mass of the bob is increased the answer to part (a) will increase. The change in gravitational potential
energy depends linearly on the mass.
5. (d) If the mass of the bob is increased the answer to part (b) will stay the same. Although the change in potential
energy will increase, the change in kinetic energy will also increase.
Insight: Another way to look at the answer to (d) is that ∆U m = mg ∆y m = g ∆y independent of mass. That means the
formula for vA in step 3 is independent of mass.
26. Picture the Problem: The motions of the masses in the Atwood’s
machine are depicted in the figure at right:
Strategy: Mechanical energy is conserved since there is no friction. Set Ei = Ef and solve for vf . The speeds of each mass must always be the same because they are connected by a rope.
Solution:
1. (a) Set Ei = Ef
and solve for vf :
2. (b) Use the expression
from part (a) to find vf :
Insight: The mass m2 loses more gravitational potential energy than m1 gains, so there is extra energy available to give
the system kinetic energy. We bent the rules for significant figures a bit in step 2 because by the rules of subtraction
4.1-3.7 kg = 0.4 kg, only one significant figure.
27. Picture the Problem: The motions of the masses in the Atwood’s
machine are depicted in the figure at right:
Strategy: Mechanical energy is conserved since there is no friction. Set Ei = Ef and solve for h when vf = 0 . The speeds of each mass must always be the same because they are connected by a rope.
Solution: 1. Set Ei = Ef
and then set vf = 0 :
2. Solve for h:
h=
1  m1 + m2

2 g  m1 − m2
 2
1
 vi =
2 ( 9.81 m/s 2 )

 3.7 + 4.1 kg 
2

 ( 0.20 m/s ) = 0.04 m = 4 cm
4.1
−
3.7
kg


Insight: The kinetic energy of the two-mass system is converted to a difference in potential energy between mass 2 and
mass 1. By the rules for subtraction of significant figures 4.1 − 3.7 kg = 0.4 kg, only one significant figure.
37. Picture the Problem: The skater travels up a hill (we know this for reasons given below), changing his kinetic and
gravitational potential energies, while both his muscles and friction do nonconservative work on him.
Strategy: The total nonconservative work done on the skater changes his mechanical energy according to equation 8­9. 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 ∆y.
Solution: 1. (a) The skater has gone uphill because the work done by the skater is larger than that done by friction, so
the skater has gained mechanical energy. However, the final speed of the skater is less than the initial speed, so he has
lost kinetic energy. Therefore he must have gained potential energy, and has gone uphill.
2. (b) Set the nonconservative
work equal to the change in
mechanical energy and solve
for ∆y :
Insight: Verify for yourself that if the skates had been frictionless but the skater’s muscles did the same amount of
work, the skater’s final speed would have been 4.37 m/s. He would have sped up if it weren’t for friction!
67. Picture the Problem: The physical situation is depicted at right.
Strategy:
Use equation 8-9 to set the mechanical energy loss equal
to the work done by friction. Then solve for the distance d that the block
slides along the surface of the inclined plane. Use Newton’s Second Law
to find the normal force between the block and plane, and use the relation
y = d sin θ to convert the distance along the incline into a gain of altitude.
Let y = 0 at the start of the slide and vf = 0 at the end.
r
N
r
fk
d
r
vi
y
r
W
θ
mg cos θ
Wnc = Ef − Ei = ( K f + U f ) − ( K i + U i )
Solution: 1. Write out equation 8-9 to
find Wnc :
= ( 0 + mgy ) − ( 12 mvi2 + 0 )
∑F
2. Use Newton’s Second Law to find N:
y
= N − mg cos θ = 0 ⇒ N = mg cos θ
3. Set Wnc equal to the work done
by friction, set y = d sin θ , and
solve for d:
Insight: At first glance it may seem like the µ k = 0.62 limits us to two significant figures, but by the rules of addition
sin 27.4° + ( 0.62 ) cos 27.4°  = ( 0.460 + 0.55 ) = 1.01 which still has three significant figures.
70. Picture the Problem: The physical situation is depicted in the figure.
Strategy:
Use the conservation of mechanical energy and the
geometry of the problem to find the skateboarder’s speed at point B. Then
use Newton’s Second Law in the vertical direction at point B to find the
force the track must exert to support the skateboarder’s weight and provide
his centripetal force. Let yB = 0 , yA = r , and vA = 0 .
Solution: 1. Set EB = EA to find vB :
2. Write Newton’s Second Law in the
vertical direction at B to find N:
∑F
y
= N − mg = m vB2 r
2g r 

N = m g +
 = mg ( 1 + 2 )
r 

= 3 ( 57.0 kg ) ( 9.81 m/s 2 ) = 1680 N = 1.68 kN
Insight: The skateboarder feels 3 times heavier than normal at point B, or in other words, he experiences 3g’s of force.
82. Picture the Problem: The physical situation is depicted at right.
Strategy: Since the system is frictionless, we can set the mechanical energy at configuration i equal to the mechanical energy at configuration f . Solve the resulting expression for the distance d. The work done by the rope on m2 is nonconservative and therefore changes the mechanical energy of m2 according to equation 8­9. Use this fact to find the work the rope does on m2 . Ki + U i = K f + U f
Solution: 1. (a) Set Ei = Ef and solve for d:
1
2
( m1 + m2 ) v 2 + 0 = 0 + m2 gd
( m1 + m2 ) v 2
d=
2m2 g
2. (b) The rope pulls upward on m2 , the same direction it is displaced, so we conclude the work done by the rope on
m2 is positive.
3. (c) Set the work done by the rope equal to the
change in mechanical energy for m2 . The mass m2
2
starts out with kinetic energy 12 m2 v and finishes
with gravitational potential energy m2 gd :
Wnc = E2,f − E2,i
= m2 gd − 12 m2 v 2
 ( m + m2 ) v 2  1
2
Wnc = m2 g  1
 − 2 m2 v
 2m2 g

=
1
2
( m1 + m2 ) v 2 − 12 m2 v 2 =
1
2
m1v 2
Insight: The nonconservative work done by the rope is positive, so m2 gains mechanical energy during this episode.
However, the conservative work done by gravity is negative, because m2 gains potential energy (equation 8-1,
Wc = −∆U ). Therefore we conclude that the gain in potential energy for m2 is greater than its loss of kinetic energy.