Mr. Sarver
Exercises
29
Wme = Fme dme
Wf = Ff df = 21 Fme 4dme = 2Wme
Your friend does twice as much work.
35
To have the same momentum, the golfball must move faster than the baseball. Multiplying
by velocity again means that the golfball will have the greater energy. See number 53 for a
demonstration.
36
The greatest kinetic energy of the bob of a pendulum is at the bottom of the swing. Potiental
energy is greatest at either end of the swing. As energy can not be created or destroyed,
only moved around, the maximum kinetic energy is equal to the maximum potiental energy,
asuming potiential energy is measured realtive to the bottom of the swing. When half of the
energy is in kinetic energy, then the other half of the energy is in potiential energy. Potiental
energy is half of the maximum value.
38
Work can only be done when the object moves in a direction that is partially in the direction
of the force. When the car moves down a slope, gravity does work on it because the car
moved down in the gravity field. When a car moves on level ground, the elevation of the car
does not change and so gravity does no work on it.
46
The ball will return to the same level with less kinetic energy then when it left. This does
not contridict the law of conservation of energy as the air did work on the ball to slow it
down. That energy went into heating the air.
47
Both balls hit the ground with the same speed. When the first ball goes up and comes
back down, it will have same velocity that the second ball did when it passes through the
same elevation. Calling on conservation of energy: The kinetic energy must be same when
it comes down as when it goes up. The velocity just changed direction. The speed must be
the same as the other ball as it went down. Now that they have same velocity at different
times, they will strike the ground at the same speed.
I also this answer that I liked: When the balls leave your hands, they have the same total
energy. Due to conservation of energy, they will have the same energy when they hit ground
and so will hit with the same velocity.
1
53
Momentum:
p1 = m1 v1 = 1kg × 2 ms = 2 kg×m
s
p2 = m2 v2 = 2kg × 1 ms = 2 kg×m
s
Neither has greater momentum.
Kinetic Energy:
KE1 = 12 m1 v12 = 12 1kg × (2 ms )2 = 2J
KE2 = 12 m2 v22 = 12 2kg × (1 ms )2 = 1J
The lighter object has more kinetic energy.
55
Momentum is conserved. They both have the same momentum but because they are opposite, the total momentum is zero. The final momentum is also zero as it is zero for both
lumps. Thus momentum is conserved. Kinentic energy is not conserved because it was nonzero at the start and zero at the finish. Energy on the other hand is conserved. The kinetic
energy was used to deform the two lumps of clay.
58
They have a similar use of the word ”conservation” in mind. They both use the word in the
sense that the energy is not ”lost” (definition 3 of ”conserve” on wolframalpha.com). The
practice of energy conservation is to prevent the loss of energy that is useful to man. The
law of energy conservation is that energy is never lost to the universe, it is always there.
Problems
10
W = (ma)d = 300kg × 10 sm2 × 4m = 12000kg sm2 m = 12000J
11a
E = KE1 + P E1 = KE2 + P E2
10000J = 0J + 10000J = KE2 + 1000J
10000J − 1000J = 9000J = KE2
11b
P E1 = 10000J = mgh1
P E2 = 1000J = mgh2
P E2
1000J
1
2
= 10000J
= hh21 10
= mgh
= h12 ⇒ h2 = .1
P E1
mgh1
He will be at one tenth his original height.
22a
w = mg ⇒ m =
p = mv = wv
g
w
g
2
22b
KE = 12 mv 2 =
22c
KE =
KE =
KE =
wv 2
2g
wv 2
2g
1.5N (38 m
)2
s
2×9.81 m2
s
)2
1.5N (38 m
s
m
2×9.81 2
s
2 2
m s
= 110 Nms
= 110N m = 110J
2
24a
KEf = P Ei = mgh
h = x0 + v0 t + 21 gt2
x0 = 0m, v0 = 0 ms
h = 12 gt2
KEf = mg 21 gt2 = 12 mg 2 t2
24b
KEf = 21 mg 2 t2 = 12 1.2kg(9.8 sm2 )2 (2s)2
2
KEf = 21 1.2kg(9.8 sm2 )2 (2s)2 = 230kg ms4 s2 = 230J
24c
The time over the action of the impact force is not given. If you worked at, there is a displacement that the force works over. That isn’t provided either and would be much harder
to find. Both could render a result.
25a
KEf = 21 mvf2 = P Ei = mghi
1
mvf2 = mghi
2
2
vf = 2ghi
√
vf = 2gh
25b
q
p
2
vf = 2 × 9.8 sm2 1.5m = 5.4 ms2 = 5.4 ms
26a
P = δE
= Fδtδx = F~ · ~v
δt
F = mg
P
P = F v ⇒ v = PF = mg
26b
100kW
10 m2 900kg
s
×
1kN
1000N
=
100kN m
s
9kN
≈ 11 ms
3