1) Answer: D
There is no work done because there is no motion
2) Answer: E
No work because no motion
3) Answer: C
The sign on work is based on whether it is in the same direction or opposite the direction of work.
5) Answer: C
Work is opposite the direction of motion on the way up and in the same direction as motion on the way.
7) Answer: A
KT=4m(2v)2KC
8) Answer: A
WFRICTION = ΔK  μmgd=1/2 mv2- 1/2mvo2 masses cancel out.
9) Answer: B
WFRICTION = μmgd Looking at the equation, that car with the greatest mass will have the greatest work
done by friction on it.
11) Answer: A
WFRICTION = ΔK  μmgd=1/2 mv2  mass cancels so μgd=1/2 v2 so distance is the only variable that
can chance based on velocity
12) Answer: B
ΔK is greater for 50 60, so work is greater..
13) Answer: D
Ug =mgΔy  4mgΔy will give you 4Ug. There is a directly proportional relationship.
14) Answer: D
Mass, gravity and height are the same for both, therefore the Ug is the same for both.
15) Answer: B
F=kx, there is a directly proportional relationship between force and change in length of the spring.
16) Answer: A, B, E
A – because mass has no effect on Us  so both will have the same K.
B – because the amount of force applied to both is the same, but the lighter one will have a greater
acceleration because the mass is less  Fnet=ma
E – Mass has no effect on US
17) Answer: E
h = v2  2=v2  v=√2
18) Answer: C
h = v2  h=22  h=4
19) Answer: B
Ug = K  Double Δy = double Ug = double K because Conservation of Energy
20) Answer: B
Lighter
Heavier
2
K=1/2 mv
K=1/2(2m)v2
21) Answer: C
Ug=K  mg Δy = ½ mv2 Mass cancels so it has no effect on the situations.
22) Answer: D
Ug=K  mg Δy = ½ mv2 Mass cancels so it has no effect on the situations.
23) Answer: E
Ug=K  mg Δy = ½ mv2 Mass cancels so because the velocities are the same, the Δy is the same.
24) Answer: D
With kinetic energy, direction has no affect. K+Ug=K For both, since all equal at the top (height and
velocity), speed at the bottom is equal.
25) Answer: A
The relationship is v2= Δy, (22)= Δy
26) Answer: D
Ug=K  They all start with the same height so they all end with the same velocity.
28) Answer: B, D
B) In order to have the same K the frog with less mass must have a greater velocity
D) The light frog has a greater velocity, therefore it has a greater height.
29) Answer: C, D, E
C) They started with the same K so they have the same overall M.E.
D) K=Ug+K  Grasshopper B has K at the top
E) K=Ug  Grasshopper A goes straight up, so all its initial K is converted to Ug.
32) Answer: B, D
B) Ug=K in order for their K to be the same, the velocity of the smaller mass must be greater.
D) Ug=K  mg Δy=1/2mv2  masses have no affect and since height is the same, K is the same.
33) Answer: B, D
B) The spring is applying the same amount of force to both, giving a greater acceleration to the lighter
dart.
D) Us=Ug for both. Lighter dart has to have a greater Δy to compensate for lass mass.
34) Answer: A, D
A)WF=ΔK  μmgd= -1/2mvo2  mass cancels so it has no effect on how far an object will go.
D) WF=μmgd  In order for the heavier sled to stop in the same distance there must be more work done
on the heavier sled, because it has greater inertia.
35) Answer: C, D, E
C) The lighter sled has a greater velocity, in order to have an equal amount of K, so it will go further.
D) They both start with the same K and they both come to rest so that both have zero K at the end.
E) Because the change in K is the same, the amount of work is the same.
38) Answer: A
Work is only based on force and overall displacement and since they weigh the same and the
displacement (how high up they go) is the same, they are doing the same amount of work. Since Joe
does this work in less time he has a greater amount of power.
39) Answer: E
P= W/t  (2W)/(.5t) = 4 times the amount of power
40) Answer: A
6xF=6xW
(6W)/(.5t) = 12 times the power