Name: ________________________ Class: ___________________ Date: __________ ID: A

AP Physics C - Conservation of Energy Practice Test (Multiple Choice Section)

1. A weight lifter lifts a mass m at constant speed to a height h in time t. What is the average power output
of the weight lifter?

a. mg b. mh c. mgh d. mght e. mgh/t

2. The following graphs, all drawn to the same scale, represent the net force F as a function of displacement
x for an object that moves along a straight line. Which graph represents the force that will cause the
greatest change in the kinetic energy of the object from x = 0 to x = x1?

a. c. e.

b. d.

Questions 3–4: A 2 kg
block, starting from rest,
slides 20 m down a
frictionless inclined plane
from X to Y, dropping a
vertical distance of 10 m
as shown above.

3. The magnitude of the net force on the block while it is sliding is most nearly

a. 0.1 N b. 0.4 N c. 2.5 N d. 5.0 N e. 10.0 N

4. The speed of the block at point Y is most nearly

a. 7 m/s b. 10 m/s c. 14 m/s d. 20 m/s e. 100 m/s

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Name: ________________________ ID: A

5. When 800 J of work are done on an object, its potential energy increases by 300 J and its kinetic energy
increases by 400 J. What was the work done against friction?

a. 100 J b. 700 J c. 800 J d. 1100 J e. 1500 J

6. If the kinetic energy of a given mass is to be doubled, its speed must be multiplied by

a. 8 b. 2 c. 2 d. 4 e. ½

7.
The diagram shows points A, B, and
C at or near Earth’s surface. As a
mass is moved from A to B, 100 J of
work are done against gravity. What
is the amount of work done against
gravity as an identical mass is moved
from A to C?

a. 100 J b. 150 J c. 173 J d. 200 J e. 350 J

8. For a block to slide without friction up a rise of height h, the minimum initial speed of the block must be:

a. 1
2 gh b. 1
2 gh c. 2 gh d. 2 gh e. 2 2 gh

9. A person pushes a box across a horizontal surface at a constant speed of 0.5 m/s. The box has a mass of
40 kg, and the coefficient of sliding friction is 0.25. The power supplied to the box by the person is

a. 0.2 W b. 5W c. 50 W d. 100 W e. 200 W

10. A block of mass m slides on a horizontal frictionless table with an initial speed v0. It then compresses a
spring of force constant k and is brought to rest. How much is the spring compressed from its natural
length?
2
v0 mg m m k
a. b. c. v0 d. v0 e. v0
2g k k k m

11. An example of a nonconservative force is

a. Gravitational Force c. Friction Force e. Magnetic Force

b. Elastic (spring) Force d. Electrostatic Force

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Name: ________________________ ID: A

12. A particle of mass 3.00 kg is projected vertically upward with an initial velocity of 20.0 m/s. Neglecting
air resistance, the particle’s kinetic energy when it is 20 m above the initial position is:

a. 1200 J b. 600 J c. 300 J d. 900 J e. 0J

13. What is the work done by friction on a 15 kg object pulled horizontally in a straight line for 15 meters, if
the coefficient of friction between the object and the surface is given by µk = 0.4?

a. –60 J b. –90 J c. –150 J d. –600 J e. –900 J

14.
A small block is released from a
height of 12 m as shown. Assuming a
frictionless track, compute the
maximum value of R in order for the
block to go around the loop without
losing contact with the track.

a. 12 m b. 8m c. 6m d. 4.8 m e. 4m

15. An object of mass m is traveling at constant speed v in a circular path of radius r. How much work is
done by the centripetal force during one-half of a revolution?

a. πmv2 b. 2πmv2 c. 0 d. πmv2r e. 2πmv2r

16.
Consider a region where the gravitational field of 10
N/kg is uniform and points in the –y direction. A 2 kg
object at the origin O is moved along the path y = x 2 to
point A ( x = 2 m, y = 4 m ) . The object is then moved
along the path ( x − 2 ) = ( y − 4 ) to the point B
2

(x = 6 m, y = 6 m ) , as shown. How much work is done

by gravity as the object is moved from O to B?

a. –120 J b. –80 J c. 0J d. 80 J e. 120 J

17. If L, M, and T denote the dimensions of length, mass, and time, respectively, what are the dimensions of
power?

a. ML/T2 b. L2M/T2 c. M2L/T2 d. L2M/T3 e. M2L/T3

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Name: ________________________ ID: A

18. As a rock of mass 4 kg drops from the edge of a 40-meter-high cliff, it experiences air resistance, whose
average strength during the descent is 20 N. At what speed will the rock hit the ground?

a. 8 m/s b. 10 m/s c. 12 m/s d. 16 m/s e. 20 m/s

19. A ball of mass M compresses a spring (with a constant k) a distance of 0.3 m and released. What is its
speed the instant after the ball passes the equilibrium point?

a. k b. c. d. 0 .3 k e. Not enough
0.3 0.3k M 0 .3 M k
M M info

20. A 45 kg object slides down an uneven frictionless incline from rest without rotating. What is its speed
after it travels 2.5 meters vertically?

a. 3.0 m/s b. 4.3 m/s c. 5.7 m/s d. 7.1 m/s e. 11 m/s

21.
A 0.50-kg steel ball is suspended from the ceiling by a
2.5-m-long string. The ball is pulled to the side until it is
1.0 m above its lowest point, and then released from rest.
The tension in the string when the ball passes through its
lowest position is:

a. 9N b. 5N c. 12 N d. 15 N e. 18 N

22.
In the figure shown, a small ball slides
down a frictionless quarter-circular slide
of radius R. If the ball starts from rest at a
height equal to 2R above a horizontal
surface, find its horizontal displacement, x,
at the moment it strikes the surface.

a. 2R b. 5R/2 c. 3R d. 7R/2 e. 4R

23. What is the average power necessary to move a 40 kg block up a frictionless 30o incline at 5 m/s?

a. 80 W b. 140 W c. 400 W d. 500 W e. 1000 W

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Name: ________________________ ID: A

24.
The 2.0 kg block shown to the right is released
from rest on the frictionless 30o incline. After
sliding 0.80 m it comes in contact with the spring
and, after sliding an additional 0.20 m, it comes
momentarily to rest. The spring constant is

a. 300 N/m b. 400 N/m c. 500 N/m d. 800 N/m e. 1000 N/m

25. A force of 200 N is required to keep an object sliding at a constant speed of 2 m/s across a rough floor.
How much power is being expended to maintain this motion?

a. 50 W c. 200 W e. Cannot be determined

b. 100 W d. 400 W

26. An engine provides 10 kW of power to lift a heavy load at constant velocity a distance from 20 m in 5 s.
What is the mass of the object being lifted?

a. 100 kg b. 150 kg c. 200 kg d. 250 kg e. 500 kg

27. An object falls freely from rest near Earth’s surface. Which graph best represents the relationship
between the object’s kinetic energy and its time of fall?

a. b. c. d. e.

28. A 5.00-kg cart is moving at 6.00 m/s. To increase its speed to 10.0 m/s requires that the work done on
the cart be:

a. 40 J b. 90 J c. 160 J d. 400 J e. 550 J

29.
A man pulls an 90-N crate up a frictionless 30o
slope 4 m high. If the crate is moved from rest at
the bottom to rest at the top of the slope, the work
done by the man on the crate is:

a. +360 J b. +180 J c. Zero d. –180 J e. –360 J

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Name: ________________________ ID: A

Questions 30–32: The graph is taken

from an experiment with a spring of
original length 1.0 m.

30. What is the spring constant of this spring?

a. 0.05 N/m b. 0.5 N/m c. 5 N/m d. 50 N/m e. 500 N/m

31. How much force must be applied to this spring to change its length to 1.025 m?

a. 1.025 N b. 6.0 N c. 10.0 N d. 12.5 N e. 512.5 N

32. What is the potential energy stored in the spring when it has been stretched 0.04 m from its original
length?

a. 0.4 J b. 0.8 J c. 40 J d. 80 J e. 100 J

33. A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is

perpendicular to the 4 m displacement, the work it does is equal to

a. 0J b. 60 J c. 80 J d. 600 J e. 2400 J

34. A particle is subjected to a conservative force whose potential energy function is

U ( x ) = ( x − 2 ) − 12 x
3

where U is given in joules when x is measured in meters. Which of the following represents a position of
stable equilibrium?

a. x = –4 b. x = –2 c. x=0 d. x=2 e. x=4

35. A 10 kg body is constrained to move along the x-axis. The potential energy U of the body in joules is
given as a function of its position x in meters by
U(x) = 6x2 – 4x + 3
The force on the particle at x = 3 meters is

a. 32 N in +x direction c. 45 N in +x direction e. 98 N in +x direction

b. 32 N in –x direction d. 45 N in –x direction

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Name: ________________________ ID: A

36. The potential energy of a body of mass m is given by − mgx + 12 kx 2 . The force acting on the body at
position x is

mgx 2 kx3
a. − + c. − mg + 12 kx e. mg − kx
2 6
2 3
mgx kx
b. − d. − mg + kx
2 6

37. A particle is acted upon by a net force F = F0 e − kx . If the particle has velocity equal to zero at x = 0, then
the maximum kinetic energy that the particle can attain is

F0 F0
a. b. c. kF0 d. (kF0 )2 e. kek F0
k ek

Questions 38–40

A plane 5 m in length is inclined at an angle of

37°, as shown. A block of weight 20 N is placed
at the top of the plane and allowed to slide down.

38. The mass of the block is most nearly

a. 1.0 kg b. 1.2 kg c. 1.6 kg d. 2.0 kg e. 2.5 kg

39. The magnitude of the normal force exerted on the block by the plane is most nearly

a. 10 N b. 12 N c. 16 N d. 20 N e. 33 N

40. The work done on the block by the gravitational force during the 5 m slide down the plane is most nearly

a. 20 J b. 60 J c. 80 J d. 100 J e. 130 J

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AP Physics C - Conservation of Energy Practice Test (Multiple Choice Section)

Answer Section

MULTIPLE CHOICE

1. E
Power is work over time. The work done to raise an object (against gravity) is mgh (the change in
gravitational potential energy, where m is the mass of the object and h is the height it is raised).
Therefore, work over time is mgh/t.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2. E
“Change in kinetic energy” is the long way of saying “work”. Work is the area under a force vs. distance
graph, and so the greatest work is done when the area under the graph is greatest.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
3. E
The “down the plane” force is mgsinθ. Set this equal to ma:
ma = mg sin θ Æ a = g sin θ
(We know that sinθ = 10/20 because we are given the opposite and hypotenuse.)
a = (10 )( 10 )
20 = 5 m/s
2

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
4. C
The gravitational potential energy becomes kinetic energy:
mgh = 12 mv 2 Æ v 2 = 2 gh = 2(10 )(10 ) Æ v = 200 = 14 .1 m/s
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
5. A
The 800 J of work must all go into some form of energy. There are 700 J accounted for by increases in
kinetic and potential energy, so the remaining 100 J must go to frictional heating.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
6. C
K = 12 m v{2 , so the speed is multiplied by
{ 2.
(×2 ) (× 2 )2
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
7. A
Work done against gravity is W = mgh. Point C has the same height as point B, so to move an object
from A to either B or C requires the same amount of work. Gravity is a conservative force, so the work
done by gravity only depends on the starting and ending point, not the path taken in between.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
8. C
For a block to rise a height h, it must have enough kinetic energy to turn into potential energy to rise the
height h:
K i = U f Æ 12 mv 2 = mgh Æ v 2 = 2 gh Æ v = 2 gh
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

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9. C
The surface is flat, so the normal force and weight force are equal and are 400 N (mg). The friction force
is F f = µFN = (0.25 )(400 ) = 100 N. Power is force•velocity, so multiply the 100 N force to overcome
friction with the 0.5 m/s to get 50 W.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
10. D
The initial energy is all kinetic energy. When the block comes to rest, all of its kinetic energy is
transferred to the spring’s potential energy:
2
2 mv0 = 2 kx
1 1 2

2
mv 0
x2 =
k
m
x = v0 (the squared velocity can be taken out of the radical)
k
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
11. C
A non-conservative force is one which cannot store potential energy. Friction does not convert energy
into a form of potential energy, but instead into thermal energy.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
12. E
Ki + Ui = K f + U f
2
1
2 mvi + 0 = K f + mgh f
1
2
(3)(20)2 + 0 = K f + (3)(10)(20)
600 = K f + 600
The final kinetic energy is zero—this is the maximum height the projectile will reach.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
13. E
The normal force equals the weight force (mg = 150 N) since the surface is assumed flat. The friction
force is µFN = (0.4)(150) = 60 N. The work is this force times distance which is (60)(15) = 900 J. The
work done by friction is negative because friction is acting in the direction opposite the displacement.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
14. D
At the maximum radius, the block barely stays in contact with the track at the top of the loop. This
makes the normal force at the top of the loop nearly zero:
2 2
mvtop mvtop
FN + mg = Æ mg = (weight and acceleration are both down)
r r
2
gR = vtop
Now set up an energy equation. The height at the top of the loop is 2R:
mgh = 12 mvtop + mg (2 R ) Æ mgh = 12 m( gR) + mg (2 R ) Æ mgh = 52 mgR
2

2
5 h=R (two fifths of 12 is 4.8 m)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

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15. C
No work is ever done on an object that is in uniform circular motion since the force (toward the center of
the circle) is perpendicular to the displacement (tangent to the circle) at all times.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
16. A
The work done by gravity is the same as the change in gravitational potential energy. Because the object
rises 6 m, the potential energy increases U = mgh = (2)(10)(6) = 120 J. However, the displacement (up)
is opposite the force of gravity (down), so the work is negative (also, it could be said that “the object
does the opposite of what gravity wants, so gravity does negative work”).
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
17. D
[ ]
F = ma , so the units of force are [kg ] m s 2 = kg ⋅ m/s 2 .
[ ]
P = Fv , so the units of power are kg ⋅ m/s 2 [m s ] = kg ⋅ m 2 /s 3 , which is (length2•mass/time3)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
18. E
Using conservation of energy with an energy-lost-to-friction term. The energy lost is the friction force
times the distance fallen:
mgh = 12 mv 2 + F f h Æ (4 )(10 )(40 ) = 12 (4 )v 2 + (20 )(40 ) Æ 800 = 2v 2
v = 20 m/s
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
19. A
Set up conservation of energy, where spring potential energy becomes kinetic energy:
2 kx = 2 mv
1 2 1 2

k 2
x = v 2 (cancel the ½ factors and solve for v)
m
k k
v=x = 0.3 (solving for v, taking the x2 out of the radical and plugging in 0.3)
m M
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
20. D
Ui = K f Æ mgh = 12 mv 2 Æ 2 gh = v 2 = 2(10 )(2.5) = 50
The square root of 50 is just a little above 7, since the square root of 49 is 7.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
21. A
Use energy to determine the speed of the ball at the bottom of the swing:
mgh = 12 mv 2 Æ mv 2 = 2mgh
Now employ circular motion concepts. The acceleration is centripetal and upward (toward the center of
the circle). Weight is down (against acceleration), but tension is up.
mv 2
FT − mg = (plug in mv2 from above and solve for tension)
r
2mgh 2(0.5)(10 )(1.0 )
FT = + mg Æ FT = + (0.5)(10 ) Æ FT = 4 + 5 = 9 N
r (2.5)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

3
 ID: A

22. C
As the ball slides down the quarter-circle, its height and PE decrease. This decreased potential energy
becomes kinetic energy:
2 mv = mg∆h
1 2
Æ plug in ∆h = R and solve Æ v = 2 gR
This is the horizontal speed with which the ball leaves the incline. Now use the horizontal-launch
projectile motion equations:
∆x = vt ∆y = 12 gt 2
2R
Apply ∆y = R to the vertical equation: R = 12 gt 2 Æ t=
g
2R
Plug time into the horizontal equation: ∆x = 2 gR = 2R
g
But there’s also another horizontal R from the quarter circle, for a grand total of x = 3R.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
23. E
If the speed of the block is constant (and the incline is frictionless), the force to push it up the plane is
equal to the “force down-the-plane” F|| = mg sin θ . Power is P = Fv . Putting this together:
P = F||v = mgv sin θ
P = (40 )(10 )(5)( 12 ) = 1000 W
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
24. C
Use energy. The initial energy is all gravitational potential energy, and the final energy is all spring
potential energy:
mgh = 12 kx 2
The block slides a total distance of 1 m (hypotenuse), but on a 30o incline, the height change is 0.5 m (the
opposite, using sin30o = ½). The spring compression is x = 0.2 m.
(2 )(10 )(0.5) = 12 k (0.2)2 Æ 20 = k (15 )2 Æ k = 500 N/m
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
25. D
Power = Force × Velocity = (200)(2) = 400 W
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
26. D
Power is work/time, and the work required to lift a mass is equal to its gain in potential energy mgh:
mgh m(10 )(20 )
P= Æ (10000 ) = Æ m = 250 kg
t (5)
27. D
If an object falls from rest, then v = gt (constant acceleration). Kinetic energy is K = 12 mv 2 = 12 m( gt ) .
2

Note that time is squared, so the graph must be a parabola.

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28. C
The work done to change the speed of an object is the change in the object’s kinetic energy:
2 2
W = ∆K = 12 mv f − 12 mvi
W = 12 (5)(10) − 12 (5)(6)
2 2

W = 250 − 90 = 160 J
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29. A
The slope is frictionless, and the crate starts and ends at rest, so the only work done by the man is to
increase the block’s potential energy.
W = mgh (mg = 90 N because the weight is given, not the mass, and h = 4 m)
W = (90 )(4 ) = 360 J
The force is positive because the man’s force was directed up the plane, and the block does indeed move
up the plane.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
30. E
The slope of a force vs. stretch graph is the spring constant. This slope is (5)/(0.01) = 500 N/m.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
31. D
If the spring’s length goes from 1 m to 1.025 m, the stretch is 0.025 m. Looking between 0.02 and 0.03
m on the graph, the sensible answer would be 12.5 N.
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32. A
Potential energy stored in the spring is the area under the graph of force vs. stretch. Using the triangle of
base 0.04 and height 20, the area is 0.1 J.
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33. A
No work is done if the force is perpendicular to the displacement.
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34. E
An equilibrium is a position where the derivative of potential energy is zero, and the second derivative of
potential energy is positive:
U = ( x − 2 ) − 12 x
3

dU d 2U
= 3( x − 2) ({ 1) − 12 = 0 and = 6( x − 2) ({ 1) > 0
2 1

dx 1 424 3 dx 2 1424 3
Power Rule Chain Power Rule Chain
Rule Rule

Solve the first equation: ( x − 2 ) = 4 , so x = 0 or x = +4.

2

Only x = +4 makes the second derivative greater than zero.

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35. B
The force on a particle is the negative derivative of the potential energy:
dU dU
F =− and = 12 x − 4
dx dx
F = −12 x + 4 , so at x = 3 m, F = –32 N
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36. E
dU
F =−
dx
F =−
d
dx
[ ]
− mgx + 12 kx 2
F = mg − kx (looks like an object hung from a vertical spring)
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37. A
The maximum kinetic energy attained by the particle is the work done over the infinite amount of
distance that can be traveled by the particle:
xf ∞
W = ∫ F • dx Æ W = ∫ F0 e − kx • dx
xi 0
∞
⎡ F ⎤ ⎡ F ⎤ ⎡ F ⎤
W = ⎢− 0 e −kx ⎥ Æ W = ⎢− 0 e −(∞ ) x ⎥ − ⎢− 0 e −(0 )x ⎥ (remember that e–∞ = 0 and e0 = 1)
⎣ k ⎦0 ⎣ k ⎦ ⎣ k ⎦
⎡ F ⎤ F
W = [0] − ⎢− 0 (1)⎥ = 0
⎣ k ⎦ k
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38. D
The weight of the block is 20 N and the equation for weight is FW = mg. Since g = 10 m/s2, (20) = m(10)
and m = 2 kg.
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39. C
The normal force on an object on an inclined plane is given by FN = mgcosθ or FN = FW cosθ. In this
case FW = 20 N and cosθ = 4/5 (adjacent/hypotenuse read right off of the diagram). Therefore FN =
(20)(4/5) = 16 N.
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40. B
“Work done by gravity” is the same as the change in gravitational potential energy:
U = (mg)(h) = (20)(3) = 60 J
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OTHER