Page 1 of 6
CTWork-0
An atomic bomb explodes. Is energy conserved in the nuclear reactions of the bomb?
A) Yes
B) No
Answer: Yes. Energy is always conserved.
Is matter (i.e. mass) conserved?
A) Yes
B) No
Answer: No, mass is not conserved. Mass is a form of energy and can be converted into
other forms of energy. In a chemical reaction, the total mass of the products is less than
the mass of the reactants by about 1 part in a billion – too small to measure, but there is a
mass loss.
CTWork-00
The position of a particle is given by the position vector at two different times. The initial
position vector is r1 ; the final position vector is r2 , as shown. Which arrow best represents the
vector r ?
y
A)
r1
r2
B)
x
C)
Answer: B. Notice that the vector r does NOT depend on my choice of coordinate
system. The vectors r1 and r2 do depend on my choice of coordinate system (where I put
the origin), but r does NOT.
CTWork-1. Albert Einstein lowers a book of mass m downward a distance h at constant speed
v. The work done by …
A) +
B) 
C) 0
the force of Albert's hand is.. A) +
B) 
C) 0
the net force on the book is .. A) +
B) 
C) 0
the force of gravity is..
Page 2 of 6
Answers: Wgrav + , Whand  , Wnet = 0
CTWork-2. A rock of mass m is twirled on a string in a horizontal plane. The work done by
the tension in the string on the rock is..
A) +
B) 
C) 0
Answer: Wtension = 0 , since in any small(infinitesimal) time interval, the tension force is
perpendicular to the displacement.
Page 3 of 6
CTWork-3. A 1 kg mass is moved part way around a square loop as shown. The square is 1
meter on a side and the final position of the mass is 0.5 m below its original position. Assume
that g = 10 m/s2. What is the work done by the force of gravity during this journey?
m g h = (1 kg) (10 m/s2) (h)
Up
0.5m
1m
start
finish
A) 10 J
B) 5 J
C) 0 J
D) –10 J
E) –5 J
Answer: +5 J
If instead of moving part way around a square, the mass where taken on a long and tortuous
journey to the Moon, Tibet, and Lithuania and then returned to the same finish point as before,
would the total work done by gravity be the ..
A) same, or
B) different.
Answer: The same. The total work done by gravity is independent of the path taken
between the start and finish.
Page 4 of 6
CTWork-4. A projectile is fired upward with an initial
speed v on an airless world. A short time later, it comes
back down and has a final speed v (just before it hits the
ground). What was the sign of the total work done by the
force of gravity during the flight?
v
A) +
B) –
C) total work = 0
v
Answer: Zero. Two ways to see this:
Method I. The negative work done by gravity on the way up is –mgh, and this is
cancelled by the positive work done by gravity +mgh, on the way down.
Method II. By the Work-KE Theorem, Wnet = KE = 0. Here the net force is the force
of gravity, so Wgrav = 0.
CTWork-5. A projectile is fired upward through air with
an initial speed v0. A short time later, it comes back down
and has a final speed v < v0 . Air resistance is NOT
negligible. What was the sign of the work done by friction
during the flight?
A) +
B) –
C) work done = 0
v0
v
Answer: negative. The direction of the force of friction is always opposite the direction
of motion, both on the way up and on the way down. So the work done is negative.
CTWork-6. The work done by friction WFRIC during the flight is
A) KE = KEf – KEi
B) – KE
C) Neither of these.
Answer: KE. Work-KE Theorem says that Wnet = KE. In this situation, Wnet =
Wgrav + Wfric , and Wgrav = 0 so Wnet = Wfric = KE.
Page 5 of 6
CTWork-7. Two marbles, one twice as heavy as the other, are dropped to the ground from the
roof of a building. (Assume no air resistance.) Just before hitting the ground, the heavier marble
has..
A) as much kinetic energy as the lighter one.
B) twice as much kinetic energy as the lighter one.
C) half as much kinetic energy as the lighter one.
D) four times as much kinetic energy as the lighter one.
E) impossible to determine.
Answer: ..twice as much kinetic energy as the lighter one. (They both have the same
speed on impact).
CTWork-8. A car moves of mass m moving with speed v suddenly applies the brakes and slows
to a stop. Suppose I ask "What was the work done on the car by the force of friction?" Do you
have enough information to answer the question?
A) Yes
B) No
Answer: Yes. In this case, the net force is the force of friction , since the other forces
(normal force and gravity) cancel. Work-KE Theorem says Wnet = = KE. So Wnet =
Wfric = KEf – KEi = – (1/2)mv2.
CTWork-9. One end of a spring is fixed in place. I stretch the spring by pulling on the other end.
What is the sign of the work done by the force from my hand?
A) positive B) negative
C) zero
D) answer depends on the direction in which I pull
E) answer depends on the coordinate system I choose
Answer: positive. No matter which way I pull, the force I exert is in the same direction as the
displacement. When force and displacement are in the same direction, the work done is positive.
Page 6 of 6
CTWork-10. A force is applied in the x-direction while an object moves a distance x along the
x-axis.
F = F(x)
x
The force varies with position according to this graph:
F
2F0
F0
x
x /2
0
The work done by this force is
A)0.5 F0 x
B) F0x
C) 1.5F0x
D) 2F0x
E) 2.5F0x
Answer: 1.5F0x (which is equal to Favg x )
Now suppose the force varies with position according to this graph:
F
F0
0
x
The work done by the force is
A) 0.5 F0 x
B) F0x
C) 1.5F0x
D) None of these
Answer: 0.5 F0 x (which is equal to Favg x ) Notice that the work done is the area under
the curve, both in this problem and in the previous one.