Download Conservation of Energy

Page 1 of 12
CTEnergy-1. A mass m is at the end of light (massless) rod of length R, the other end of which
has a frictionless pivot so the rod can swing in a vertical plane. The rod is initially horizontal
and the mass is pushed down with an initial speed vo . What initial kinetic energy is required for
the mass to pivot 270o to the vertical position?
m
R
vo
KEINITIAL =
1
mv o2  A) mgR
2
B) mg(2R)
C) mg(3R)
D) None of these
Answer: mgR Begin by writing KEi + PEi = KEf + PEf . You get the same answer
regardless of where you place the zero of PE.
CTEnergy-2. A projectile is fired with an initial speed vo at an angle  from the horizontal.
What is the KE of the projectile when it is on the way down at a height h above the ground?
(Assume no air resistance.)
KE = ?
vo
h

A) (1/2)mvo2 + mgh
B) mgh
C) (1/2)mvo2 – mgh
D) Impossible to tell
Answer: (1/2)mvo2 – mgh. Begin by writing KEi + PEi = KEf + PEf
Page 2 of 12
CTEnergy-3. A block initially at rest is allowed to slide down a frictionless ramp and attains a
speed v at the bottom. To achieve a speed 2v at the bottom, how many times higher must the
new ramp be?
h
A)
v
2  14
.
B) 2
C) 3
D) 4
E) none of these.
Answer: 4
CTEnergy-4. A mass slides down a rough ramp (with friction) of height h. Its initial speed is
zero. Its final speed at the bottom of the ramp is v.
vi = 0
v
While the mass is descending , its KE A) increases
B) decreases
C) remains constant.
While the mass is descending, its PE A) increases
B) decreases
C) remains constant.
While the mass is descending, its (KE + PE) = total mechanical energy
A) increases
B) decreases
C) remains constant.
Answers: The KE increases, the PE decreases, and the (KE+PE) decreases because some
of the mechanical energy is converted into thermal energy.
Page 3 of 12
CTEnergy-5. What is the work done by the two forces together in moving the object a distance x
to the right as shown in the diagram? The magnitude of each force is F and the directions of the
two forces are 90o apart.
A) 2Fx


D) 1  2 F x E) None of these.
C) 2 F x
B) Fx
F
F
x
Answer: Fx
The vertical force does no work.
CTEnergy-6. A mass is oscillating back and forth on a spring as shown. Position 0 is the
unstretched position of the mass.
E
0
M
At which point is the magnitude of the force on the mass maximum?
A) 0 B) M C) E
At which point is the magnitude of the acceleration of the mass maximum?
At which point is the elastic potential energy maximum?
A) 0
+E
B) M
A) 0 B) M C) E
C) E
Answers: force is max at E, acceleration is max at E, PE is max at E.
CTEnergy-7. A spring-loaded dart gun shoots a dart straight up into the air, and the dart reaches
a maximum height of h. The same dart is shot straight up a second time from the same gun, but
this time the spring is compressed twice as much before firing. How far up does the dart go this
time, neglecting friction and assuming an ideal spring?
A) 2h
B) 4h
Answer: 4h
C) 8h
D) h/4
E) h/2
Page 4 of 12
CTEnergy-8. A spring-loaded gun fires a dart at an unknown angle  above the horizontal. No
air resistance. The spring constant k and the initial compression x of the spring are known. Can
you solve for the maximum height h above the starting point using Conservation of Energy?
A)Yes
B) No
h=?
k, x

Answer: No. The KE at the top of the trajectory is unknown, so you can’t solve for h in
terms of known quantities.
CTEnergy-9. A hockey puck slides without friction along a frozen lake toward an ice ramp and
plateau as shown. The speed of the puck is 4m/s and the height of the plateau is 1m. Will the
puck make it all the way up the ramp?
v = 4m/s
A) Yes
B) No
h = 1m
C) impossible to determine from the info given
Answer: No. The initial KE is (1/2)mv2 = 8m (in SI units). The PE at the top is mgh =
9.8m (in SI units). There is not enough initial KE to convert into PE.
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CTEnergy-10. A small mass, starting at rest, slides without friction down a rail to a loop-de-loop
as shown. The maximum height of the loop is the same as the initial height of the mass.
Will the ball make it to the top of the loop?
A) Yes, the ball will make it to the top of the loop.
B) No, the ball will not make it to the top.
C) Not enough info to say
Answer: This was a tough one! No, the ball will not make it to the top. The ball needs to
have significant KE at the top of the loop in order to remain in contact with the rail. If,
instead of an open rail, the ball was in a tube, as shown below, then the ball would make
it to the top.
Same question, but now the loop-de-loop track is replaced by a frictionless tube.
Answer: Yes, now the ball will make it to the top.
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CTEnergy-11. A cart rolls without friction along a track. The graph of PE vs. position is shown.
The total mechanical energy (KE + PE) is 45kJ. The top of the hill at x = 34 m is round, and the
bottom of the valley near x = 55 m is flat.
50
Etot
40
PE
(kJ) 30
20
10
0
10
30
50
70 x(m)
To within 5 kJ, what is the maximum KE over the stretch of track shown?
A) 55kJ
B) 10kJ
C) 45kJ
D) 35 kJ
E) None of these
Answer: 35 kJ
When the KE is a minimum (over this stretch of track), what is the direction of the acceleration?
A) up
B) down
C) right
D) Left
E) some other direction (at an angle) or zero.
Answer: down . When the KE is min, it is on the top of the round hill. For circular motion with
constant speed, the acceleration is toward the center of the circle.
When the KE is max, what is the direction of the acceleration?
A) up
B) down
C) right
D) Left
E) some other direction (at an angle) or zero.
Answer: no direction, the acceleration is zero . When the KE is max, it is in the flatbottomed valley, moving with constant velocity.
Page 7 of 12
CTEnergy-12. A cart rolls without friction along a track. The graph of PE vs. position is shown.
The total mechanical energy (KE + PE) is 0 kJ.
10
Etot
0
PE
(kJ) -10
-20
-30
0
10
30
50
70 x(m)
What is the MAX KE of the cart during its journey (to within 5kJ)?
A) 30 kJ
B) 50 kJ
C) -30 kJ
D) 10 kJ
E) None of these
Answer: 30 kJ
Suppose the cart is at position x = 20 m, is moving right, and has total energy Etot = –15 kJ. Will
the cart make it over the hill at x=34 m?
A) Yes.
B) No.
Answer: No. It need a total energy of at least 10 kJ to make it over the hill.
Page 8 of 12
CTEnergy-13. Suppose an unusual (non-Hooke's Law) spring has a force which is a constant,
independent of positive stretch length x : Fspring = –C , where C is a positive constant. (Instead
of Fsping = – kx). The potential energy associated with this spring is (for positive stretch x) is
A) PE  C
B) PE  C x
C) PE 
1
1
C x 2 D) PE  C x 3 E) None of these
3
2
Answer: PE  C x Remember the definition of PE: PEF = WF = +Wext
CTEnergy-14. Elevator 1 can carry a load of mass m up a distance h in a time t, before the
engine overheats. Its power output is P1.
Elevator 2 can carry the same load up twice the distance (2h) in twice the time (2t). What is its
power output P2 ? A) P2 = P1
B) P2 = 2P1
C) P2 = 4P1
D) None these
Answer: P2 = P1
P = W/t = mgh/t
Elevator 3 can carry twice the load (2m) up twice the distance (2h) in the same time (t) as
elevator 1. What is its power output P3 ?
A) P3 = P1
B) P3 = 2P1
C) P3 = 4P1
D) None these
Answer: P3 = 4P1
2h
time 2t
2h
time t
h
time t
m
m
1
2m
2
3
Page 9 of 12
CTEnergy-15. Student A climbs a flight of stairs 10 m high in 8 s. Student B climbs a flight 20
m high in 16 s. Which student is producing the greater mechanical power output?
A) A
B) B
C) Tie
16 s
20 m
8s
10 m
Answer: Tie. The mechanical power output is the power that is going into increasing the
gravitational PE = mgh. Power = mgh/t
Page 10 of 12
CTEnergy-16. A hockey puck sliding on an ice rink is moving at 1 m/s when it slides onto a
carpet that someone left on the ice. The puck comes to rest after moving 1m on the carpet. How
far along the carpet would the puck go, if its initial speed was 2m/s?
A) 1.5 m
B) 2 m
C) 3 m
D) 4 m
E) Impossible to determine.
[Hint: Apply the Work-Energy Theorem. If the puck slides twice as far, the friction does twice
as much (negative) work.]
Answer: 4 m
(Warm-up question) How much more KE does the puck have moving at 2 m/s compared to
moving at 1 m/s?
A)
2 times as much
Answer: 4 times as much
B) twice as much
C) 3 times as much
D) 4 times as much
Page 11 of 12
CTWork-17. A “system” consists of a mass m hanging from a spring (spring constant k) and the
Earth as shown. In situation A, the mass is hanging freely, at rest, so the spring is stretched
beyond its relaxed length. In situation B, the mass is also stationary, but the spring is now at its
relaxed length, because an alien from Planet X has lifted the mass and is holding it up.
B
A
In which situation is the total energy of the mass/spring/earth system bigger?
A) A
B) B
C) same energy in both.
Answer: B. The Alien visitor did work on the system when he lifted it, and this
increased the energy of the system. The energy of a system is only constant if the system
is isolated .
Page 12 of 12
CTWork-18. Three equal masses are tied together (massless strings) and pulled to the left by a
force of magnitude F over a frictionless, horizontal surface. The free-body diagrams for the 3
masses are shown.
Some of the 11 forces shown
in the free-body diagrams are
Newton’s 3rd Law ActionReaction pairs. How many
action-reaction pairs can you
count among the 11 forces
shown?
A) 2 B) 3 C) 4 D) 5
E) None of these
Answer: 2 pairs
F