RII-1. The free-body diagram for a car rounding a banked curve is shown. The car is
going so fast that it is about to slip. The coordinate system has been chosen.
y
N

x
a

sN
mg
What is the correct Y-equation:
A: N sin + sN cos – mg = 0
B: N – sN sin – mg cos = 0
C: N – mg cos = 0 D: N cos – sN sin – mg = 0
E: None of these/ don't know.
Answer: N cos – sN sin – mg = 0
See web lecture notes, Review Exam II, page 1.
RII-2. A rider in a "barrel of fun" finds herself stuck with her back to the wall. The freebody diagram is shown.
Is the following statement definitely true for this situation? sN=mg
A: Yes, it is certainly true. B: No, it might not be true.
Answer: NO! It might not be true. The expression sN is the maximum possible force of
static friction. The force of static friction can be any values less than this maximum:
Fstatfric  sN. If the barrel is turning so slowly that the rider is just about to slip down the
wall, then Fstatfric = sN. If the barrel is turning very rapidly, then N is very large (since
N=mv2/R) and sN > mg. Her weight mg is a constant, independent of the speed of
rotation.
RII-3. The tip of the nose of a pogo stick rider moves along the path shown. The
maximum compression of the pogo stick spring is shown.
A
B
C
D
x
At what point(A, B, C or D) is the elastic potential of the spring energy a maximum?
Answer: D, when the spring is fully compressed
At what point is the gravitational potential energy a maximum?
Answer: A, at the maximum height.
At what point is the kinetic energy a maximum?
Answer: C, which is when the bottom tip of the stick has just touched the ground and the
spring is just starting to compress.
RII-4. A projectile is fired straight up and then it comes back down to its original height.
There is air resistance as the projectile is moving. During the flight of the projectile, the
work done by the force of gravity is..
A: zero.
B: positive
C: negative
Answer: Zero. The work done by gravity is negative on the way up, positive on the way
down. The total work done is zero.
During the flight of the projectile, the work done by the force of air resistance is...
A: zero.
B: positive
C: negative
Answer: Negative. The force of friction always opposes the motion, so the work done by
friction is negative on the way up and on the way down.
During the flight of the projectile, the work done by the net force is...
A: zero.
B: positive
C: negative
Answer: Negative. Two ways to see this:
1. Work done by net force = Work done by gravity and by friction, so Wnet =
Wgrav+Wfric. Wgrav=0 and Wfric<0 (from above questions) so Wnet<0.
2. Appeal to the Work-Energy Theorem: Wnet = KE. Due to air resistance, the final
speed of the projectile, just before it hits the ground is less than the initial upward
velocity. Hence, the KE decreased and KE < 0. Therefore, Wnet < 0.
RII-5. A new "constant force spring" is invented which has the remarkable property that
the force exerted by the spring is independent of the stretch of the spring, Fspr = -q, where
q is a constant. What is the potential energy contained in this spring when it is stretched
a length x.
A: (1/2)qx2
B: q x
C: q
D: None of these.
Answer: PE = qx. Apply the definition of PE: PE = –WField = +Wext = force  distance
= q x.
This is just like our expression for gravitational potential energy near the Earth's surface.
(ignoring signs) PE = Work = Force  distance = mg h .
RII-6 A planet is in elliptical orbit around the Sun. The zero of potential energy U is
chosen at r = , so U( r )  
GMm
.
r
How does the magnitude of U (= -U) compare to the KE?
A: -U > KE
B: -U < KE
C: -U=KE
D: depends on the position in the orbit.
Answer: -U > KE
When the velocity of the planet is the escape velocity, the total energy Etot=0, since the
condition for escape velocity is
KE i  PE i  KE f  PE f
GMm
2
1
 0  0  E tot  0
2 mv esc 
R
When v=vesc, Etot=0. When v>vesc, Etot>0. When v<vesc, Etot<0.
If Etot is greater than or equal to zero, the planet
cannot be in a bound orbit; it will escape to
infinity. Only when the Etot is negative is the
planet in a bound orbit. In order for Etot to be
negative, the negative U must be larger in
magnitude than the positive KE. |U|>KE or U>KE.
U
r
Etot
|U|
KE
RII-7. The gravitational potential energy of a rock near a star is shown in the diagram.
When the rock is at the position shown, it has a kinetic energy of 25MJ. Will the rock
escape to infinity or is it bound in orbit about the star?
A: Escape
B: in bound orbit
U(r)
0
C: impossible to tell
r
-10MJ
-20MJ
-40MJ
Answer: the rock is in a bound orbit. The total energy is Etot = PE + KE = -30MJ + 25MJ
= -5MJ. The rock cannot escape the "gravity well" The maximum possible r the rock
can possibly attain is where the Etot line intersects the U(r) curve.
U(r)
0
-10MJ
-20MJ
-40MJ
r
Etot
PE(-) KE(+)
CTRII-8. A pendulum is launched in two different ways. During both launches, the bob
is given an initial speed of 3.0 m/s and the same initial angle from the vertical. On launch
1, the speed is upwards, on launch 2, the speed is downwards.
1
2
Which launch will cause the pendulum to swing the largest angle from the equilibrium
position on the left side?
A: Launch 1
B: Launch 2
C: Both launches give the same maximum displacement.
Answer: Both launches give the same maximum displacement. Apply conservation of
energy, Etot = KEi + PEi = KEmax = PEmax. For both launch 1 and 2, the initial KE and
the initial PE are the same. So the Etot is the same for both launches.
CT9-9. A block of mass m with initial speed v slides up a frictionless ramp of height h
inclined at an angle  as shown. Assume no friction.
m
v

h
True A or False B:
Whether the block makes it to the top of the ramp depends on the mass of block and on
the angle .
Answer: false. The block will make it to the top if (1/2)mv2 > or = mgh. The m’s
cancel. Whether the block makes it to the top depends only on v and h.
Suppose now that there is friction between the block and the ramp surface.
True A or False B:
Whether the block makes it to the top of the ramp depends both on the mass of block and
on the angle .
Answer: False. Whether the block makes it to the top depends on the angle  but not on
the mass m. To see this, write out conservation of energy equation. Suppose the block
slides up a distance x along the ramp before stopping, then we can write
E init ial = E final
KE i = PE f + Q (heat )
1
2
mv 2 = mgh + mk Nx
1
2
mv 2 = mgh + mk mg cos qx
Recall the heat Q = magnitude of work done by friction.