Physics 121
Examination #2
PLEASE DO NOT DISCUSS THIS EXAM WITH ANYONE
(except a Physics 121 TA)
UNTIL AFTER ITS DEADLINE.
INSTRUCTIONS:
Work all the problems. Problems 1-5 are worth 16 points each, questions 6-10 are worth 4 points each.
The maximum possible score is therefore 100. You will be graded on your method of solution as well as on your final answer.
We can only read legible solutions. If you think that we won't be able to follow a step, please explain it. An "A" solution will
follow the 3D BE SNUB homework outline or a nearly equivalent format, making it possible for the graders to follow your work.
Please do not write on this exam. Write your answers on paper to be handed in with an Exam
Cover—ask for one if you forgot to bring it. There is no time limit (except Testing Center closing), but you should
complete the exam in about 2 hours. The exam is closed book except for the material on the Exam Cover. You should have
brought paper for your solutions. Ask at the desk if you forgot. You may use a non-programmed electronic calculator for
numerical calculations. When you are finished, place your solutions and this copy of the exam inside the exam cover and return it
to the exam proctors.
1. A block of mass m = 4 kg is being pulled across a uniform
F
horizontal surface by a constant force F = 20 N which is
θ
directed at an angle θ above the horizontal as shown in the
accompanying figure. This force causes an acceleration, a.
m
a
The coefficient of kinetic friction between the block and the
supporting surface is μ = 0.32. (a) Find a symbolic
expression for the acceleration, a, as a function of m, F, θ, μ and g. (b) Solve symbolically for the
value of θ which will produce the maximum acceleration. (c) What is the numerical value of θmax for
the given data? (d) What is the numerical value of the resulting maximum acceleration, amax? (e)
What would be the numerical value of the acceleration for θ = 0°?
We begin by drawing a free-body diagram for the block. The net force on the block is
Fnet = (Fnetx, Fnety) = (F cos θ - μN, N+F sin θ – mg). But by Newton’s 2nd law,
Fnet = ma = (ma, 0).
The y-force component is therefore 0 = N+F sin θ –mg from which
we infer N = mg – F sin θ.
(a) Since the total acceleration is in the x-direction, we can write
ma = F cos θ – μN = F cos θ – μ(mg – F sin θ), or
a = (F /m)(cos θ + μ sin θ) – μg. (4)
F
N
θ
f = -uNî
y
(b) Let θmax be the value of θ which maximizes a. We solve for
θmax by setting 0 = a/θ = (F /m)(μ cos θ – sin θ) which implies
that θmax = tan-1 μ. (3)
mg
x
(c) θmax = tan-1 0.32 = 17.7° . (3)
(d) amax = (F /m)(cos θ + μ sin θ) – μg =(20 N/4 kg)(cos 17.7° + 0.32 sin 17.7°) – 0.32(9.8 m/s2) = 2.11 m/s2. (3)
(e) a(θ=0°) = (20 N/4 kg)(cos 0° + 0.32 sin 0°) – 0.32(9.8 m/s2) = 1.86 m/s2. (3)
2. Two masses, M and m where M > m, are attached to the ends of a massless cord
which is suspended from a massless, frictionless pulley. The system is initially
(t =0) at rest, with y0 = h, where y0 is the initial position of the mass M in the given
coordinate system. Using this coordinate system find expressions for (a) a, the
acceleration of mass M, (b) T, the tension in the cord, (c) y(t), the y-position of
mass M as a function of time during its descent for t  0s, (d) K(t) the system
kinetic energy as a function of time during the descent of M.
y
(a) Here we need two free-body diagrams, one for each suspended mass.
a
T
T
a
mg
Mg
a
m
M
Using the designated
coordinate system for M,
h
and requiring sign
consistency for a requires
x
that upward be positive in
the M free-body diagram and downward in
the m free-body diagram. Hence the
equations of motion are: Ma = T – Mg and
ma = mg – T. Adding these two equation
yields (M + m)a = (m – M)g, or
(m  M ) g
. (4)
mM
(b) Since T = M(a+g) it follows that
T
2 Mmg
. (4)
M m
(c) Since the acceleration is constant and negative, the constant acceleration kinematic relation x = x0+v0t +½at2
becomes
y  h
( M  m) 2
t . (4)
2( M  m )
(d) Since v(t) = dy(t)/dt and K = ½mtotalv2, it follows that
 ( M  m )t
( M  m) 2 t 2
v (t ) 
. So, since mtotal = (M+m), K (t ) 
. (4)
( M  m)
2( M  m)
3. A ball is hanging by a short string from the roof of a ferris-wheel
gondola 15 meters from the axis of rotation. It is observed that
the string makes an angle of 20 with the vertical when the
gondola is at the same height as the axis of rotation (i.e., when
the line from the axis to the gondola is horizontal). What is the
revolution period (the time it takes to complete one cycle) for
the ferris wheel? (Hint: think of the gondola as a small box,
on the end of a rigid rod, being whirled in a vertical circle at
constant speed. Be sure to construct an accurate free-body
diagram of the ball at the end of the string.)
We set up the free-body diagram at the right at the instant the
gondola is at the same level as the ferris-wheel axis. We take the
x-axis to be directed toward that axis. This means that the
x-component of acceleration is also the centripetal component.
Therefore it follows that
y
θ
x
T
Ferris wheel axis
mg
(6 for free-body diagram)
Fy = may = 0 = T cos θ – mg. Hence T = mg/cos θ.
FX= maX = Tsin θ = Fcentripetal = mrω2 = mr(2π/P)2 (where P is the period). It follows then that
P  2
mr
r
15 m
 2
 2
 12.9 s. (10)
T sin 
g tan 
(9.8m / s 2 ) tan( 20  )
v=0
4. A 5-kg block is projected along a horizontal surface toward a curved
incline, as shown in the accompanying figure. The block has an initial
speed of 15 m/s. It slides up the incline and momentarily stops a height
h=5m
h = 5 meters above its initial position.
During this time it has slid a total
v0=15m/s
distance of 18 meters along the
m =5kg
surface. Calculate the average
d=18m
frictional force acting on the block
during this motion.
This problem is simply solved as an energy problem. The initial energy of the system which is all kinetic is
Ei = ½mv02. The final energy of the system which is all gravitational potential energy, Ef = mgh, must equal the initial
energy plus the negative work done by friction, Wf = -<f>d. Hence the average, frictional force, <f> is
 f 
W f
d

Ei  E f
d
1
mv 02  mgh
m 1
5 kg 1
2

 ( v 02  gh) 
[ (15 m/s) 2  (9.8 m/s 2 )(5 m)]  17.6 N. (16)
d
d 2
18 m 2
5. 5.
A 1200-kg car starts from rest on a horizontal road and accelerates (nonuniformly) to a speed of
30 km/hr in 20 seconds. (a) What is the final kinetic energy of the car? (b) What is the average net
power, in watts, delivered to the car during its acceleration? (c) If the speed of the car is then held
constant at 30 km/h and the engine delivers 75 horsepower, what net frictional force is acting on the
car? (1 hp = 746 W.)
(a) The final kinetic energy is K = ½ mv 2 = ½(1200 kg)[(30 km/hr)(1000 m/km)/(3600 s/hr)]2 = 41.7 kJ. (5)
(b) The average net power is <P> = ΔK/Δt = K/Δt = (4.17  104 J)/(20 s) = 2.08 kW. (5)
(c) If the speed remains constant, then the power delivered by the engine must equal the power
dissipated by friction, i.e., P = Fv, or
F = P/v = (75 hp)(746W/hp)/ [(30 km/hr)(1000 m/km)/(3600 s/hr)] = 6710 N. (6)
___ 6. As you sit motionless on a chair, the chair exerts a constant upward force on you. Which of the
following is the reaction force associated with that force? (a) the upward force exerted on the chair by
the floor, (b) the upward force exerted on the earth by your gravitational pull, (c) the downward force
exerted on you by the earth’s gravitational pull, (d) the downward force exerted by you on the chair,
(e) the downward force exerted on the floor by the chair.
___ 7. If the chair in the last question is a real chair with non-zero mass, which of the described upward
forces is greatest in magnitude? (a) the originally described force, (b) the upward force exerted by
the floor on the chair, (c) the upward gravitational force exerted by you on the earth, (d) (a) and (b);
(e) (a) and (c); (f) (b) and (c); (g) (a), (b) and (c).
___ 8. A particle which is constrained to remain on the x-axis has potential energy given by
U(x) = ka2 + ax sin(bx), where k, a and b are positive constants. The particle is in equilibrium at the
origin since F(0) = -(dU/dx)x=0 = 0. The nature of that equilibrium point is (a) stable, (b) unstable, (c)
neutral, (d) not enough information is provided to make a determination possible.
___ 9. In the preceding question possible SI units for the constant k are (a) joules, (b) newtons, (c)
joules/meter, (d) meters/newton, (e) joules/newton.
___10. A farmer whirls a pail of milk in a vertical circle at a constant speed, just fast enough that the milk
doesn’t spill as the upside-down pail passes over his head. At the bottom of its arc, the force exerted
by the farmer on the pail of milk (which has weight W) is (a) 0, (b) W/2, (c) W, (d) 2W. (e) It cannot be
determined since it depends on the length of the farmer’s arm which is unspecified.