Version 063 – Test 2 – swinney – (58535)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
Consider a force pulling 3 blocks along a rough
horizontal surface, where the masses are a
multiple of a given mass m, as shown in the
figure below. The coefficient of the kinetic
friction is µ. The blocks accelerate at a rate
a = 4 µ g.
6m
T1
T2
3m
9m
F
1
Basic Concept: Newton’s 2nd law.
Solution: Since all three blocks must have
the same acceleration (no block can “speed
ahead” or “lag behind”) the Momentum Principle applies to the whole system of three
blocks. (Take right to be positive.)
Ff
m1 + m2 + m3
F
µ (m1 + m2 + m3 ) g
F − Ff = (m1 + m2 + m3 ) a = 18 m a ,
where Ff = µ (m1 + m2 + m3 ) g = 18 µ m g .
µ
F = 18 µ m g + 18 m a
= 18 µ m g + 18 m (4µ g)
= 18 µ m g (1 + 4)
Determine the force F .
1. F = µ m g
= 90 µ m g .
2. F = 27 µ m g
3. F = 18 µ g
002 10.0 points
An object with mass 120 kg, moved in outer
space from location
4. F = 4/3 µ m g
~ri = h12, −23, −6i m,
5. F = 90 µ m g correct
to location
6. F = 126 µ m g
~rf = h19, −26, −9i m.
7. F = 40 µ m g
A single constant force
8. F = 4 µ m g
~ = h250, 440, −220i N
F
9. F = 72 µ m g
Explanation:
m1
T1
T2
m2
m3
µ
Given : m1 = 6 m ,
m2 = 3 m ,
m3 = 9 m ,
m1 + m2 + m3 = 18 m , and
a = 4 µ, g .
F
acted on the object while the object moved.
It’s final speed was 12 m/s. What was the
speed of the object at the initial location?
1. 12.3491
2. 9.83714
3. 11.78
4. 10.884
5. 11.6619
6. 12.5191
7. 12.746
8. 11.5181
9. 12.1381
10. 11.2175
Version 063 – Test 2 – swinney – (58535)
Correct answer: 11.2175 m/s.
Explanation:
We begin by finding the work done on the
object, and then we can use the update form
of the energy principle. However, since W =
~ · ∆~r , we need ∆~r to find the work done.
F
∆~r = ~rf − ~ri
= h19, −26, −9i m
− h12, −23, −6i m
= h7, −3, −3i m .
Now we can find the work done:
~ · ∆~r
W =F
= h250, 440, −220i N · h7, −3, −3i m
= 1090 J .
Knowing the work, we can use the update
form of the energy principle,
A beaker filled with oil rests on a table. It
contains a mass mo of oil with a density ρo .
A block of iron of mass mi and density ρi is
suspended from a spring scale and partially
immersed in the oil, such that one third of the
block is in air.
Neglecting the buoyancy of the air, the
weight measured by the scale W is
1
3
2
2. W = mi g +
3
2
3. W = mi g −
3
2
4. W = mi g −
3
1. W = mi g −
ρi
ρo
ρo
ρi
ρi
ρo
ρo
ρi
6. W = mi g −
7. W = mi g +
8. W = mi g −
to find the initial kinetic energy. We write
9. W = mi g −
1
m v2
2
r
2(KEi)
⇒ vi =
s m
KE =
=
2(7550 J)
120 kg
= 11.2175 m/s .
003
10.0 points
mi g .
mo g .
mi g . correct
10. W = mi g +
1 ρo
mi g .
3 ρi
1 ρi
mi g .
3 ρo
ρo
mi g .
ρi
ρi
mi g .
ρo
ρi
mo g .
ρo
Explanation:
The buoyant force is B = ρo Vsubmerged g.
2
2 mi
In our case, Vsubmerged = Vi =
, so
3
3 ρi
B=
Rearranging the equation for kinetic energy, we can find the initial speed:
mo g .
5. W = mi g .
Ef = Ei + W ⇒ KEf = KEi + W,
KEi = KEf − W
1
= m vf2 − W
2
1
= (120 kg)(12 m/s)2 − (1090 J)
2
= 8640 J − (1090 J)
= 7550 J .
2
2 ρo
mi g .
3 ρi
The forces acting on the block of iron are
the tension measured by the spring W , buoyant force and the the force of gravity so
W + B − mi g = 0
2 ρo
mi g
W = mi g −
3 ρi
004 10.0 points
If you were designing a circular race track
for F1 cars that travel 250 km/hr on a track
banked at 36 degrees all the way around the
Version 063 – Test 2 – swinney – (58535)
circuit, what track radius would you use so
that no frictional force would be required for
a car to make the turn? Use g = 9.8 m/s2 .
1. 790.015
2. 704.674
3. 1041.43
4. 761.026
5. 677.311
6. 1180.15
7. 1289.87
8. 849.618
9. 819.546
10. 1075.3
Correct answer: 677.311 m.
Explanation:
Looking at a free body diagram on which
we have drawn the normal and gravitational
forces, we see that the normal force must be
decomposed in order to write the equations
of motion in the x and y directions. The x
direction is a natural choice for writing the
momentum principle because it happens to
be the centripetal direction in this problem.
Writing the equations of motion along both
directions,
3
After coming down a steep hill at a constant
speed of 19 m/s, a car travels along the circumference of a vertical circle of radius 316 m
until it begins to climb another hill.
r
x
What is the magnitude of the net force on
the 120 kg driver of the car at the lowest point
on this circular path?
1. 379.438
2. 712.088
3. 969.423
4. 137.089
5. 1006.01
6. 385.722
7. 2982.86
8. 5935.61
9. 1323.44
10. 51.7986
Correct answer: 137.089 N.
y dir :
dpy
= FN cos θ − m g
dt
0 = FN cos θ − m g
mg
⇒ FN =
.
cos θ
m v2
r
(120 kg)(19 m/s)2
=
(316 m)
= 137.089 N
Fr = m ar =
We can also write
dpx
= FN sin θ
dt
Putting these together, we have
x dir :
m v2
= m g tan θ
R
v2
⇒R=
g tan θ
(250 km/hr)2
=
(9.8 m/s2 ) tan 36◦
= 677.311 m .
005
Explanation:
10.0 points
006 10.0 points
The 8 N weight is in equilibrium under the
influence of the three forces acting on it. The
F force acts from above on the left at an angle
of α with the horizontal. The 5.1 N force
acts from above on the right at an angle of
56 ◦ with the horizontal. The force 8 N acts
straight down.
Version 063 – Test 2 – swinney – (58535)
4
= −(5.1 N) cos 56 ◦
= −2.85188 N and
56
◦
5. 1
F
N
α
X
F1y = −F2 sin α2 − F3
= −(5.1 N) sin 56 ◦ − (−8 N)
= 3.77191 N , and
8N
What is the magnitude of the force F ?
1. 4.72869
2. 6.10251
3. 4.47708
4. 3.80626
5. 6.00622
6. 4.11685
7. 4.29166
8. 5.46715
9. 6.2757
10. 3.9107
Correct answer: 4.72869 N.
Explanation:
Standard angular measurements are from
the positive x-axis in a counter-clockwise direction.
Let :
F1
F2
α2
F3
Fy = F1y + F2y + F3y = 0
q
2 + F2
F1 = F1x
1y
q
= (−2.85188 N)2 + (3.77191 N)2
= 4.72869 N .
007 10.0 points
Two blocks of the same size, shape and mass
m are one atop the other. A constant horizontal force F is applied to the lower block.
The coefficient of kinetic friction between the
lower block and the level tabletop on which
the blocks slide is µk .
1
=F,
= 5.1 N ,
= 56 ◦ , and
= −8 N .
m
2
F
Consider the free body diagram. The green
vectors are the components of the slanted
forces.
α
F1
F2
α2
µk
Find the force of static friction that must
act on the upper block in order that the blocks
accelerate together.
F
2
F + µk m g
2. fs =
2
F
3. fs = + µk m g
2
1. fs =
F3
The weight is is equilibrium, so
X
Fx = F1x + F2x + F3x = 0
F1x = −F2 cos α2 − 0
m
4. fs = µk m g
5. fs = F − 2µk m g
Version 063 – Test 2 – swinney – (58535)
Correct answer: −0.625 J.
6. fs = F + 2µk m g
F − µk m g
2
F
8. fs = − µk m g correct
2
Explanation:
a
a
1
Explanation:
We take x to be the displacement with respect to the equilibrium position of the spring.
Then the spring force is given by F = −k x. In
the initial and final states we have xi = 0 and
xf = d − L0 = 0.175 m − 0.3 m = −0.125 m,
respectively. Then the amount of work that
the spring does is given by:
7. fs =
m
2
m
F
fs
fs
fk
fk = µk (2 m) g since the table supports
both blocks.
Apply Newton’s second law horizontally to
each block. For the upper block m a = fs and
for the lower block m a = F − fs − fk .
Adding the two equations,
2 m a = F − 2 µk m g
F
a=
− µk g .
2m
Thus
fs = m a =
5
F
− µk m g .
2
008 10.0 points
A ball traveling horizontally runs into a horizontal spring, whose stiffness is k = 80 N/m.
As the spring is compressed, the ball slows
down. If the relaxed length of the spring is
L0 = 0.3 m, and the spring is compressed until its length is d = 0.175 m, how much work
does the spring do on the ball?
1. -7.2
2. -1.89063
3. -0.4
4. -4.9
5. -11.25
6. -3.6
7. -0.625
8. -5.625
9. -6.25
10. -7.89688
W =
=
Z
xf
F dx
Zxixf
−k x dx
xi
x2
x2i 2
2
(−0.125 m)2
−0
= −80 N/m ·
2
= −0.625 J
= −k
f
−
009 10.0 points
An object moves in a circle at a constant
speed v. The radius of the circle is r.
A force F acts toward the center of the
circle. How much work does this force do as
the object moves?
1. W =
m v2
.
r
2. W = F (π r 2 ) .
3. More information is needed.
4. W = m v 2 .
5. W = F (2 π r).
6. Zero correct
m v2
.
2
m v2
8. W =
.
4
Explanation:
7. W =
Version 063 – Test 2 – swinney – (58535)
The velocity is always tangent to the circle,
so when the force acts toward the center of
the circle, the force always acts perpendicular
to the direction of motion, and the force does
no work. This is also obvious from the fact
1
that the kinetic energy K =
m v 2 of the
2
object does not change.
6
Fcar
5.
Fgrav, car F
grav, man
Fcar
cor-
6.
010 10.0 points
A man stands in an elevator car as it is being
lifted upward, as shown in the figure.
Fgrav, man
rect
Elevator
Cable
Fcable
7.
Fgrav, man Fgrav, car
Choose the correct free body diagram for
the man, where Fcable is the force of the cable
on the elevator car. Fgrav, man is the force of
gravity on the man. Fgrav, car is the force of
gravity on the elevator car. Fcar is the force
of the car on the man.
Note that some parallel vectors are offset
horizontally for clarity.
Fcable
1.
Explanation:
Only the forces acting directly on the man
are to be in the free body diagram. Therefore,
the force from the cable should be omitted,
while those from gravity and from the floor’s
normal force should be included.
011 10.0 points
Two satellites A and B orbit the Earth in
the same plane. Their masses are 5 m and
6 m, respectively, and their radii 4 r and 5 r,
respectively.
Fgrav, car
2.
Fcable
5r
4r
Fgrav, man Fgrav, car
3.
Fcar Fcable
Fgrav, man Fgrav, car
4.
Fcar Fcable
Fgrav, man
B
6m
A
5m
What is the ratio of the orbital speed vB of
Satellite B to the orbital speed vA of Satellite
A?
Version 063 – Test 2 – swinney – (58535)
vB
1.
=
vA
2.
3.
4.
vB
=
vA
vB
=
vA
vB
=
vA
r
r
r
r
5
6
2. he stands with both feet flat on the
ground.
25
24
3. he stands flat on one foot.
3
2
4. he lies down on his back.
6
5
5. he sits on his butt.
6. The pressure on the ground is the same,
however he stands, sits or lies.
5. None of these
r
vB
4
6.
=
correct
vA
5
r
24
vB
=
7.
vA
25
r
5
vB
=
8.
vA
4
r
vB
2
=
9.
vA
3
Explanation:
F
A
The forces exerted on the earth in different
cases are the same while when the man stands
on the toes of one foot, the area is the smallest
and the pressure is the greatest.
P =
Explanation:
The force of gravity is responsible for holding a satellite in its orbit, so the orbital centripetal force is equal to the force of gravity:
Fr = m
v2
ME m
=G
,
r
r2
where ME is the mass of the Earth, m is the
mass of the satellite, and r is the radius of
the orbit (from the Earth’s center). Thus the
tangential speed v of an orbit at distance r is
r
r
GM
1
v=
∝
.
r
r
Since
7
4r
4
rA
=
= , the ratio
rB
5r
5
vB
=
vA
r
rA
=
rB
r
4
.
5
012 10.0 points
The pressure exerted on the ground by a man
is greatest when
1. he stands on the toes of one foot. correct
013 10.0 points
A small block of mass m is on a low-friction table and is attached to a horizontal spring. The
spring has stiffness ks and a relaxed length L.
The other end of the spring is fastened to a
fixed point at the center of the table. The
block slides on the table in a circular path of
radius R > L. How long does it take for the
block to go around once?
r
4πmR
1. T =
k L
s s
4 π2 m R
2. T =
k (L − R)
s s
3 π 2 m L2
3. T =
k R2
s s
4 π 2 m R2
4. T =
ks L
s
4 π 2 m R2
5. T =
k (R − L)2
s s
4πmR
6. T =
ks (L − R)
s
2πmR
7. T =
ks (R − L)
Version 063 – Test 2 – swinney – (58535)
4πmR
ks (R − L)
9. T =
s
3 π 2 m R2
ks (R − L)
10. T =
s
4 π2 m R
correct
ks (R − L)
O
r
8. T =
s
is
2
~ m |~v |
.
Fnet =
R
1. |~a| =
2. |~a| =
3. |~a| =
4. |~a| =
5. |~a| =
~
Fspring = ks s
~ and Fnet = |Fspring |
6. |~a| =
m |~v|2
⇒
= ks (R − L)
R
r
ks R(R − L)
⇒ |~v | =
.
m
For circular motion, |~v | = 2 π R/T .
substitute:
θ
The magnitude of a, the total acceleration
Explanation:
The net force on the ball is equal to
the force by the spring on the ball, which
~
is F
spring = ks s. The distance the spring
stretches is s = R − L. Because the speed of
the ball is constant and is much less than the
speed of light,
So we have
8
We
s
s
s
s
s
s
T
m
T
m
T
m
T
m
T
m
T
m
2
2
2
2
2
2
+ g2 .
+ g 2 sin2 θ .
+ 2 g cos θ
T
+ g2 .
m
sin2 θ + g 2 cos2 θ .
+ g 2 cos2 θ .
cos2 θ + g 2 sin2 θ .
7. None of these
s
2
T
− g2 .
8. |~a| =
m
s
2
T
T
9. |~a| =
− 2 g cos θ + g 2 . corm
m
rect
Explanation:
r
ks R(R − L)
m
s
4 π2 m R
⇒T =
.
ks (R − L)
2πR
=
T
O
T
θ
θ
014 10.0 points
A small sphere of mass m is connected to
the end of a cord of length r and rotates
in a vertical circle about a fixed point O. The
tension force exerted by the cord on the sphere
is denoted by T .
mg
The centripetal force is
m v2
.
r
This centripetal force is provided by the tenFc =
Version 063 – Test 2 – swinney – (58535)
sion force and the radial component of the
weight. In this case, they are in opposite
directions so,
m v2
Fc =
= T − m g cos θ .
r
Then, we see that the radial acceleration is
T
− g cos θ .
ar =
m
The tangential acceleration is caused by the
tangential component of the weight, so
at = g sin θ .
Thus the total acceleration is
q
a = a2r + a2t
" 2
T
T
− 2 g cos θ
=
m
m
21
2
2
2
+ g (sin θ + cos θ)
=
s
T
m
2
− 2 g cos θ
T
+ g2 .
m
015 10.0 points
At what angular velocity ω would the Earth
have to spin for someone of mass m standing
on the equator to be flung off?
s
Gm2
1. ω =
2
ME RE
s
GME
correct
2. ω =
3
RE
s
GME2
3. ω =
3
m RE
s
GME2
4. ω =
3
2m RE
s
GME
5. ω =
2
2RE
s
GME
6. ω =
3
3RE
7. ω =
s
GME m
3
4RE
8. ω =
s
GME
3
2RE
9
Explanation:
First let us write the momentum principle
in the centripetal direction.
mv 2
= Fgrav − FN = Fgrav
RE
If a person is to be flung off, her normal
force will go to zero as the Earth spins faster.
Solving for her tangential velocity,
mv 2
GME m
=
2
RE
RE
s
GME
v=
RE
From this tangential velocity, the angular
velocity of the Earth will be
s
v
GME
ω=
=
3
RE
RE
016 10.0 points
A high-speed lifting mechanism supports a
746 kg object with a steel cable 24.3 m long
and 4.09 cm2 in cross sectional area.
Determine the elongation of the cable if
the object is accelerated upward at a rate of
3.1 m/s2 .
The acceleration of gravity is 9.8 m/s2 and
Young’s modulus for steel is 2 × 1011 Pa .
Express your answer in mm. Be careful
with units in this problem.
1. 3.11958
2. 3.3803
3. 2.85878
4. 2.60191
5. 2.64714
6. 3.05394
7. 2.99482
8. 3.43968
9. 2.80764
10. 4.06137
Version 063 – Test 2 – swinney – (58535)
Correct answer: 2.85878 mm.
1. v =
Explanation:
Let :
correct
m = 746 kg ,
L0 = 24.3 m ,
A = 4.09 cm2 ,
g = 9.8 m/s2 , and
Y = 2 × 1011 Pa .
and
Let : ay = 3.1 m/s2 .
When the load is accelerating upward,
F − m g = m ay
F = m (g + ay ) ,
so the elongation of the cable is
m (g + ay )
F L0
=
AY
A Y
(746 kg) (9.8 m/s2 + 3.1 m/s2 )
=
(4.09 cm2 ) (2 × 1011 Pa)
(24.3 m)
+
(4.09 cm2 ) (2 × 1011 Pa)
2
1000 mm 100 cm
×
1m
1m
= 2.85878 mm
017 10.0 points
A block of mass m is pushed a distance D up
an inclined plane by a horizontal force F . The
plane is inclined at an angle θ with respect to
the horizontal. The block starts from rest and
the coefficient of kinetic friction is µk .
3. v =
4. v =
5. v =
6. v =
7. v =
r
r
r
r
r
r
2
(F cos θ − m g sin θ − µk N ) D
m
2
(F cos θ − µk N ) D
m
2
(F sin θ + µk N ) D
m
2
(F cos θ + m g sin θ) D
m
2
(F cos θ + m g sin θ − µk N ) D
m
2
(F sin θ − µk N ) D
m
2
(F cos θ − m g sin θ) D
m
r
2
(F cos θ − m g sin θ + µk N ) D
m
Explanation:
The force of friction has a magnitude
Ff riction = µk N . Since it is in the direction opposite to the motion, we get
Wf riction = −Ff riction D = −µk N D.
The normal force makes an angle of 90◦
with the displacement, so the work done by it
is zero.
The work done by gravity is
Wgrav = m g D cos(90◦ + θ)
= −m g D sin θ .
The work done by the force F is
WF = F D cos θ .
From the work-energy theorem we know that
Wnet = ∆K ,
D
m
2. v =
8. v =
∆L =
F
r
10
µk
WF + Wgrav + Wf riction =
θ
1
m vf2 .
2
Thus
If N is the normal force, the final speed of
the block is given by
vf =
r
2
(F cos θ − m g sin θ − µk N ) D .
m
Version 063 – Test 2 – swinney – (58535)
018
10.0 points
The density of ice is smaller then the density of water but larger than the density of oil.
Consider a completely submerged ice cube
floating in a glass at the interface of water
and oil.
When the ice melts what happens to the surface levels of the water and oil? Assume there
is no evaporation and the water and oil don’t
mix.
1. The answer depends on the actual density
differences.
11
The ice cube is floating so the weight of
the fluid it displaces must equal the ice cube’s
weight. The buoyant force on the ice cube
is due to both the displaced water and the
displaced oil. Thus, the weight of the ice
cube is greater than the weight of the water
it displaces. Therefore, the volume of the
melted ice cube is larger than the volume of
water the ice cube initially displaced. Hence
the water level rises.
However, the melted ice cube’s total volume
is smaller than the initial ice cube’s volume.
Hence the oil level falls.
019 10.0 points
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball
is then set in motion in a horizontal circle
at a distance h from the ceiling so that the
thread’s trajectory describes a cone.
The acceleration of gravity is 9.8 m/s2 .
2. The oil level falls and the water level
falls.
3. The oil level falls and the water level rises.
correct
θ
g
h
5. The oil level stays the same and the water
level falls.
6. The oil level stays the same and the water
level stays the same.
7. The oil level stays the same and the water
level rises.
8. The oil level rises and the water level
falls.
9. The oil level rises and the water level stays
the same.
10. The oil level rises and the water level
rises.
Explanation:
The oil level falls and the water level rises.
ℓ
4. The oil level falls and the water level stays
the same.
r
v
m
What is the period of rotation in terms of
g, ℓ, h and θ.
s
(ℓ2 − h2 )
1. Tperiod = 2 π
ℓg
s
h
2. Tperiod = 2 π
cos2 θ
g
s
ℓ
3. Tperiod = 2 π
g
p
4. Tperiod = π ℓ g h
s
ℓ
5. Tperiod = 2 π
sin2 θ
g
Version 063 – Test 2 – swinney – (58535)
s
h
6. Tperiod = 2 π
correct
g
s
2 ℓ2 tan θ
7. Tperiod = π
gh
s
ℓ
tan θ
8. Tperiod = 2 π
g
s
ℓ sin θ
9. Tperiod = 2 π
g
s
ℓ2 cos θ
10. Tperiod =
gh
12
d = vt.
Because the tangential speed of the ball
around the circle is constant, we have
vball =
s
Tperiod
.
s is the distance the ball travels in one revolution, which is the perimeter of the circle of
radius ℓ sin θ, therefore, we have
s = 2 π ℓ sin θ .
Explanation:
Use the free body diagram below.
Equating both expressions for vball , we have
p
2 π ℓ sin θ
= vball = g ℓ tan θ sin θ
Tperiod
s
2 π ℓ sin θ
g ℓ sin2 θ
=
Tperiod
cos θ
θ
T
mg
The tension on the string can be decomposed into a vertical component which balances the weight of the ball and a horizontal
component which causes the centripetal acceleration, acentrip that keeps the ball on its
horizontal circular path at radius r = ℓ sin θ.
If T is the magnitude of the tension in the
string, then
Tvertical = T cos θ = m g
Thoriz = m acentrip
T sin θ =
2
m vball
.
ℓ sin θ
(2)
Solving (1) for T yields
T =
mg
cos θ
and substituting (3) into (2) gives
m g tan θ =
2
m vball
.
ℓ sin θ
Solving for v yields
p
v = g ℓ tan θ sin θ .
ℓ cos θ
g
= 2π
s
h
.
g
as sin θ cancels and we can substitute cos θ =
h/ℓ.
(1)
and
or
Tperiod = 2 π
s
(3)
020 10.0 points
A(n) 79 g particle moving with an initial speed
of 34 m/s in the positive x direction strikes
and sticks to a(n) 160 g particle moving with
150 m/s in the positive y direction.
How much kinetic energy is lost in this
collision?
1. 222.956
2. 403.44
3. 653.297
4. 543.979
5. 529.44
6. 149.806
7. 32.1789
8. 382.654
9. 434.72
10. 625.548
Version 063 – Test 2 – swinney – (58535)
Correct answer: 625.548 J.
Explanation:
This is an inelastic collision. Momentum is
conserved.
m1 v1 = (m1 + m2 )vx
m2 v2 = (m1 + m2 )vy
Final velocity of both particles is
q
v = vx2 + vy2 .
Energy before collision is
1
1
KEi = m1 v12 + m2 v22 .
2
2
Energy after collision is
1
KEf = (m1 + m2 )v 2 .
2
Energy lost is
∆KE = KEi − KEf .
13