# Test 2-solutions (1) ```Version 058 – Test 2 – florin – (57850)
This print-out should have 20 questions.
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001
10.0 points
A tennis ball at the end of a string is released from rest and swings through a vertical
position, reaching its maximum speed there,
then comes to momentary rest on the other
side.
1
where a = 10 J/m2 and b = 7.4 J/m4 . What is
the force on the object when it is at r = 3.8 m?
1. -530.0
2. -82.084
3. -223.704
4. -34.775
5. -330.568
6. -195.16
7. -141.317
8. -24.116
9. -127.217
10. -631.0
Explanation:
The force exerted on an object is given by
F = −dU/dr so taking the derivative of the
potential energy function gives
rest
rest
1.
−→
What is the direction of the ball’s acceleration when it is at its lowest point and at
maximum speed?
correct
and inserting r = 3.8 m gives
F = −[(10 J/m2 ) + 3(7.4 J/m4 )(3.8 m)2 ]
F = −330.568 N
003 10.0 points
A satellite in a low circular earth orbit is
subjected to a very small constant frictional
force f , due to the resistance of the thin atmosphere at that altitude. As the satellite loses
energy, the radius of the orbit r slowly decreases, and the satellite spirals downwards.
Calculate the change in total energy ∆E of
the satellite per revolution. Assume the orbit
is approximately circular over a revolution.
2. −→
−→
3.
F = −dU/dr = −(a + 3b r 2 )
4. −→
5. ←−
−→
6.
Explanation:
1. ∆E = 4 f r 3 /3 π
dv
At the lowest point
= 0 so the entire
dt
acceleration is ar and is directed toward the
center of the circle.
2. ∆E = 2 π r f
002 10.0 points
Suppose the potential energy of an object
is given by:
U = a r + b r3
3. ∆E = f 2 /r 2
4. ∆E = −f /2 π r
5. ∆E = −2 π r f correct
6. ∆E = −3 π 2 r f
Version 058 – Test 2 – florin – (57850)
7. ∆E = −f
2
8. W = 2 G Me M Re
5 G Me M
2 Re
Explanation:
If we consider the system to be you plus the
earth,the gravitational potential energy is
8. ∆E = f /2 π r
9. W =
9. ∆E = −f r
10. ∆E = −π 2 f
Explanation:
The energy lost per revolution is just the
work done by the frictional force along the
approximately circular orbit of radius r. We
know that the frictional force acts in a direction opposite to that of instantaneous direction of motion, so the work done by this nonconservative force is Wnc = −f s = −2 π r f ,
so from conservation of energy,
G Me M
.
r
The initial energy is Ei = Ug,i and the
final energy is Ef = Ug,f . Thus the Energy
Principle gives:
∆E = Ef − Ei = Wnc = −f (2 π r) .
Alternately, if you consider the system to
just be you, the net work is zero and both you
and the earth do work on the system.
Z 2 RE
G ME m
W = Wg =
dr
r2
RE
004
10.0 points
Imagine a thin tower 4000 miles tall; ı.e.
as tall as the earth’s own radius Re , placed
at the north pole of the earth. Suppose you
start at rest at the base of the tower and climb
to the top (wearing a spacesuit and carrying
supplies you need, so that your total mass is
M ).
How much work would you have done by
the time you are at the top of the tower and
at rest again? Neglect the mass of the tower
compared to the mass of the earth.
1. W = G Me M Re
G Me M
correct
2 Re
2 G Me M
3. W =
5 Re
G Me M
4. W =
3 Re
2. W =
5. W = 4 G Me M Re
G Me M
4 Re
3G Me M
7. W =
2 Re
6. W =
Ug (r) = −
W = ∆Ug = Ug (2 Re ) − Ug (Re ) =
G Me M
.
2 Re
005 10.0 points
A 12 g bullet is accelerated in a rifle barrel
76.5 cm long to a speed of 455 m/s.
Use the work-energy theorem to find the
average force exerted on the bullet while it is
being accelerated.
1. 7025.87
2. 1623.73
3. 3047.76
4. 6173.34
5. 1029.2
6. 1676.06
7. 1749.51
8. 7786.28
9. 4034.87
10. 1512.17
Explanation:
Let : m = 12 g = 0.012 kg ,
x = 76.5 cm , and
v = 455 m/s .
Version 058 – Test 2 – florin – (57850)
The kinetic energy of the bullet before it
is fired is 0 J since it is not moving. At
the moment that it leaves the rifle barrel, its
kinetic energy is
1
m v2
2
1
= (0.012 kg) (455 m/s)2
2
= 1242.15 J .
K=
7. 138.202
8. 96.3393
9. 217.339
10. 163.76
Explanation:
Let :
Since the work done on the bullet is equal to
its change in kinetic energy, this is also the
work that is done on the bullet:
W =
1
m v 2 = 1242.15 J .
2
Also, the work is given by
W = Favg x
W
1242.15 J
Favg =
=
x
0.765 m
= 1623.73 N .
006 10.0 points
There was once a plan to roll cars partway
through the English Channel and tow them
the rest of the way. Imagine that a car of mass
m = 3938 kg rolls down a slope of length
ℓ = 51 km and at an angle θ = 6.8 ◦ to
the horizontal as shown in the picture. Also
imaging there is a small amount of friction
on the car as it rolls given by coefficient of
friction &micro;k = 0.1.
3
m = 3938 kg ,
g = 9.8 m/s2 ,
ℓ = 51 km = 51000 m ,
θ = 6.8◦ .
and
The change in potential energy is
∆U = Uf − Ui
= 0 − mgy
= −m g ℓ sin θ
= −(3938 kg) (9.8 m/s2 ) (51 km) sin 6.8◦
= −2.33044 &times; 108 J
with a magnitude of 2.33044 &times; 108 J .
Let :
&micro;k = 0.1 .
From the Energy Principle, while rolling
downhill we have
∆K + ∆U = Wsurr
m v2
= −∆U + Wf
2
r
2 (−∆U − Wf )
.
v=
m
Since
What would the speed of the car be when
it reaches the level part of the track, if the car
starts from rest? The acceleration of gravity
is 9.8 m/s2 .
1. 225.407
2. 148.622
3. 242.247
4. 145.058
5. 258.824
6. 233.491
−∆U − W = −∆U − &micro;k m g ℓ cos θ
= 2.33044 &times; 108 J
− 0.1(3938 kg)(9.8 m/s2 )
&times; (51000 m) cos 6.8◦
= 3.76074 &times; 107 J , then
v=
s
2(3.76074 &times; 107 J)
= 138.202 m/s .
3938 kg
Version 058 – Test 2 – florin – (57850)
4. vmin = 0
s
007 10.0 points
A little boy and little girl in a playground are
sitting on a large metal disk which is rotating
freely at a constant number of revolutions per
minute. The boy is at the outer rim, and the
girl is about halfway between the rim and the
center.
How do their accelerations compare?
4. The girl’s acceleration is greater.
5. None of these
Explanation:
The speed of a point on a rotating disk is
proportional to the distance from the center of
the disk: v ∝ r. But the acceleration toward
the center of the disc is determined by the
v2
quantity
∝ r in this case. Thus the boy’s
r
acceleration is about twice that of the girl.
008 10.0 points
Consider a car of mass m moving without
slipping on a circular track of radius R. The
track is inclined at an angle θ with respect to
the horizontal direction. Suppose the track
has a coefficient of static friction &micro;s . What
is the minimum speed vmin at which the car
must move to avoid slipping up or down the
track?
s
cos θ − &micro;s sin θ
1. vmin = g R
sin θ + &micro;s cos θ
s
cos θ − &micro;s sin θ
2. vmin = g R
sin θ − &micro;s cos θ
s
cos θ + &micro;s sin θ
3. vmin = g R
sin θ + &micro;s cos θ
6. vmin =
s
sin θ + &micro;s cos θ
cos θ − &micro;s sin θ
gR
sin θ − &micro;s cos θ
correct
cos θ + &micro;s sin θ
7. vmin =
s
gR
cos θ + &micro;s sin θ
sin θ − &micro;s cos θ
gR
sin θ − &micro;s cos θ
cos θ − &micro;s sin θ
gR
sin θ + &micro;s cos θ
cos θ + &micro;s sin θ
8. vmin = ∞
s
9. vmin =
2. The boy’s acceleration is greater. correct
3. Neither is accelerating since the disk has
a constant speed of rotation.
gR
5. vmin =
1. Both boy and girl have the same acceleration.
4
10. vmin =
s
Explanation:
Call the horizontal axis the x-axis and
choose it to be the radially inward direction
at a certain instant of time. Similarly call the
vertical axis the y-axis and choose it to be upward. Therefore, the car experiences a net
centripetal acceleration along the x-axis and
zero net acceleration along the y-axis. Looking for the minimum velocity implies that
friction is directed up the track - if the speed
was zero the car would tend to slide down the
track and static friction opposes the direction
that motion would tend.
From The Momentum Principle,
ΣFy = N cos θ + &micro;s N sin θ − m g = 0
N=
mg
cos θ + &micro;s sin θ
ΣFx = N sin θ − &micro;s N cos θ =
N (sin θ − &micro;s cos θ) =
2
m vmin
R
2
m vmin
R
.
Substituting the expression for N from the
y-equation gives,
2
m vmin
sin θ − &micro;s cos θ
= mg
R
cos θ + &micro;s sin θ
Version 058 – Test 2 – florin – (57850)
Finally, solving for vmin gives
vmin =
s
gR
sin θ − &micro;s cos θ
cos θ + &micro;s sin θ
009 10.0 points
An ice cube is held at a depth h by a string
attached to the bottom of a bucket of water.
h
5
on the ice). Then mathematically
X
F = ma
FB − T − W = m a .
The ice has no acceleration and so the
right-side of the above equation becomes zero.
Plugging-in the formulas for both the buoyant
force and the weight, gives
T = ρw V g − ρi V g
= V (ρw − ρi ) g
The same analysis works for after the ice
melts. Notice that T is proportional to V so
putting T ′ in for T and V ′ in for V gives
As time passes the ice melts. Suppose initially the ice has a volume V and the magnitude of tension in the string is T . When the
volume of the ice has decreased to V ′ = 1/3 V ,
what is the magnitude of tension T ′ ?
1. T ′ = 3 T
2. T ′ = 2/3 T
3. T ′ = 5/6 T
4. T ′ = 1/3 T correct
5. T ′ = T
T ′ = V ′ (ρw − ρi ) g
= 1/3 V (ρw − ρi ) g
= 1/3 T
010 10.0 points
A block of styrofoam with mass m has a lead
weight with mass 1/2 m glued on top. When
the lead-styrofoam piece is placed in water,
the styrofoam piece is partially submerged in
the water. The water has a density ρw , and
the density of the styrofoam ρs = 1/3 ρw . If
the styrofoam has a total volume V , what is
the fraction of the submerged volume Vsub to
the total volume V ?
6. T ′ = 3/2 T
1. Vsub /V = 1
′
7. T = 3/4 T
2. Vsub /V = 1/2 correct
′
8. T = 1/6 T
′
9. T = 2 T
10. T ′ = 5/3 T
4. Vsub /V = 3/2
Explanation:
First create a free-body diagram of the scenario. From this it is evident there are three
forces which act on the ice: the buoyant force
FB (which acts upward), the weight W of the
ice (which of course acts downward), and the
tension T of the rope (which pulls downward
5. Vsub /V = 1/3
6. Vsub /V = 3/4
7. depends on the density of lead
8. Vsub /V = 2/3
Version 058 – Test 2 – florin – (57850)
9. Vsub /V = 0 by the Momentum Principle
6
4. Cannot be determined with knowing the
relative angles of the ramps.
10. Vsub /V = 1/4
Explanation:
If we consider the system to be the lead
and styrofoam combination, the total mass is
3/2 m, which can be expressed in terms of the
density of styrofoam as 3/2 ρs V , where V is
the volume of the styrofoam block.
The buoyant force depends only on the volume displaced by the object and the density
of the fluid, i.e.,
FB = ρw Vsub g
The forces on our system are just the gravitational force and the buoyant force, which
must sum to zero since the system is not moving
3/2 mg − FB
3/2 mg
3/2 ρsV g
Vsub /V
Vsub /V
011
=0
= FB
= ρw Vsub g
= (3/2) (ρs/ρw )
= 1/2
10.0 points
5. v1 &gt; v2 &gt; v3 = v4
6. v4 &gt; v3 &gt; v2 &gt; v1
7. v4 &gt; v1 = v2 = v4
8. v1 &gt; v2 &gt; v3 &gt; v4
Explanation:
Using a system that includes the block
and the Earth, there is no work done by
the surroundings so the initial energy in
all four cases is all potential energy with
U = m g h and the final energy is all kinetic
then 1/2 m vf2 = m g h. Thus the kinetic energies and speeds of all four blocks are the same
when they are at y = 0.
012 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 so
that the thread’s trajectory describes a cone.
The acceleration of gravity is 9.8 m/s2 .
θ
g
ℓ
The figure shows four situations - one in
which an initially stationary block is dropped
and three in which the block is allowed to
slide down frictionless ramps again beginning
from rest. Rank the situations according to
the speed of the block at y = 0 from fastest to
slowest.
1. v1 = v2 = v3 &gt; v4
2. v1 &gt; v2 = v3 &gt; v4
3. v1 = v2 = v3 = v4 correct
r
v
m
What is the angular frequency ω of the ball
when it is in circular motion? Answer in terms
of g, ℓ and θ
q
1. ω = g sin2 θ/ℓ
p
2. ω = g/ℓ tan θ
p
3. ω = g/ℓ cos θ
p
4. ω = g/ℓ sin θ correct
Version 058 – Test 2 – florin – (57850)
5. ω =
6. ω =
7. ω =
8. ω =
9. ω =
10. ω =
q
p
7
and solving for ω yields
g/ℓ tan2 θ
g cos θ/ℓ
p
g tan θ/ℓ
q
g/ℓ cos2 θ
p
g sin θ/ℓ
q
g/ℓ sin2 θ
ω = v/r
q
=
.
Explanation:
Use the free body diagram below.
θ
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
2
m vball
.
ℓ cos θ
(2)
Solving (1) for T yields
mg
T =
sin θ
and substituting (3) into (2) gives
2
m vball
m g cos θ
=
.
sin θ
ℓ cos θ
Therefore,
v=
r
g ℓ cos2 θ
.
sin θ
m
&micro;s
2
F
m
&micro;k
Find F the magnitude of the maximum
force that can be applied so that the blocks
accelerate together.
Thoriz = m acentrip
T cos θ =
1
(1)
and
or
rℓ cos θ
g
=
ℓ sin θ
013 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 &micro;k , and the coefficient of
static friction between the two blocks is &micro;s
T
Tvertical = T sin θ = m g
g ℓ cos2 θ
sin θ
1. F = m g (2 &micro;s − &micro;k )
2. F = m g (2 &micro;s + &micro;k )
(3)
3. F = m g (&micro;s + 2 &micro;k )
4. F = m g (&micro;s − 2 &micro;k )
5. F = 1/2 m g (&micro;s + &micro;k )
6. F = 2 m g (&micro;k − &micro;s )
7. F = 2 m g (&micro;s + &micro;k ) correct
Version 058 – Test 2 – florin – (57850)
8. F = 1/2 m g (&micro;k − &micro;s )
Explanation:
a
1
a
2
m
m
F
fs
fs
fk
The only force on the top block is static
friction therefore the maximum value for m a
is given by m a = &micro;s m g Thus it must also
be true that the maximum value for F will
satisfy F − fs − fk = &micro;s m g. Solving for F
gives F = 2 m g (&micro;s + &micro;k )
014 10.0 points
A car of mass 519 kg goes over a hill with
radius 138 m at the top.
What is the maximum speed that the car
can go over the hill without leaving the road?
The acceleration of gravity is 9.8 m/s2 .
1. 27.5056
2. 35.1397
3. 41.4126
4. 37.6962
5. 29.1321
6. 38.2126
7. 36.775
8. 40.2119
9. 25.49
10. 31.7711
Explanation:
m v2
= mg − N
r
where N is the normal force acting on the
car from the ground. The car will fly off the
ground just when N = 0 so the maximum
speed allowed will be
√
gr
q
= (9.8 m/s2 )(138 m)
vmax =
= 36.775 m/s .
8
015 10.0 points
An outfielder throws a 1.23 kg baseball at a
speed of 40 m/s and an initial angle of 11.2◦ .
What is the kinetic energy of the ball at the
highest point of its motion?
1. 1743.75
2. 946.877
3. 1853.75
4. 2286.23
5. 8689.73
6. 3703.39
7. 2026.06
8. 1088.82
9. 1068.85
10. 1896.83
Explanation:
Given :
m = 1.23 kg ,
θ = 11.2◦ , and
v = 40 m/s .
At its highest point, the baseball has only a
horizontal velocity vx = v cos θ contributing
to its kinetic energy, so the kinetic energy is
1
m vx2
2
1
= m (v cos θ)2
2
1
= (1.23 kg) [(40 m/s) cos 11.2◦ ]2
2
= 946.877 J .
K=
016 10.0 points
Consider a coin which is tossed straight up
into the air. After it is released it moves
upward, reaches its highest point and falls
back down again.
What is its acceleration when the coin is
at its highest point, i.e. when its velocity is
zero? Take up to be the positive direction.
1. The acceleration is in the negative direction and constant. correct
Version 058 – Test 2 – florin – (57850)
2. The acceleration is zero.
3. The acceleration is in the positive direction and increasing.
4. The acceleration is in the negative direction and increasing.
5. The acceleration is in the negative direction and decreasing.
6. The acceleration is in the positive direction and decreasing.
7. The acceleration is in the positive direction and constant.
Explanation:
Gravity supplies the acceleration so it is a
constant negative value.
017 10.0 points
A block of mass 2 m has a velocity vector
given by ~v = v0 &lt; 3, 4, −2 &gt;. What is K the
kinetic energy of the mass?
1. K = 29 m v02 correct
2. K = 22 m v02
3. K = m v02 &lt; 18, 32, −8 &gt;
4. K =
54 m v02
5. K = m v02 &lt; 18, 32, 8 &gt;
6. K = 44 m v02
7. K = m v02 &lt; 9, 16, 4 &gt;
8. K = 50 m v02
9. K = 25 m v02
10. K = m v02 &lt; 9, 16, −4 &gt;
Explanation:
If M is the mass and ~v is the velocity, kinetic
energy is given by K = 1/2 M (vx2 + vy2 + vz2 )
9
which in this case is K = (1/2) 2 m v02 ((3)2 +
(4)2 + (−2)2 ) so K = 29 m v02
018 10.0 points
A metal sphere of radius r = 0.12 m is
dropped into the ocean and sinks to a depth
of d = 3510 m .
What is the total force exerted on the outside of the sphere at this depth? You may
assume the pressure is constant at all points
around the sphere. Atmospheric pressure is
1.013 &times; 105 Pa, the density of seawater is
ρ = 1024 kg/m3 and the acceleration due
to gravity is g = 9.8 m/s2 .
1. 11299400.0
2. 26440600.0
3. 5930400.0
4. 6392230.0
5. 7182320.0
6. 16113900.0
7. 8909660.0
8. 23535300.0
9. 22524000.0
10. 19013000.0
Correct answer: 6.39223 &times; 106 N.
Explanation:
Let :
r = 0.12 m ,
d = 3510 m
ρ = 1024 kg/m3 ,
g = 9.8 m/s2 .
and
The total pressure on the sphere is due to both
the atmospheric and the hydrostatic pressure
of the water:
P = P0 + ρ g d
= 1.013 &times; 105 Pa+
(1024 kg/m3 )(9.8 m/s2 )(3510 m)
= 3.53249 &times; 107 N/m2
Now that we have the pressure we can easily
calculate the total force on the outside of the
sphere as the product of the pressure times
the surface area of the sphere A, where
A = 4π r 2
Version 058 – Test 2 – florin – (57850)
F =P ∗A
= (3.53249 &times; 107 N/m2 )(4π)(0.12 m)2
= 6.39223 &times; 106 N
019 10.0 points
What is the minimum kinetic energy needed
to launch a payload of mass m to an altitude
that is one Earth radius, RE , above the surface of the Earth (the payload will then fall
back to Earth)? (Note that ME is the mass of
the Earth.)
G m ME
correct
RE
G m ME
2.
RE
G m ME
3. 2
RE
G m ME
4. 0.25
RE
G m ME
5. 0.1
RE
Explanation:
Energy is conserved:
1. 0.5
10
020 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
has velocity is vf when the block is at D.
D
F
m
&micro;k
θ
The coefficient of kinetic friction &micro;k is
1. &micro;k =
m g D sin θ + 1/2m vf2 − F D cos θ
K i + Ui = K f + Uf
5. &micro;k =
F D sin θ − m g cos θ − 1/2m vf2
The kinetic energy after the payload has
been launched to an altitude RE above the
Earth’s surface is
6. &micro;k =
Kf = 0 .
7. &micro;k =
The potential energy at Earth’s surface before
the launch is
Ui = −
G ME m
,
RE
and the potential energy at altitude RE above
the Earth’s surface after the launch is
Ui = −
G ME m
.
2RE
So, the minimum energy needed to launch
the payload to RE above Earth’s surface is
Ki =
G ME m G ME m
G m ME
−
=
RE
2RE
2 RE
m g D cos θ
F D cos θ − m g sin θ
2. &micro;k =
m g cos θ + F sin θ
F cos θ − m g sin θ − 1/2m vf2
3. &micro;k =
m g cos θ + F sin θ
F D cos θ − m g D cos θ − 1/2m vf2
4. &micro;k =
m g D sin θ + F D cos θ
correct
8. &micro;k =
m g sin θ
F D sin θ − m g sin θ − 1/2m vf2
m g cos θ
F D cos θ − m g D sin θ − 1/2m vf2
m g D cos θ + F D sin θ
F D cos θ − m g D sin θ − 1/2m vf2
m g D cos θ
Explanation:
The force of friction has a magnitude
Ff riction = &micro;k N . Since it is in the direction opposite to the motion, we get
Wf riction = −Ff riction D
= −&micro;k N D.
= −&micro;k (m g cos θ + F sin θ) D
Version 058 – Test 2 – florin – (57850)
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 sin θ .
The work done by the force F is
WF = F D cos θ .
From the Energy Principle we know that
Wnet = ∆K ,
WF + Wgrav + Wf riction =
1
m vf2 .
2
Thus
muk =
F D cos θ − m g D sin θ − 1/2m vf2
m g D cos θ + F D sin θ
11
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