Version 011 – midterm2 – shih – (58505)
This print-out should have 16 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
Two satellites have circular orbits about the
same planet. Their masses are m and 3 m,
respectively, with orbits of radii r and 2 r,
respectively.
GM m
, where
lite is the force of gravity F =
r2
M is the planet’s mass and m is the mass of
the satellite itself. Consequently, the satelGM
lite is in free fall with acceleration g = 2
r
directed towards the planet’s center.
For a circular orbit, the free-fall acceleration equals the centripetal acceleration
v2
ac =
, so
r
GM
v2
=
r2
r
r
2r
r
m
3m
What is the ratio of the orbital speeds of
the two satellites?
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
1
2
1
=√
3
=
v=
=2
1
= √ correct
2
1
,
r
R
r
r = 3R
Find the energy required to launch the
satellite from the Earth into its orbit. Make
use of conservation of energy.
1. Ki =
2. Ki =
3. Ki =
4. Ki =
Explanation:
The only force acting on an orbiting satel-
r
002 10.0 points
Denote the mass and the radius of the Earth
by M and R. A satellite of mass m is circulating around the Earth at a radius of r = 3R .
=9
√
= 2
1
9
1
=
3
√
= 3
GM
∝
r
regardless of the satellite’s mass m.
Consequently, for two satellites in circular
orbits around the same planet,
r
r
v3m
rm
r
1
=
=
= √ .
vm
r3m
2r
2
=3
=
1
5. Ki =
3
7
7
3
7
6
4
3
6
7
GM m
R
GM m
R
GM m
R
GM m
R
GM m
R
Version 011 – midterm2 – shih – (58505)
3 GM m
6. Ki =
5
R
5 GM m
7. Ki =
3
R
2 GM m
8. Ki =
3
R
5 GM m
9. Ki =
correct
6
R
6 GM m
10. Ki =
5
R
Explanation:
From conservation of energy,
K A + UA = K B + UB .
For the satellite in orbit,
KB =
1
Mm
m v2 = G
2
2r
with the gravitational force responsible for
the centripetal force:
m v2
r
=
GM m
.
r2
The potential energy at the planet’s surface A
Mm
is UA = −G
, and the potential energy
R
Mm
, so
at the orbit B is UB = −G
r
E B = K B + UB
Mm
Mm
Mm
=G
−G
= −G
.
2r
r
2r
Mm
Mm
+G
2r
R
Mm
Mm
= −G
+G
2 (3R)
R
GM m
1
5 GM m
=
1−
=
.
R
6
6
R
KA = EB − UA = −G
003 10.0 points
The Joule and the kilowatt-hour are both
units of energy.
11.6 kW · h is equivalent to how many
Joules?
2
1. 27000000.0
2. 24840000.0
3. 63360000.0
4. 41760000.0
5. 49320000.0
6. 46080000.0
7. 50760000.0
8. 33480000.0
9. 67320000.0
10. 56880000.0
Correct answer: 4.176 × 107 J.
Explanation:
1 W = 1 J/s
1000 W 3600 s
·
kW
1h
= 4.176 × 107 W · s
11.6 kW · h = (11.6 kW · h) ·
= 4.176 × 107 J .
004 (part 1 of 2) 10.0 points
The two blocks are connected by a light
string that passes over a frictionless pulley
with a negligible mass. The block of mass
m1 lies on a rough horizontal surface with a
constant coefficient of kinetic friction µ. This
block is connected to a spring with spring
constant k. The second block has a mass m2 .
The system is released from rest when the
spring is unstretched, and m2 falls a distance
h before it reaches the lowest point.
Note: When m2 is at the lowest point, its
velocity is zero.
h
k
m1
m1
µ
m2
h
m2
Consider the moment when m2 has descended by a distance s, where s is less than h.
At this moment the sum of the kinetic energy
for the two blocks K is given by
Version 011 – midterm2 – shih – (58505)
1. K = −(m1 + m2 ) g s +
1 2
k s + µ m1 g s .
2
1 2
k s − µ (m1 + m2 ) g s .
2
1
3. K = −m2 g s + k s2 + µ (m1 + m2 ) g s .
2
1
4. K = −m2 g s + k s2 + µ m1 g s .
2
1
5. K = (m1 + m2 ) g s + k s2 − µ m1 g s .
2
1
6. K = (m1 + m2 ) g s − k s2 − µ m1 g s .
2
1
7. K = m2 g s − k s2 − µ m1 g s . correct
2
1
8. K = (m1 + m2 ) g s − k s2 + µ m1 g s .
2
Explanation:
Basic Concepts:
Work-Energy Theorem
Spring Potential Energy
Frictional Force according to the WorkEnergy Theorem
Solution:
2. K = m2 g s −
3
g (m2 − µ m1 )
.
k
2 g [(m1 + m2 ) − µ m1 ]
.
4. h =
k
g [(m1 + m2 ) − µ m1 ]
5. h =
.
2k
g (m2 − µ m1 )
6. h =
.
2k
Explanation:
Based on the last equation of Part 1, at
s = h, KB = 0, so
1
µ m1 g h = m2 g h − k h2
2
In turn,
3. h =
h=
2 g (m2 − µ m1 )
.
k
006 10.0 points
A long string attached to a mass M forms
a simple pendulum. The string, however, is
weak enough so that it is likely to break at
some point in the oscillation if you let it swing.
You pull the mass back and start it oscillating.
ext
WA→B
= (KB − KA ) + (UBg − UAg )
K = KB
dis
= (UAg − UBg ) + (UAsp − UBsp ) − WA→B
= m2 g s −
1 2
k s − µ m1 g s .
2
005 (part 2 of 2) 10.0 points
The falling distance h where m2 stops is given
by
2 g (m2 − µ m1 )
. correct
1. h =
k
g [(m1 + m2 ) − µ m1 ]
2. h =
.
k
θ
φ
ar
For the present case, the external work
ext
WA→B
= 0, A corresponds to the initial state
and B the state where m2 has descended by a
distance s. The sum of the kinetic energy of
m1 plus that of m2 at B is given by
r
dis
+ (UBsp − UAsp ) + WA→B
g
m
a
v 6= 0 at
At what point in the cycle is the string most
likely to break?
1. Just when the mass passes through the
point where the string is vertical. correct
2. Just after release.
3. It is equally likely to break at all positions.
4. Just after the mass returns to the starting
point.
Version 011 – midterm2 – shih – (58505)
5. Just after the mass turns around to return.
Explanation:
Since the string is weak, we must examine
the tension in the string to determine at what
point it will break. The tension T is given
by T = M g cos θ. When theta is zero, the
tension is a maximum, so that the string will
be most likely to break when it is vertical.
007 10.0 points
A newly discovered planet has twice the mass
of the Earth, but the acceleration due to gravity on the new planet’s surface is exactly the
same as the acceleration due to gravity on the
Earth’s surface.
What is the radius Rp of the new planet in
terms of the radius R of Earth?
1
R
2
√
2. Rp = 2 R correct
1. Rp =
3. Rp = 4 R
4. Rp = 2 R
√
2
5. Rp =
R
2
Explanation:
From Newton’s second law and the law of
universal gravitation, the gravitational force
near the surface is
Mm
Fg = m g = G 2
r
GM
g= 2 .
r
Mp = 2 Me and gp = ge , so
G Mp
G Me
2 G Me
=
=
2
2
R
Rp
Rp2
1
2
= 2
2
R
Rp
√
Rp = 2 R .
008
10.0 points
4
An ore car of mass 42000 kg starts from rest
and rolls downhill on tracks from a mine. At
the end of the tracks, 21 m lower vertically, is
a horizontally situated spring with constant
5.9 × 105 N/m.
The acceleration of gravity is 9.8 m/s2 .
Ignore friction.
How much is the spring compressed in stopping the ore car?
1. 5.41298
2. 5.79019
3. 5.63231
4. 6.18223
5. 6.45509
6. 5.57193
7. 8.0829
8. 7.94607
9. 5.88048
10. 7.11316
Correct answer: 5.41298 m.
Explanation:
Energy is conserved, so the change of potential energy from when the car is at rest to
when it just hits the spring is
mgh =
1
m v2 .
2
The kinetic energy is then converted to potential energy in the spring as the cart comes
to rest. When the spring is fully compressed
by an amount d, all of the kinetic energy has
been converted to potential energy so
1
1
m v 2 = k d2 .
2
2
Thus,
1
k d2 = m g h ,
2
and solving for d we have
r
2mgh
d=
s k
=
2 (42000 kg) (9.8 m/s2 ) (21 m)
(5.9 × 105 N/m)
= 5.41298 m .
Version 011 – midterm2 – shih – (58505)
5
y directions, we have
009 10.0 points
A(n) 71 kg block is pushed along the ceiling
with a constant applied force of 1100 N that
acts at an angle of 63 ◦ with the horizontal,
as in the figure. The block accelerates to the
right at 5.6 m/s2 .
The acceleration of gravity is 9.8 m/s2 .
y:
x:
F sin θ − m g − N = 0
F cos θ − µ N = m a .
(1)
(2)
Solving Eq. 1 for N , we have
N = F sin θ − m g .
(3)
◦
2
= (1100 N) sin 63 − (71 kg) 9.8 m/s
= 284.307 N .
µ
which is positive since the normal force in Eq.
1 has a minus sign in front of it because it is
pointing downward.
71 kg
5.6 m/s2
110
0N
◦
63
What is the magnitude of the normal force
the ceiling exerts on the block?
1. 99.588
2. 97.4955
3. 419.539
4. 265.657
5. 104.161
6. 248.656
7. 284.307
8. 491.422
9. 118.48
10. 531.151
Correct answer: 284.307 N.
010 10.0 points
Given: Each block has masses m1 = m2 =
m3 = m and the coefficient of kinetic friction
is µ. The magnitude of F equals twice the
total frictional force.
Apply a horizontal force F in pushing an array of three identical blocks in the horizontal
plane (see sketch).
F
m1
m2
m3
µ
Find the force F23 with which the second
block is pushing the third block.
1. F23 = 3 m g
Explanation:
2. F23 = 4 m g
µ
µN
3. F23 = m g
F cos θ
m
N
F sin θ
mg
4. F23 = 2 m g
5. F23 = µ m g
6. F23 = 2 µ m g correct
◦
Given : θ = 63 ,
a = 5.6 m/s2 ,
m = 71 kg , and
F = 1100 N ,
Take up to be the positive y direction and
to-the-right to be the positive x direction.
Applying Newton’s second law in the x and
7. F23 = 3 µ m g
8. F23 = 4 µ m g
Explanation:
Basic Concepts: Newton’s 2nd law:
X
~ = m~a
F
Version 011 – midterm2 – shih – (58505)
Solution: Since all three blocks must have
the same acceleration (no block can “speed
ahead” or “lag behind”) we can write down
Newton’s 2nd Law for the whole system of
three blocks: (take right to be positive)
6
gravity on you is a maximum; i.e., at which
point would you weigh the most?
1. None of these
2. C; A
F − Ff = (m + m + m) a .
Call the force from the second block on
the third block F23 . The force of friction
Ff is equal to µN where N is the normal
force. Furthermore, the blocks are horizontal
~ is opposite in direction but equal in
so N
magnitude to the force of gravity 3 m g . This
means Ff = µ (3 m g). Now, F is given as
twice Ff , so
F = 6µmg
5. B and C; C
6. B and C; D
7. D; D
9. A and B; D
6µmg − 3µmg = 3ma.
or
a = µg.
The force of friction on the third block is
again µ N , but now this equals µ m g, since
we are concentrating on block #3 only. The
equation of motion (1) applied to the third
block gives (with positive still being taken to
be to the right)
F23 − µ m g = m a ,
or using a = µ g,
F23 = 2 µ m g .
011 10.0 points
Two planets with the same diameter are close
to each other, as shown. One planet has twice
the mass as the other planet.
m
4. B; D
8. A and D; A
and we have
A
3. A and D; D correct
B
C
2m
D
At which locations would both planets’
gravitational force pull on you in the same
direction? From among these four locations,
where would you stand so that the force of
Explanation:
At A and D, the planets are on the same
side of you, so they pull you in the same
direction, causing you to weigh more. At
B and C, you would weigh less because the
planets pull on you in opposite directions.
The closer you stand to the more massive
one, the more you weigh, so you would feel a
maximum weight at D.
012 10.0 points
A 19 kg child sits in a swing supported by two
chains, each 2.8 m long.
The acceleration of gravity is 9.8 m/s2 .
If the tension in each chain at the lowest
point is 149 N, find the child’s speed at the
lowest point.
Note: Neglect the mass of the seat.
1. 2.26127
2. 2.79285
3. 2.67238
4. 2.07394
5. 4.05904
6. 5.74108
7. 2.96606
8. 1.17288
9. 3.05382
10. 4.64217
Version 011 – midterm2 – shih – (58505)
7
Correct answer: 4.05904 m/s.
Explanation:
Let :
m = 19 kg ,
ℓ = 2.8 m ,
t = 149 N , and
g = 9.8 m/s2 .
At the lowest point, Newton’s second law
states
X
m v2
Fy = 2 T − m g −
= 0.
ℓ
Solving for v,
r
ℓ
(2 T − m g)
v=
m
2.8 m
[2 (149 N)
=
19 kg
1/2
2
−(19 kg) 9.8 m/s
= 4.05904 m/s .
013 10.0 points
A child pushes horizontally on a box of mass
m which moves with constant speed v across
a horizontal floor. The coefficient of friction
between the box and the floor is µ.
At what rate does the child do work on the
box?
1. P =
v
µmg
vi = 0
y
s
vf
θ
Use energy methods to determine the speed
of the crate when it reaches the bottom of the
ramp.
1. 4.192
2. 4.51889
3. 5.70318
4. 5.29384
5. 3.02214
6. 4.3879
7. 3.5963
8. 5.41842
9. 4.4691
10. 4.33276
Correct answer: 3.02214 m/s.
2. P = m g v
3. P = µ m g v correct
4. P =
014 10.0 points
A 6 kg crate slides down a ramp at a loading
dock. The ramp is 1 m in length and inclined
at an angle of 30◦ , as shown in the figure.
The crate starts from rest at the top, experiences a constant frictional force of magnitude
2 N, and continues to move a short distance
on the flat floor.
The acceleration of gravity is 9.8 m/s2 .
µmg
v
Explanation:
Since vi = 0, the initial kinetic energy is
zero. If the y coordinate is measured from
the bottom of the ramp, then yi = s sin θ =
0.5 m . Therefore, the total mechanical energy
of the system at the top is all potential energy:
5. P = µ m v 2
Explanation:
Since the box moves horizontally with constant speed, the force exerted by the child is
equal to the frictional force on the box µ m g.
Thus, the power generated by the child is
Ui = m g y i
= 29.4 J .
P =Fv
= µmgv.
When the crate reaches the bottom, the potential energy is zero because the elevation to
Version 011 – midterm2 – shih – (58505)
the crate is yf = 0. Therefore, the total mechanical energy at the bottom is all kinetic
energy:
1
Kf = m vf2 .
2
However, we cannot say that Ui = Kf in this
case, because there is an external nonconservative force that removes mechanical energy
from the system: the force of friction. In this
case, Wext = −f s, where s is the displacement along the ramp. (Remember that the
forces normal to the ramp do not do work
on the crate because they are perpendicular
to the displacement.) With f = 2 N and
s = 1 m , we obtain
Wext = −f s
= −2 J .
This says that some mechanical energy is lost
because of the presence of the retarding frictional force. So, we have
∆K + ∆U = Wext ,
or
1
m vf2 − m g yi = −f s ,
2
which gives us
s f
vf = 2 s g sin θ −
m
s
2N
2
◦
= 2(1 m) (9.8 m/s ) sin 30 −
6 kg
8
When the block is in equilibrium, each
spring is stretched an additional 0.22 m.
The force constant of each spring is most
nearly
1. k ≈ 98 N/m correct
2. k ≈ 33
3. k ≈ 41
4. k ≈ 39
5. k ≈ 21
6. k ≈ 40
7. k ≈ 29
8. k ≈ 70
9. k ≈ 12
10. k ≈ 140
Explanation:
Let : m = 4.4 kg ,
x = 0.22 m , and
g = 9.8 m/s2 .
= 3.02214 m/s .
kx
015 10.0 points
Two identical massless springs are hung from
a horizontal support. A block of mass 4.4 kg is
suspended from the pair of springs, as shown.
The acceleration of gravity is 9.8 m/s2 .
kx
m
mg
Due to equilibrium of the forces,
k
k
4.4 kg
2kx = mg
(4.4 kg) (9.8 m/s2 )
mg
=
k=
2x
2 (0.22 m)
= 98 N/m .
Version 011 – midterm2 – shih – (58505)
m
ℓ
8.8 m
016 10.0 points
A pendulum made of a string and a sphere
is able to swing in a vertical plane. The
pendulum is released from a position of 60 ◦
from vertical as shown in the figure below.
The string hits a peg located a distance d
below the point of suspension and rotates
about the peg.
The acceleration of gravity is 9.8 m/s2 .
9
r
θ
g
The radius of the sphere about the peg is
11
m
(1)
Ui = −m g ℓ cos θ
(2)
Relative to the the point of suspension, the
initial potential energy is
d
7 kg
r = ℓ −d.
and at the top of the circle, the final potential
energy is
60 ◦
9.8 m/s2
Find the smallest value of d (highest peg
position) in order for the sphere to swing in a
full circle centered on the peg.
1. 8.8
2. 8.29431
3. 7.10255
4. 8.40726
5. 7.66446
6. 7.8128
7. 6.8998
8. 8.0325
9. 9.24799
10. 8.51728
Correct answer: 8.8 m.
Explanation:
Let : m = 7 kg ,
ℓ = 11 m ,
θ = 60 ◦ , and
g = 9.8 m/s2 .
Uf = −m g (d − r) = −m g (ℓ − 2 r) .
From this, Kf =
energy, we find
−m g (ℓ − 2 r) +
(3)
m v2
, and conservation of
2
m v2
= −m g ℓ cos θ . (4)
2
T
mg
From a free-body diagram taken near the
top of the circle centered on the peg, we have
m v2
.
(5)
r
The sphere will be at its maximum height
at the top of the circle of radius r. For the
highest possible peg position, when the sphere
is at the top of the circle, the tension in the
string is zero, T = 0 . This condition just
barely lets the sphere complete a full circle.
Thus, Eq. 5 can be rewritten
−T − m g = −
mgr
m v2
=
.
(6)
2
2
m v2
in the Eq. 4,
Using Eq. 6 to eliminate
2
we obtain
mgr
−m g (ℓ − 2 r) +
= −m g ℓ cos θ
2
−2 ℓ + 4 r + r = −2 ℓ cos θ
Version 011 – midterm2 – shih – (58505)
5 r = 2 ℓ (1 − cos θ) .
(7)
(8)
Therefore
2ℓ r=
1 − cos θ
5
2 (11 m) =
1 − cos 60 ◦
5
= 2.2 m , so
d= ℓ−r
= (11 m) − (2.2 m)
= 8.8 m .
10