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Answer, Key – Homework 9 – David McIntyre – 45123 – May 10, 2004
1
This print-out should have 16 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before making your selection. The due time is
Central time.
Chapter 8 problems.
where M = me + ml = 1740 kg is the total
mass (elevator plus load). Therefore,
001 (part 1 of 1) 0 points
A commonly used unit is the kilowatt hour.
The physical quantity measured in kilowatt
hours is
Using the equation
T =f +Mg
= 4050 N + (1740 kg) (9.8 m/s2 )
= 21102 N .
P =
dW
~ · ~v
=F
dt
~ is in the same direction as
and the fact that T
~v gives
1. None of these
2. current.
~ · ~v = T v
P =T
= (21102 N) (2.86 m/s)
= 60351.7 W .
3. power.
4. force.
5. work (energy). correct
Explanation:
A Watt, which measures power, is a Joule
per second. When power is multiplied by
time, the result is in Joules, which measures
energy or work.
¶
103 J
(3600 s)
1 kilowatt hr =
s
= 3600 × 103 J
µ
003 (part 2 of 2) 0 points
What power must the motor deliver at a instantaneous speed of 2.86 m/s if the elevator
is designed to provide an upward acceleration
of 1.11 m/s2 ?
Correct answer: 65875.5 W.
Explanation:
Applying Newton’s second law to the elevator gives
T − f − M g = M a.
Thus
002 (part 1 of 2) 0 points
An elevator has a mass of 1040 kg and carries a maximum load of 700 kg. A constant
frictional force of 4050 N retards its motion
upward.
The acceleration of gravity is 9.8 m/s2 .
What must be the minimum power delivered by the motor to lift the elevator at a
constant speed of 2.86 m/s?
Correct answer: 60351.7 W.
Explanation:
~ that
The motor must supply the force T
pulls the elevator upward. From Newton’s
second law and from the fact that the acceleration is zero (since v is constant), we obtain
T − f − M g = 0,
T = M (a + g) + f
= (1740 kg) (1.11 m/s2 + 9.8 m/s2 )
+ 4050 N
= 23033.4 N .
and the required instantaneous power is
P =Tv
= (23033.4 N) (2.86 m/s)
= 65875.5 W .
004 (part 1 of 3) 4 points
A 1680 kg car accelerates uniformly from rest
to 10 m/s in 3.08 s.
Answer, Key – Homework 9 – David McIntyre – 45123 – May 10, 2004
Find the work done on the car in this time.
Correct answer: 84000 J.
Explanation:
The work is given by the change in kinetic
energy of the car
W = ∆K
1
= m v2
2
1
= (1680 kg) (10 m/s)2
2
= 84000 J .
005 (part 2 of 3) 3 points
Find the average power delivered by the engine in the first 1.0472 s.
Correct answer: 9272.73 W.
Explanation:
The acceleration is
v
a=
t
10 m/s
=
3.08 s
= 3.24675 m/s2 .
The instantaneous power is given by
~ · ~v .
P=F
The force can be expressed as
F = m a.
The acceleration is constant. The velocity is
v = a t, therefore
P=mav
= m a (a t)
= m a2 t
= (1680 kg) (3.24675 m/s2 )2 (1.54 s)
= 27272.7 W .
007 (part 1 of 1) 0 points
Use the potential energy vs. position plot
shown below to answer the following question.
A particle is released from point A and
moves in the potential U (x). Suppose the
mechanical energy of the system is conserved.
Since v = a t, the work is
1
W = m v2
2
1
= m a 2 t2
2
1
= (1680 kg) (3.24675 m/s2 )2 (1.0472 s)2
2
= 9710.4 J .
The average power is given by the work divided by the time.
W
t
9710.4 J
=
1.0472 s
= 9272.73 W .
P avg =
T
A
U (x)
x
S
Z
At which position(s) will the kinetic energy
of the particle have its maximum value?
1. The particle remains stationary at point
A.
2. Point Z. correct
3. Point S.
006 (part 3 of 3) 3 points
Find the instantaneous power delivered by the
engine at t2 = 1.54 s.
Correct answer: 27272.7 W.
Explanation:
2
4. Point T .
5. Points S and Z.
Explanation:
Answer, Key – Homework 9 – David McIntyre – 45123 – May 10, 2004
µ
¶
1
Since the total energy is constant, the max2
−f d = (0 − 0) + 0 − k L
imum kinetic energy will occur when the po2
tential energy is a minimum. Thus point Z is
1
µ m g d = k L2 .
the correct answer.
2
008 (part 1 of 1) 0 points
Solving for d gives
A(n) 1.2 kg block is pushed by an external
force against a spring with spring constant
2217 N/m until the spring is compressed by
0.2 m from its uncompressed length (x = 0).
The block rests on a horizontal plane that has
a coefficient of kinetic friction of 0.5. The
external force is then rapidly removed so that
the compressed spring can push the mass.
The acceleration of gravity is 9.8 m/s2 .
Remember: The block is not attached to
the spring.
d=
0.2 m
2217 N/m
x
1.2 kg
µ = 0.5
x=0
After the block is released, how far along
the plane will the block move before coming
to a stop?
Correct answer: 7.54082 m.
Explanation:
Basic Concepts: Spring Potential Energy.
Frictional Forces.
Work-Energy Theorem.
Let :
L = 0.2 m ,
m = 1.2 kg ,
k = 2217 N/m ,
µ = 0.5 .
k L2
2µmg
(2217 N/m) (0.2 m)2
=
2 (0.5) (1.2 kg) (9.8 m/s2 )
= 7.54082 m .
009 (part 1 of 1) 10 points
A 3.95 kg mass starts from rest and slides a
distance d down a frictionless 13.4 ◦ incline,
where it contacts an unstressed spring of negligible mass as shown in the figure. The
mass slides an additional 0.313 m as it is
brought momentarily to rest by compressing
the spring of force constant 669 N/m.
The acceleration of gravity is 9.8 m/s2 .
m
d
k
θ
Find the initial separation d between mass
and spring.
Correct answer: 3.33997 m.
Explanation:
From the conservation of energy:
Ki + U i = K f + U f
and
Solution: Examining the vertical forces,
we observe that the normal force is N = m g
and so the friction force is f = µ m g . Call the
distance the block travels d . Then the work
W done by the non-conservative friction force
is
W = ∆K + ∆U
3
we obtain
mg(d + `) sin θ =
k`2
2
Therefore
k`2
−`
2mg sin θ
(669 N/m)(0.313 m)2
=
− 0.313 m
2(3.95 kg)(9.8 m/s2 ) sin 13.4◦
= 3.33997 m
d=
Answer, Key – Homework 9 – David McIntyre – 45123 – May 10, 2004
010 (part 1 of 1) 10 points
A 13 kg block is released from point A at
height h = 7.1 m as shown. The track is frictionless except for the portion BC, of length
4.2 m. The block travels down the track,
hits a spring of force constant k = 1632 N/m,
and compresses it 0.19 m from its equilibrium
position before coming to rest momentarily.
The acceleration of gravity is 9.8 m/s2 .
A
h
B
C
Determine the coefficient of kinetic friction
between surface BC and block.
Correct answer: 1.63542 .
Explanation:
From the conservation of energy:
∆E = Wf
or
kx2
− mgh = −µmgd
2
Hence,
kx2
2
µ=
mgd
(1632 N/m)(0.19 m)2
7.1 m −
2 (13 kg)(9.8 m/s2 )
=
4.2 m
= 1.63542
4
What is the total energy of the satellite in
orbit?
Correct answer: −2.14428 × 1010 J.
Explanation:
For a satellite orbiting the earth in a circular
orbit, its total energy is the sum of the potential energy and kinetic energy.
GM m 1
+ m vo2
Re + A 2
But since the required centripetal force is supplied by the gravitational force,
Eorbit = −
m vo2
GM m
=
(Re + A)2
Re + A
Use this relation to eliminate the vo in the
Eorbit formula, we have
1 GM m
2 Re + A
1
= − × 6.67259 × 10−11 N m2 /kg2
2
5.98 × 1024 kg × 746 kg
×
6.37 × 106 m + 571000 m
= −2.14428 × 1010 J
Eorbit = −
mgh −
011 (part 1 of 3) 4 points
012 (part 2 of 3) 3 points
What is the total energy of the satellite just
before it hits the ground?
Correct answer: −3.74049 × 1010 J.
Explanation:
The total energy just before the satellite his
the ground is the sum of potential energy and
kinetic energy.
Given:
Mearth = 5.98 × 1024 kg
Rearth = 6.37 × 106 m
A satellite of mass 746 kg is in a circular orbit at an altitude of 571 km above the earth’s
surface. Because of air friction, the satellite
eventually is brought to the earth’s surface, it
hits the earth with a velocity of 5 km/s. Let
the gravitational potential energy be zero at
r = ∞. The universal gravitational constant
G = 6.67259 × 10−11 N m2 /kg2 .
GM m 1
+ m v2
Re
2
= −6.67259 × 10−11 N m2 /kg2
5.98 × 1024 kg × 746 kg
×
6.37 × 106 m
1
+ (746 kg)(5000 m/s)2
2
= −3.74049 × 1010 J
Eground = −
013 (part 3 of 3) 3 points
Answer, Key – Homework 9 – David McIntyre – 45123 – May 10, 2004
What is the work done by friction?
Correct answer: −1.59621 × 1010 J.
Explanation:
From work energy theorem
5
Solve for the kinetic energy
K=
1
G Me m
m v2 =
2
6 Re
∆W = Ef − Ei
In this case, it is
∆W = Eground − Eorbit
= (−3.74049 × 1010 J)
− (−2.14428 × 1010 J)
= −1.59621 × 1010 J
014 (part 1 of 3) 0 points
What is the kinetic energy of a satellite of
mass m which is in a circular orbit of radius
3 Re about the earth?
015 (part 2 of 3) 0 points
What is the total energy of the satellite?
1. E =
2. E = 3 G Me m
4. E
6. E
7. E
8. E
4. K = 3 G Me m
5. K =
6. K =
7. K =
8. K =
G Me m
correct
6 Re
G Me m
Re2
G Me m
3 Re
m v2
6 Re
9. K = 3 m g Re
10. K = m g Re
Explanation:
The acceleration of the satellite in circular
orbit of radius 3 Re is
v2
ac =
3 Re
So the force on the satellite is
F = m ac =
G Me m
m v2
=
3 Re
(3 Re )2
1
m v2
2
G Me m
= 3 m g Re +
3 Re
G Me m
=−
3 Re
G Me m
correct
=−
6 Re
G Me m
=−
6 Re2
G Me m
=
3 Re
Gm
=
Re
3. E = m g Re +
5. E
G Me m
1. K = −
6 Re2
m v2
2. K =
3 Re
G Me m
3. K = −
3 Re
G Me m
Re
9. E
Explanation:
The potential energy of the satellite is
U =−
G Me m
3 Re
So the total energy is
E =K +U
G Me m G Me m
−
=
6 Re
3 Re
G Me m
=−
.
6 Re
016 (part 3 of 3) 0 points
Given: The universal gravitational constant
G = 6.67 × 10−11 N m2 /kg2 , the mass of the
Earth Me = 5.98 × 1024 kg and its radius
Re = 6.37 × 106 m.
Answer, Key – Homework 9 – David McIntyre – 45123 – May 10, 2004
How much work must an external force do
on the satellite to move it from a circular orbit
of radius 2 Re to 3 Re , if its mass is 2000 kg?
Correct answer: 1.04361 × 1010 J.
Explanation:
The work done by an external force to move
the satellite from the closer orbit to the further orbit will be the work against gravity (a
positive number which yields the change in
potential energy) plus the change in kinetic
energy (a negative number since the kinetic
energy is smaller in the orbit with the greatest
radius).
W = E f − Ei
¶
µ
G Me m
G Me m
=−
− −
6 Re
4 Re
G Me m
=
12 Re
1
=
(6.67 × 10−11 N m2 /kg2 )
12µ
¶
(5.98 × 1024 kg)(2000 kg)
×
6.37 × 106 m
= 1.04361 × 1010 J
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