SCI-111 Assignment #3 Solutions
Text Problems:
Chapter Number
3
Questions for Thought
Group B Problems
1. How much work is done when a force of 800.0 N is exerted while pushing a crate across a
level floor for a distance of 1.5 m?
This problem is a simple application of the definition of work, where the applied force is 800 N
and the displacement along the direction of the force is 1.5 m.
W  F  x  800 N 1.5 m  1,200 N  m  1,200 J
There was 1,200 J of work done.
2. A force of 400.0 N is exerted on a 1,250 N car while moving it a distance for 3.0 m. How
much work was done on the car?
This problem is also simple application of the definition of work, where the applied force is 400
N and the displacement along the direction of the force is 3.0 m. The weight of the car (1,250 N)
is irrelevant to the problem
W  F  x  400 N  3.0 m  1,200 N  m  1,200 J
There was 1,200 J of work done on the car.
3. A 5.0 kg textbook is raised a distance of 30.0 cm as a student prepares to leave for school.
How much work did the student do on the book?
This problem is also simple application of the definition of work, where the applied force is the
weight (mg) of the textbook and the displacement along the direction of the force is 0.30 m.
W  F  x  mg  x  5.0 kg  9.8
m
m
 0.30 m  14.7 kg  2  m  14.7 N  m  14.7 J
2
s
s
There was 14.7 J of work done by the student in lifting the textbook.
4. An electric hoist does 196,000 J of work in raising a 250.0 kg load. How high was the load
lifted?
This problem is also simple application of the definition of work, where the applied force is the
weight (mg) of the t250 kg load. The displacement along the direction of the force is unknown,
but the work done during that displacement is known to be 196,000 J. We need to rearrange the
definition of work W  F  x  mg  x to solve for the displacement x.
W  F  x  mg  x
W
196,000 J
J
Nm
 x 

 80
 80
 80 m
kg  m
mg 250 kg  9.8 m
N
s2
s2
The hoist lifted the 250 kg load a distance of 80 meters.
6. (a) How many horsepower is a 250 Watt light bulb? (b) How many Watts is a 230 hp car?
Use the conversion 1 hp = 746 Watts for both parts of this problem.
 1 hp

250 Watts  250 Watts  
  0.335 hp
 746 Watts 
 746 Watts 
  171,580 Watts  171 kW
230 hp  230 hp  
 1 hp 
A 250 Watt light bulb is equivalent to 0.335 hp and a 230 hp car is equivalent to 171 kilowatts.
7. What is the kinetic energy of a 30-gram bullet that is traveling at 200 m/s?
This is a simple KE calculation. Just use the definition of Kinetic Energy.
1
1
m
kg  m 2

KE  mv 2   0.030kg   200   600
 600 J
2
2
s
s2

The bullet has 600 Joules of kinetic energy.
2
8. How much work will be done by a 30-gram bullet that is traveling at 200 m/s?
From energy conservation, the amount of work the bullet can do is at most equal to its kinetic
energy. So the most work the bullet can do is 600 Joules.
9. A force of 50 pounds is used to push a box 10.0 ft across a level floor.
a. How much work was done on the box?
I prefer that you use metric units to answer this question. The 50 lbs of force becomes
222.4 N, and the 10.0 ft becomes 3.05 m in the metric system. The work done is just as
before:
W  F  x  222.4 N  3.05 m  678.3 N  m  678.3 J
The work done on the box is 678.3 Joules, or 500.6 ft-lbs.
b. What is the change in potential energy as a result of this move?
There is not change in the potential energy of the box since there is not change in height
pushing across a level floor. Where did the work go in pushing the box? It turned into heat
in the floor and the bottom of the box from friction.
10. How much work is done in raising a 50 kg crate a distance of 1.5 m above a storeroom
floor? What is the change of potential energy as a result of this move? How much kinetic
energy will the crate have as it falls and hits the floor?
This problem is also simple application of the definition of work, where the applied force is the
weight (mg) of the crate and the displacement along the direction of the force is 1.5 m.
W  F  x  mg  x  50 kg  9.8
m
m
1.5 m  735 kg  2  m  735 N  m  735 J
2
s
s
There was 735 J of work done in lifting the crate. This work effected a equal change in the
potential energy in that the PE of the crate increased by 735 J. If the crate were to fall to the
floor it would have 735 J of KE, as PE transformed into KE, just before it hit the floor.
11. What is the kinetic energy of a 60.0 g tennis ball approaching a tennis racket at 20.0 m/s?
This is a simple KE calculation. Just use the definition of Kinetic Energy.
1 2 1
kg  m 2
 m
mv   0.060kg   20   12
 12 J
2
2
s2
 s
2
KE 
The tennis ball has 12 Joules of kinetic energy.
12. What is the kinetic energy of a 1,500.0 kg car with a velocity of 72.0 km/hr? How much
work must be done on this car to bring it to a complete stop?
This is another simple KE calculation. Just use the definition of Kinetic Energy, but you must
change the speed into m/s. 72 km/hr = 20 m/s (See conversions in front of text)
1 2 1
kg  m 2
 m
KE  mv  1,500kg   20   300,000
 300,000 J
2
2
s2
 s
2
The car has 300,000 Joules of kinetic energy. As a result the car will be brought to a complete
stop only after this 300,000 J kinetic energy has be transformed into work hopefully by heating
the brakes, or less fortunately, by breaking a telephone pole.
13. The driver of an 800.0 kg car decides to double his speed from 20.0 m/s to 40.0 m/s. What
effect would this have on the amount of work required to stop the car, that is, on the kinetic
energy of the car?
This problem does not need a calculation done to answer it. It asks you to recognize that since
the kinetic energy depends of the square of the velocity, that doubling the velocity will NOT
double the kinetic energy, but quadruple the kinetic energy. Algebraically, it works as shown
below:
KE  
1
1
1
1
2
m  v2  m  2v   m  4v 2  4  m  v 2  4  KE
2
2
2
2
The new KE (primed) is the KE after doubling the velocity. You can see that the new KE is four
times the original KE.
14. Compare the kinetic energy of an 800.0 kg car moving at 20.0 m/s to the kinetic energy of a
1,600 kg car moving at an identical speed.
This problem is very similar to the previous problem. In this case, we note that the Kinetic
energy of an object depends linearly on its mass; if the mass is increased the KE increases
proportionately. Mathematically it looks like:
KE  
1
1
1
m  v 2  2m   v 2  2  m  v 2  2  KE
2
2
2
The new KE (primed) is the KE after doubling the mass. You can see that the new KE is twice
the original KE.
15. A 175.0 lb hiker is able to ascend a 1,980 ft high slope in 1 hour and 45 minutes. How
much work did the hiker do? What was the average power output in hp?
I prefer that you use metric units to answer this question. The 175 lbs of force becomes 778.4 N,
the 1,980 ft becomes 603.9 m in the metric system and the 1 hr 45 min time is 6,300 seconds.
The work done is just as before, where the force applied is the hiker’s weight:
W  F  x  778.4 N  603.9 m  4.70 105 N  m  4.70 105 J
The hiker did 4.70 x 105 J or 3.47 x 105 ft-lb of work climbing the slope.
The average power output is simply found using the definition of power
W 4.70 105 J
P

 74.61Watts
t
6,300s
The hiker’s average power output was 74.6 Watts or 0.1 horsepower, using the conversion table
from the front of the text.
16) (a) How many seconds will it take a 10.0 hp motor to lift a 2,000.0 pound elevator a distance of
20 feet? (b) What was the average speed of the elevator?
Let’s do the algebra first, then worry about the numbers. By definition, Power 
Work
.
Time
W F d

t
t
F d
2,000 lb  20 ft
t 

 7.27 s
ft  lb
P
10  550 
s
P
The average velocity is given by v 
d
20 ft
ft

 2.75 .
t 7.27 s
s
It will take 7.27 seconds for a 10.0 hp motor to lift a 2,000.0 pound elevator a distance of 20 feet
with an average speed of 2.75 ft/s?
17) A ball is dropped from 20 ft above the ground.
A) At what height will the ball have half kinetic energy and half potential energy?
The figure to the right illustrates a similar problem
with a ball dropped from 20 meters (rather than feet).
As the ball falls its PE decreases while its KE grows.
Half way down, the PE will be reduced to half its
original value. Where did that PE go? Into KE. Thus
at the halfway point, the energy will be half the
original PE and half KE. Note that the velocity of the
ball will not be half at the middle point, since KE depends on v2.
B) Using energy considerations only, what is the velocity of the ball just as it hits the ground?
It is convenient to express 20 ft as 6.1m. At the bottom of the fall all the PE will be converted
into KE. Thus
PE Top  KEBottom
mgh 
1 2
mv
2
 v  2 gh  2  9.8
m
m
 6.1 m  10.84
2
s
s
The speed of the ball will be 10.84 m/s.
18) What is the velocity of a 60.0 kg jogger with a kinetic energy of 1,080.0 Joules?
1 2
mv
2
2  KE
v 

m
KE 
2 1,080 J
 6.00 m/s
60.0 kg
The velocity of the jogger is 6.00 m/s.
19) A small sports car and a pickup truck start coasting down a 10.0 m hill together, side-by-side.
Assuming no friction, what is the velocity of each vehicle at the bottom of the hill?
Both vehicles have a potential energy at the top of the hill given by mgh where m is the mass of each
respective vehicle. The pickup truck has more potential energy due to its greater mass. At the
bottom of the hill, assuming no friction of any kind, both vehicles will have a KE of ½ mv2. Again,
the pickup will have more KE since it started with more PE. However, if you examine the energy
balance equation
PE  KE
mgh 
1 2
mv
2
1 2
v
2
v  2 gh
gh 
You see that the final velocity does not depend on mass. Both vehicles will have the same velocity,
in the absence of friction, at the bottom of the hill.
20) A 70.0 kg student runs up the stairs of a football stadium to a height of 10.0 m above the
ground in 10.0 s. What is the power of the student in kW?
The work done by the student is his/her weight times the change in height:
W  F  x  mg  x  70 kg  9.8
m
10.0 m  6,860 J
s2
The student did 6,860 Joules of work in climbing the stairs.
The average power output is simply found using the definition of power
P
W 6,860 J

 686 Watts
t
10 s
The student’s power output was 686 Watts or 0.686kW.
Instructor Assigned Topic: In a long paragraph, explain fully why space shuttle astronauts
i) Cannot be in an environment with truly “zero gravity”,
ii) Are in free fall while they orbit the Earth, and
iii) Why they appear weightless.
Points to include
1) Space shuttle astronauts cannot be in an environment with truly “zero gravity” because
A) Gravity is present much farther away from the Earth to hold the Moon in orbit, so it must be even
stronger closer to the Earth that is the source of gravity in this situation, and
B) Although gravity weakens rapidly with distance (1/r2), it will only vanish if you are infinitely
away from every other body.
2) Space shuttle astronauts are in free fall while they orbit the Earth because
A) As Newton’ demonstrated in his cannon on the mountain experiment, objects in orbit are really
falling to the Earth.
B) If objects in orbit were not falling toward Earth, they would continue in a straight-line motion off
into space.
C) The reason the Space Shuttle and its astronauts do not hit the Earth is that they have a very high
horizontal (tangential) velocity of about 8 km/s, so that the Spherical Earth effectively curves out
of the path of the Shuttle as it falls.
3) Space shuttle astronauts appear weightless because
A) The astronauts are falling at the same rate as the Space Shuttle around them.
B) We know water, that has weight, will spill out of a hole in the bottom of a cup holding it.
C) However, if the cup is dropped while full of water, the water will cease spilling from the hole for
the duration of the fall, as if the water had no weight.
D) When two objects fall together, there is no relative motion.
E) So the Space Shuttle with the astronauts inside it fall together, thus there is no relative motion
between them and the astronauts appear weightless.