B. Power
C. Kinetic Energy
D. Potential Energy
E. Interconversion of KE and PE;Conservation of Energy
Review Problems for Chapter 5
Previous chapters
emphasized forces and their role in changing or maintaining motion. Although the motion of any object can be
completely described if the forces acting on it are known, this method of
predicting motion may not be practical.
When the forces acting on an object are constant, the acceleration is
constant and a simple set of equations completely describe the motion. However, if the forces are the least bit
complicated or if the forces are unknown, it may be too difficult, or even
impossible, to explain and predict the motion.
In this chapter a powerful new principle, conservation of energy, is
introduced. This principle simplifies
the description of motion when the forces are complicated. It may even be employed when the forces are
unknown. In developing the principle of
conservation of energy, we first introduce the concept of work, which will lead
is to the concept of energy.
There are many ways in
which the term “work” is used in our daily lives. A dictionary will give over thirty definitions of this word. In physics the term “work” is a technical
term with a precise meaning and a narrow definition. We define work in the following way:
The
work done by a force acting on an object is equal to the product of the
component of this force in the direction of the displacement and the
displacement of the object.
Or
(5.1)
Where Fd =
component of the force in the direction of the displacement, and d =
displacement.
Note: Many textbooks will use “W” to represent
work when using Equation 5.1. This inevitably leads to confusion since the
letter “W” is also used to represent weight.
To avoid this problem we will always write out the word “Work” in this
textbook.
When we examine Equation
5.1 we see that there are three requirements for doing work as we have narrowly
defined this technical term. The first
is that work requires force. So using
this definition you do no work when you sit and study, not matter how fatiguing
the assignment may be. The second
requirement for work is that there must be a displacement.
As a result no work is being done if you hold a heavy load of books
on your arms. This is because, except to the extent that
you muscles expand and contract in maintaining your position, the force is
not exerted through any distance at all.
The final requirement, that there must be a force in the direction
of displacement, tells us that even if you place your books in a backpack
and carry them across campus no work is involved in the technical sense we
have introduces. In this case there
is a force and a displacement, but the book moves no distance in the direction
of force.
The units in which we
measure work are obtained from its defining equation. They are as follows:
Work = Fd ´ d
MKS Joule = Newton ´ meter
CGS erg = dyne ´ centimeter
English ft-lb = pound ´ foot
We should never mix
the units in this equation. All
quantities must be in a single system of units.
Examples:
1. Three
students push their car which has run out of gas. If each student pushes with a
force
of 500 N and the car is pushed a distance of 40 meters, find the work done by
the students.
This
problem is simply solved by using the Eq (5.1). The total force in the direction of the 40 m displacement is 1500
N.
Work
= Fdd = 1500 N x 40m = 60,000 Nm
Work = 60,000 J
2. The
distance from the ground floor of a building to the second floor is 10 ft. If a 160 lb. Student climbs these stairs,
how much work does he do?
3. A
student pushes on a heavy box with a force of 400 N and moves it a distance of
5 meters. If the force is applied at an
angle of 37 degrees as shown below, find the work done by the student.
The definition of work
is independent of time. This means that
a student walking up a flight of stairs does the same amount of work if he runs
up the stairs. In each case the force
exerted and the distance the student moves is the same. The difference is the time taken to do the
work. The rate of doing work is called
power, and it is defined as the work done per unit of time.
(5.2)
The units of power can
be any unit of work divided by any unit of time. The most common units of power are:
MKS system: 
English system: 
Other units are also
used frequently for power. For example,
a kilowatt is 1000 times larger than a Watt.
The unit horsepower (hp) is frequently used in the British system. This
quantity is defined
1 horsepower =
550 ft-lb/sec = 746 Watt
Examples:
1. Calculate
the work done by a 50 kg student who climbs one flight of stairs (height = 3
meters). Then determine the rate at
which she works if she walks in 5 seconds, or runs in at a time of 1 second.
The
work done is determined using Eq. (5.1).
Using
Eq. (5.2) the rate at which work is done is easily determined. If the time required to go up the stairs is
5 seconds then
If
she runs up in one second we have
2. A
manually operated winch line is used to lift a 200 kg mass to the roof of a
building. Assuming you can work at a
steady rate of 200 Watts how long does it take to lift the weight 10 meters?
First
determine the amount of work done.
Now
find the time required by using Eq (5.2) written as
Note: A person in excellent condition can develop a power of 200
Watts with his arms or 1500 Watts with her legs for a brief time (10-15
sec). For most people either task would
be too demanding.
Consider a pitcher
throwing a baseball to a catcher. While
the ball of mass m is in his hand, the pitcher exerts a force on it over some
distance and releases the baseball with velocity v. Clearly the pitcher has done work on the ball, and this work has
resulted in a change in the motion of the baseball. The ball then travels to the catcher who stops it by exerting a
force with his mitt as the mitt is displaced.
The simple example illustrates an obvious relation between work and
motion: The pitcher does work on the ball and changes the state of motion of
the ball. Then the ball does work on
the mitt when its motion is stopped.
Work
must be done to set any object in motion, and any moving object can do
work. We define energy as the
ability to do work.
It should be clear
that there must be some relation between the motion of an object and the
concepts of work and energy. To see how
these are related, consider the special case of a constant force acting on an
object of mass m which is initially at rest.
Suppose the object is subjected to this force for a time t and it is
displaced by an amount d.
During the
displacement, the force does work given by
We wish to relate the
work to the change in motion of the object caused by this force. Since the force is related to the
acceleration of the object by F=ma. The
above equation for work can then be written
Now the distance an
object moves while accelerating from rest is
The work equation may
now be written as
Finally, recall the
kinematics equation
Substitute this result
into the work equation above
This result is very
important. It tells us that the work
required to set an object into motion is given by the simple expression ½mv2. This quantity is called kinetic energy
(KE). Because kinetic energy involves
simple quantities to measure (mass and velocity) it allows us to calculate the
work done on an object and the ability of an object to do work without using
forces. The importance of this result
can be illustrated by examining the simple task of throwing a ball. It would be very difficult to measure the
force exerted on a ball while you are in the act of throwing it. As a result, Equation (5.1) could not be
easily used to determine the work done on the ball. However, if we know the ball’s mass and velocity after it is
released, it is easy to determine its kinetic energy.
We define the
following:
Kinetic
energy (KE) is 1/2mv2, and it represents the amount of work done to
set a body in motion. It also
represents the amount of work a body of mass m can do as a result of its motion
with speed v.
(5.3)
Note: We have defined kinetic energy in terms of the amount of work
required to set a body in motion. The
unit of energy is therefore the same as the unit of work (Joule or foot-pound).
Examples: 1. How much
work must a baseball pitcher do to throw a baseball (mass = 0.01 slug) at a
velocity of 90 mph (132 ft/sec)?
Since
the work required to set an object in motion is simply the kinetic energy it
has when it is moving we have
2. How much work must an 80 kg runner do to
accelerate from rest to a velocity of 5 m/sec?
3. Determine the kinetic energy of a Volkswagon
(m = 1000 kg) moving at 10 m/sec (22 mph) and at 20 m/sec (44 mph).
Eq 5.3 defines kinetic energy. If the Volkswagon is traveling at 10 m/sec
we have
and if it moves at 20 m/sec we have
Questions:
1.
How much work is required to accelerate a Volkswagon
from rest to 22 mph?
2.
How much work is required to accelerate a Volkswagon
from rest to 44 mph?
3.
How much work is required to accelerate a Volkswagon
from 22 mph to 44 mph?
In the previous
section we saw how a body is put in motion by doing work on it. As this body is slowed to a stop it can then
do work. Consider now what happens when
we raise an object and release it. Work
is required to raise the object; if dropped, it can crush another object or
drive a nail into a board. Potential
energy (PE) refers to the ability of a system to do work by virtue of its
position.
The
potential energy (PE) possessed by an object is the amount of work the object
is capable of doing because of its position.
For an object on the
earth, the force required to lift it is equal to the body’s weight = mg. If we lift this object a height = h, then we
do work. As a result, we have
(5.4)
where m is the mass of
the body and g is the acceleration due to gravity. Notice that h is a difference between two heights. It measures how far the object has been
lifted above a certain level. The units
of potential energy are the same as those for work (Joules or foot-pounds).
Examples:
1.
Calculate the increase in potential energy when a weightlifter raises a 250 lb
weight a height of 6 ft.
2. In
the 1988 Winter Olympics a Soviet couple, Ekaterina Gordeeva and Sergei
Grinkov, won the gold medal in pairs figure skating. If Ekaterina has a mass of 45 kg, calculate the work done by
Sergei when he lifted her 2 meters in the air.
Work done by Sergei = Increase in Potential Energy of
Ekaterina
E. Interconversion
of KE and PE; Conservation of Energy
Consider throwing a
ball into the air as an example of the interchange of KE and PE. When the ball is released it has a large
amount of KE. As it rises, the KE will
decrease and the PE will increase. At
its highest level the PE is a maximum and KE is zero. When the ball falls down KE increases at the expense of PE.
In this simple example
of throwing a ball upward it is implicit that kinetic energy can be transformed
into potential energy and vice versa, without any loss of energy. This is, in fact, a general result embodied
in the principle of conservation of energy. This, together with momentum conservation, is once of the
fundamental laws of nature. Many
problems in which complicated forces are involved, and therefore the solutions
to which are extremely difficult to construct by using Newton’s laws, can be
solved in a simple way by using the law of conservation of energy.
Examples:
1. The car shown below has just run out of gas
and is coasting with velocity v. What
must this velocity be if the car is to just make it to the top of the hill?
To
solve this problem using forces would be very difficult but it is simplified
greatly using conservation of energy.
Note
that the final result is independent of the mass of the car.
2. Suppose you can throw a baseball with a speed
of 30 m/sec (67 mph). If you threw the
ball vertically upward with this velocity how high would it rise?
This
would not be a difficult problem to solve using the equations of motion for
constant acceleration. However, it is
even easier to solve using the principle of conservation of energy.
3.
The world record for the pole vault is approximately 20 feet. Assuming that a pole vaulter simply changes
kinetic energy into potential energy, calculate the approach velocity of a pole
vaulter required to make a 20 ft vault.
If
you think about this example the result seems unreasonable. In Chapter 1 we calculated the top speed of
a world class sprinter to be about 33 ft/sec.
If a pole vaulter can run 36 ft/sec while carrying a seventeen foot pole
he should obviously be running in the dashes.
Can you think of any refinements to our initial assumptions that will
alter our results?
Summary:
Work Kinetic
Energy
Power Potential
Energy
Energy Conservation
of Energy
1.
Determine the work done by a 70 kg (154 lb) student who
does one pull-up. Assume the distance
he lifts himself is 0.5 m. Compare the
power required if he does a pull-up in one second to the power required if he
takes 10 seconds.
2.
What work is done by a weightlifter in the
clean-and-jerk when he lifts 200 kg to a height of 2 meters? Do short weight lifters have an advantage?
3.
In the football stadium there are 50 steps. Each step is 8 in (0.2 m) high. How much work would a 90 kg (198 lb)
football player do if he climbed every step?
Give the answer in Joules and foot-pounds. If the football player worked at a rate of 1000 watts, how long
would it take him to reach the top?
4.
A baseball has a mass of 0.01 slugs. Let us examine what happens when a pitcher
throws a 90 mph (132 ft/sec) fastball.
Assume the catcher’s mitts “gives” 2 inches (0.167 ft) as the ball is
caught.
a.
What is the initial kinetic energy of the ball?
b. How
much work is done by the ball on the catcher’s mitt?
c.
What is the average force exerted on the catcher’s
mitt?
5.
Joan Benoit is the only American to win a gold medal in
the marathon. If she has a mass of 50
kg,
a.
How much work must she do to accelerate from rest to a
aped of 5 m/sec?
b. How
much work must she do to increase her speed from 5 m/sec to 8 m/sec?
6.
A car moving 25 mph will skid 50 feet with the brakes
locked. How far will the car skid with
brakes if it is initially traveling 50 mph?
7.
A baseball is thrown upward with an initial velocity of
60 mph (88 ft/sec). Find the height the
ball rises.
8.
World class high jumpers can clear a height of 7
ft. Make some reasonable assumptions
and use conservation of energy to fine the vertical takeoff velocity of such a
high jumper.
9.
If you were to fall from the 5 meter high diving
platform with what downward velocity would you strike the water? Would the answer change if you jumped up
first? What if you jumped out?
10. A
child is on a swing at the park. At the
lowest point the speed is 3 m/sec. How
high above this low point will the child rise?