A. Work
B. Power
C. Kinetic Energy
D. Potential Energy
E. Interconversion of KE and PE; Conservation of Energy
Problems
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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.
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 degreesas 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.
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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 objectis 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 Volkswagen (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 Volkswagen 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 Volkswagen from rest to 22 mph?
2. How much work is required to accelerate a Volkswagen from rest to 44 mph?
3. How much work is required to accelerate a Volkswagen
from 22 mph to 44 mph?
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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
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
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