Chapter 5: Work and Energy
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 t 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, conrvation 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 conrvation of energy, we first introduce the concept of work, which will lead is to the concept of energy.
A. Work
There are many ways in which the term “work” is ud 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 preci meaning and a narrow definition. We define work in the following way:
邂逅的意思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)
Work F d d =
Where F d = component of the force in the direction of the displacement, and d = displacement.
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Note: Many textbooks will u “W” to reprent work when using Equation 5.1. This inevitably leads to confusion since the letter “W” is also ud to reprent weight. To avoid this problem we will always write out the word “Work” in this textbook.胜利的反义词
When we examine Equation 5.1 we e 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 cond 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 becau, 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 n we have introduces. In this ca 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 = F d × 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 F d N m N m
Work J d ==×=⋅=15004060006000
2.
The distance from the ground floor of a building to the cond floor is 10 ft. If a 160 lb. Student climbs the stairs, how much work does he do?
W ork F d lb ft ft lb d ==×=⋅160101600
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.冉伯牛
Work F d F F N N Work F d N m Joules d d d ======×=cos ()(.)374000832032051600o
B. Power
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 ca 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.
Power Work done Time taken to do work
=
Power Work t = (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: 11 Joule Watt c
= English system: 1 1356ft lb Watt ⋅=c
.
Other units are also ud frequently for power. For example, a kilowatt is 1000 times larger than a Watt. The unit horpower (hp) is frequently ud in the British system. This quantity is defined
1 horpower = 550 ft-lb/c = 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 conds, or runs in
at a time of 1 cond.
The work done is determined using Eq. (5.1).
Wo Work F d mg d
rk kg m m Work Joules d ===××=()(/c )5010315002
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 conds then
Power Work t Joules J Watts ====150********c c
If she runs up in one cond we have
Power Work t Joules J Watts ====1500115001500c c
2.
A manually operated winch line is ud 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.郑源个人资料
Wo Work F d mg d rk kg m m Work Joules d ===××=()(/c ),2001010200002
Now find the time required by using Eq (5.2) written as
Time
Work
Power
Joules
Watt
===
20000
200
100
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c
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 c). For most people either task would be too
demanding.
C. Kinetic Energy
Consider a pitcher throwing a baball to a catcher. While the ball of mass m is in his hand, the pitcher exerts a force on it over some distance and releas the baball 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 bab
all. 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 t 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 e how the are related, consider the special ca of a constant force acting on an object of mass m which is initially at rest. Suppo 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
Wo rk Fd
=
We wish to relate the work to the change in motion of the object caud 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
Wo
rk Fd ma d ==()
Now the distance an object moves while accelerating from rest is
d v t at at d at I =+=+=122122
120
The work equation may now be written as
Work Fd ma d ma at Work m at ====()()(()12212)
提高政治素养
青蛙吃害虫Finally, recall the kinematics equation
v v at at
v at F I F =+=+=0
Substitute this result into the work equation above
Work m at m v Work mv F F ===12
2122122()()
This result is very important. It tells us that the work required to t an object into motion is given by the simple expression ½mv 2. This quantity is called kinetic energy (KE). Becau 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 ud to
determine the work done on the ball. However, if we know the ball’s mass and velocity after it is relead, it is easy to determine its kinetic energy.
We define the following:
Kinetic energy (KE) is 1/2mv 2, and it reprents the amount of work done to t a body in motion. It also reprents the amount of work a body of mass m can do as a result of its
motion with speed v.
KE mv =122 (5.3) Note: We have defined kinetic energy in terms of the amount of work required to t 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 baball pitcher do to throw a baball (mass = 0.01 slug) at a
velocity of 90 mph (132 ft/c)?
Since the work required to t an object in motion is simply the kinetic energy it has
when it is moving we have
Work KE mv Work slug ft ft lb ====1221200113287(.)(/c)⋅
