Hf2nech05hquqhqhqhqhq PDF

Preview

Summary

Hqyqhqhqhqhqhqhqqq...

Description

Copyright © by Holt, Rinehart and Winston. All rights reserved.

Copyright © by Holt, Rinehart and Winston. All rights reserved.

CHAPTER 5

Work and Energy PHYSICS IN ACTION This whimsical piece of art is a type of audiokinetic sculpture. Balls are raised to a high point on the curved blue track. As the balls roll down the track, they turn levers, spin rotors, and bounce off elastic membranes. As each ball travels along the track, the total energy of the system remains unchanged. The types of energy that are involved—whether associated with the ball’s motion, the ball’s position above the ground, or friction—vary in ways that keep the total energy of the system constant. In this chapter, you will learn about work and the different types of energy that are relevant to mechanics.

•

How many different kinds of energy are used in this sculpture?

•

How are work, energy, and power related in the functioning of this sculpture?

CONCEPT REVIEW Kinematics (Section 2-2) Newton’s second law (Section 4-3) Force of friction (Section 4-4)

Copyright © by Holt, Rinehart and Winston. All rights reserved.

Work and Energy

167

5-1

Work 5-1 SECTION OBJECTIVES

DEFINITION OF WORK

•

Recognize the difference between the scientific and ordinary definitions of work.

•

Define work by relating it to force and displacement.

•

Identify where work is being performed in a variety of situations.

•

Calculate the net work done when many forces are applied to an object.

Many of the terms you have encountered so far in this book have meanings in physics that are similar to their meanings in everyday life. In its everyday sense, the term work means to do something that takes physical or mental effort. But in physics, work has a distinctly different meaning. Consider the following situations: • A student holds a heavy chair at arm’s length for several minutes. • A student carries a bucket of water along a horizontal path while walking at constant velocity. It might surprise you to know that under the scientific definition of work, there is no work done on the chair or the bucket, even though effort is required in both cases. We will return to these examples later.

A force that causes a displacement of an object does work on the object

work the product of the magnitudes of the component of a force along the direction of displacement and the displacement

Imagine that your car, like the car shown in Figure 5-1, has run out of gas and you have to push it down the road to the gas station. If you push the car with a constant force, the work you do on the car is equal to the magnitude of the force, F, times the magnitude of the displacement of the car. Using the symbol d instead of ∆x for displacement, we can define work as follows: W = Fd Work is not done on an object unless the object is moved because of the action of a force. The application of a force alone does not constitute work. For this reason, no work is done on the chair when a student holds the chair at arm’s length. Even though the student exerts a force to support the chair, the chair does not move. The student’s tired arms suggest that work is being done, which is indeed true. The quivering muscles in the student’s arms go through many small displacements and do work within the student’s body. However, work is not done on the chair.

Work is done only when components of a force are parallel to a displacement Figure 5-1

This person exerts a constant force on the car and displaces it to the left. The work done on the car by the person is equal to the force times the displacement of the car.

168

Chapter 5

When the force on an object and the object’s displacement are in different directions, only the component of the force that is in the direction of the object’s displacement does work. Components of the force perpendicular to a displacement do not do work.

Copyright © by Holt, Rinehart and Winston. All rights reserved.

For example, imagine pushing a crate along the ground. If the force you exert is horizontal, all of your effort moves the crate. If your force is other than horizontal, only the horizontal component of your applied force causes a displacement and does work. If the angle between the force and the direction of the displacement is q, as in Figure 5-2, work can be written as follows:

d θ F

W = Fd(cos q ) If q = 0°, then cos 0° = 1 and W = Fd, which is the definition of work given earlier. If q = 90°, however, then cos 90° = 0 and W = 0. So, no work is done on a bucket of water being carried by a student walking horizontally. The upward force exerted to support the bucket is perpendicular to the displacement of the bucket, which results in no work done on the bucket. Finally, if many constant forces are acting on an object, you can find the net work done by first finding the net force.

W = Fd(cos θ) Figure 5-2

The work done on this crate is equal to the force times the displacement times the cosine of the angle between them.

NET WORK DONE BY A CONSTANT NET FORCE

Wnet = Fnet d(cos q ) net work = net force × displacement × cosine of the angle between them Work has dimensions of force times length. In the SI system, work has a unit of newtons times meters (N • m), or joules (J). The work done in lifting an apple from your waist to the top of your head is about 1 J. Three push-ups require about 1000 J.

The joule is named for the British physicist James Prescott Joule ( 1 8 1 8 –1 889). Joule made major contributions to the understanding of energy, heat, and electricity.

SAMPLE PROBLEM 5A

Work PROBLEM

How much work is done on a vacuum cleaner pulled 3.0 m by a force of 50.0 N at an angle of 30.0° above the horizontal? SOLUTION

Given:

F = 50.0 N

Unknown:

W=?

q = 30.0°

d = 3.0 m

Use the equation for net work done by a constant force: W = Fd(cos q ) Only the horizontal component of the applied force is doing work on the vacuum cleaner. W = (50.0 N)(3.0 m)(cos 30.0°) W = 130 J

Copyright © by Holt, Rinehart and Winston. All rights reserved.

Work and Energy

169

PRACTICE 5A

Work 1. A tugboat pulls a ship with a constant net horizontal force of 5.00 × 103 N and causes the ship to move through a harbor. How much work is done on the ship if it moves a distance of 3.00 km? 2. A weight lifter lifts a set of weights a vertical distance of 2.00 m. If a constant net force of 350 N is exerted on the weights, what is the net work done on the weights? 3. A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25° downward from the horizontal. Find the work done by the shopper on the cart as the shopper moves along a 50.0 m length of aisle.

E RA NT

PHYSICS TU

IVE • CT

I

4. If 2.0 J of work is done in raising a 180 g apple, how far is it lifted?

TO R

Module 5 “Work” provides an interactive lesson with guided problem-solving practice to teach you about calculating net work.

The sign of work is important Work is a scalar quantity and can be positive or negative, as shown in Figure 5-3. Work is positive when the component of force is in the same direction as the displacement. For example, when you lift a box, the work done by the force you exert on the box is positive because that force is upward, in the same

Negative (–) work

Positive (+) work

90°

F

F

d

d q

180° F

0° F

Figure 5-3

Depending on the angle, an applied force can either cause a moving car to slow down (left), which results in negative work done on the car, or speed up (right), which results in positive work done on the car.

170

Chapter 5

d

d 270°

Copyright © by Holt, Rinehart and Winston. All rights reserved.

direction as the displacement. Work is negative when the force is in the direction opposite the displacement. For example, the force of kinetic friction between a sliding box and the floor is opposite to the displacement of the box, so the work done by the force on the box is negative. If you are very careful in applying the equation for work, your answer will have the correct sign: cos q is negative for angles greater than 90° but less than 270°. If the work done on an object results only in a change in the object’s speed, the sign of the net work on the object tells you whether the object’s speed is increasing or decreasing. If the net work is positive, the object speeds up and the net force does work on the object. If the net work is negative, the object slows down and work is done by the object on another object.

NSTA

TOPIC: Work GO TO: www.scilinks.org sci LINKS CODE: HF2051

Section Review 1. For each of the following statements, identify whether the everyday or the scientific meaning of work is intended. a. Jack had to work against time as the deadline neared. b. Jill had to work on her homework before she went to bed. c. Jack did work carrying the pail of water up the hill. 2. If a neighbor pushes a lawnmower four times as far as you do but exerts only half the force, which one of you does more work and by how much? 3. For each of the following cases, indicate whether the work done on the second object in each example will have a positive or a negative value. a. The road exerts a friction force on a speeding car skidding to a stop. b. A rope exerts a force on a bucket as the bucket is raised up a well. c. Air exerts a force on a parachute as the parachutist slowly falls to Earth. 4. Determine whether work is being done in each of the following examples: a. a train engine pulling a loaded boxcar initially at rest b. a tug of war that is evenly matched c. a crane lifting a car 5. A worker pushes a 1.50 × 103 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220. a. How much work is done by the worker on the crate? b. How much work is done by the floor on the crate? c. What is the net work done on the crate? 6. Physics in Action A 0.075 kg ball in a kinetic sculpture is raised 1.32 m above the ground by a motorized vertical conveyor belt. A constant frictional force of 0.350 N acts in the direction opposite the conveyor belt’s motion. What is the net work done on the ball? Copyright © by Holt, Rinehart and Winston. All rights reserved.

Work and Energy

171

5-2

Energy 5-2 SECTION OBJECTIVES

KINETIC ENERGY

•

Identify several forms of energy.

•

Calculate kinetic energy for an object.

•

Apply the work–kinetic energy theorem to solve problems.

•

Distinguish between kinetic and potential energy.

Kinetic energy is energy associated with an object in motion. Figure 5-4 shows a cart of mass m moving to the right on a frictionless air track under the action of a constant net force, F. Because the force is constant, we know from Newton’s second law that the particle moves with a constant acceleration, a. While the force is applied, the cart accelerates from an initial velocity vi to a final velocity vf . If the particle is displaced a distance of ∆x, the work done by F during this displacement is

•

Classify different types of potential energy.

•

Calculate the potential energy associated with an object’s position.

Wnet = F∆ x = (ma)∆ x However, in Chapter 2 we found that the following relationship holds when an object undergoes constant acceleration: vf2 = v2i + 2a∆ x vf2 − v2i a∆x = ᎏ 2

kinetic energy the energy of an object due to its motion

Substituting this result into the equation Wnet = (ma)∆x gives

冢

冣

2 2 vf − v i Wnet = m ᎏ 2

1 2 1 2 Wnet = 2mvf − 2mv i

Kinetic energy depends on speed and mass The quantity 12 mv 2 has a special name in physics: kinetic energy. The kinetic energy of an object with mass m and speed v, when treated as a particle, is given by the expression shown on the next page.

vi

F = ma

vf

F

Figure 5-4

The work done on an object by a constant force equals the object’s mass times its acceleration times its displacement.

172

Chapter 5

∆x

Copyright © by Holt, Rinehart and Winston. All rights reserved.

KINETIC ENERGY 1

KE = 2 mv 2 1

kinetic energy = 2 × mass × (speed)2 Kinetic energy is a scalar quantity, and the SI unit for kinetic energy (and all other forms of energy) is the joule. Recall that a joule is also used as the basic unit for work. Kinetic energy depends on both an object’s speed and its mass. If a bowling ball and a volleyball are traveling at the same speed, which do you think has more kinetic energy? You may think that because they are moving with identical speeds they have exactly the same kinetic energy. However, the bowling ball has more kinetic energy than the volleyball traveling at the same speed because the bowling ball has more mass than the volleyball.

I

PHYSICS TU

IVE • CT

ERA NT

TO R

Module 6 “Work–Kinetic Energy Theorem” provides an interactive lesson with guided problem-solving practice.

SAMPLE PROBLEM 5B

Kinetic energy PROBLEM

A 7.00 kg bowling ball moves at 3.00 m/s. How much kinetic energy does the bowling ball have? How fast must a 2.45 g table-tennis ball move in order to have the same kinetic energy as the bowling ball? Is this speed reasonable for a table-tennis ball? SOLUTION

Given:

The subscripts b and t indicate the bowling ball and the table-tennis ball, respectively. mb = 7.00 kg

Unknown:

mt = 2.45 g

vb = 3.00 m/s

KEb = ? vt = ?

Use the kinetic energy equation: 1

1

KEb = 2mb v b2 = 2(7.00 kg)(3.00 m/s)2 = 31.5 J 1

KEt = 2mt v t2 = KEb = 31.5 J vt =

冪莦莦莦莦 冪莦 2KEb  = mt

(2)(31.5 J)  2.45 × 10−3 kg

vt = 1.60 × 102 m/s This speed is much too high to be reasonable for a table-tennis ball. Copyright © by Holt, Rinehart and Winston. All rights reserved.

Work and Energy

173

PRACTICE 5B

Kinetic energy 1. Calculate the speed of an 8.0 × 104 kg airliner with a kinetic energy of 1.1 × 109 J. 2. What is the speed of a 0.145 kg baseball if its kinetic energy is 109 J? 3. Two bullets have masses of 3.0 g and 6.0 g, respectively. Both are fired with a speed of 40.0 m/s. Which bullet has more kinetic energy? What is the ratio of their kinetic energies? 4. Two 3.0 g bullets are fired with speeds of 40.0 m/s and 80.0 m/s, respectively. What are their kinetic energies? Which bullet has more kinetic energy? What is the ratio of their kinetic energies? 5. A car has a kinetic energy of 4.32 × 105 J when traveling at a speed of 23 m/s. What is its mass?

work–kinetic energy theorem the net work done on an object is equal to the change in the kinetic energy of the object

The equation Wnet = 12m v 2f − 12m v 2i derived at the beginning of this section says that the net work done by a net force acting on an object is equal to the change in the kinetic energy of the object. This important relationship, known as the work–kinetic energy theorem, is often written as follows: WORK–KINETIC ENERGY THEOREM

Wnet = ∆KE net work = change in kinetic energy

Figure 5-5

A moving hammer has kinetic energy and so can do work on a nail, driving it into the wall.

174

Chapter 5

It is important to note that when we use this theorem, we must include all the forces that do work on the object in calculating the net work done. From this theorem, we see that the speed of the object increases if the net work done on it is positive, because the final kinetic energy is greater than the initial kinetic energy. The object’s speed decreases if the net work is negative, because the final kinetic energy is less than the initial kinetic energy. The work–kinetic energy theorem allows us to think of kinetic energy as the work an object can do as it comes to rest, or the amount of energy stored in the object. For example, the moving hammer on the verge of striking a nail in Figure 5-5 has kinetic energy and can therefore do work on the nail. Part of this energy is used to drive the nail into the wall, and part goes into warming the hammer and nail upon impact. Copyright © by Holt, Rinehart and Winston. All rights reserved.

SAMPLE PROBLEM 5C

Work–kinetic energy theorem PROBLEM

On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does the sled move if the coefficient of kinetic friction between the sled and the ice is 0.10? SOLUTION 1. DEFINE

Given:

m = 10.0 kg vi = 2.2 m/s vf = 0 m/s mk = 0.10

Unknown:

d=?

Diagram:

vi Fk

2. PLAN

d Choose an equation(s) or situation: This problem can be solved using the definition of work and the work–kinetic energy theorem. Wnet = Fnet d(cos q) Wnet = ∆KE The initial kinetic energy is given to the sled by the person. 1 KEi = 2mv 2i

Because the sled comes to rest, the final kinetic energy is zero. KEf = 0 ∆KE = KEf − KEi = − 2 mv i2 1

The net work done on the sled is provided by the force of kinetic friction. Wnet = Fnet d(cos q ) = mk mgd(cos q ) The force of kinetic friction is in the direction opposite d. q = 180° 3. CALCULATE

Substitute values into the equations: Wnet = (0.10)(10.0 kg)(9.81 m/s2) d (cos 180°) Wnet = (−9.8 N)d 1 ∆KE = −KEi = −( 2)(10.0 kg)(2.2 m/s)2 = −24 J Use the work–kinetic energy theorem to solve for d. Wnet = ∆KE (−9.8 N)d = −24 J

continued on next page

d = 2.4 m

Copyright © by Holt, Rinehart and Winston. All rights reserved.

CALCULATOR SOLUTION Your calculator should give an answer of 2.44898, but because the answer is limited to two significant figures, this number should be rounded to 2.4.

Work and Energy

175

4. EVALUATE

Note that because the direction of the force of kinetic friction is opposite the displacement, the net work done is negative. Also, according to Newton’s second law, the acceleration of the sled is about −1 m/s2 and the time it takes the sled to stop is about 2 s. Thus, the distance the sled traveled in the given amount of time should be less than the distance it would have traveled in the absence of friction. 2.4 m < (2.2 m/s)(2 s) = 4.4 m

PRACTICE 5C

Work–kinetic energy theorem 1. A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 45 N. How far must the student be pushed, starting from rest, so that her final kinetic energy is 352 J? 2. A 2.0 × 103 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by traction between the wheels and the road. The other is a 950 N resistive force due to various frictional forces. Use the work–kinetic energy the...