chapter 7 work and energy - wordpress.com · 7-4 kinetic energy and the work-energy principle...

18/04/2012

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Lecture PowerPoints

Chapter 7

Physics for Scientists and

Engineers, with Modern

Physics, 4th Edition

Giancoli


Chapter 7

Work and Energy

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Units of Chapter 7

• Work Done by a Constant Force

• Scalar Product of Two Vectors

• Work Done by a Varying Force

• Kinetic Energy and the Work-Energy Principle


7-1 Work Done by a Constant Force

The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement:

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In the SI system, the units of work is the joule:

As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion.



Example 7-1: Work done on a crate.

A person pulls a 50 kg crate 40 m along a horizontal floor by a constant force FP = 100 N, which acts at a 37°angle as shown. The floor is smooth and exerts no friction force. Determine (a) the work done by each force acting on the crate, and (b) the net work done on the crate.

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Example 7-1: Work done on a crate.m = 50 kg, FP = 100 N, x = 40 m(a) the work done by each force acting on the crate.(b) the net work done on the crate.

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=

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m

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NB: This is NOT the same Example 7-1 Work done on a crate.As in text Giancoli. The example in Giancoli takes friction into account.m = 50 kg, FP = 100 N, Ffr = 50 N, x = 40 m(a) the work done by each force acting on the crate.(b) the net work done on the crate.

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NB: This is NOT the same Example 7-1 Work done on a crate.As in text Giancoli. The example in Giancoli takes friction into account.m = 50 kg, FP = 100 N, Ffr = 50 N, x = 40 m(a) the work done by each force acting on the crate.(b) the net work done on the crate.

J 12002000- 3200

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J 2000-180cos4050cos

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ofcomponent horizontalby (i) done isWork

work.no do they sont displaceme lar toperpendicu are and (a)

fr

==

=°××==

=°××==

θ

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xFW

m

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N

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gFr

rr



Solving work problems:

1. Draw a free-body diagram.

2. Choose a coordinate system.

3. Apply Newton’s laws to determine any unknown forces.

4. Find the work done by a specific force.

5. To find the net work, either

a) find the net force and then find the work it does, or

b) find the work done by each force and add.

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Example 7-2: Work on a backpack.

(a) Determine the work a hiker must do on a 15.0 kg backpack to carry it up a hill of height h = 10.0 m, as shown. Determine also

(b) the work done by gravity on the backpack, and

(c) the net work done on the backpack. For simplicity, assume the motion is smooth and at constant velocity (i.e., acceleration is zero).

Note that only the vertical component of the hiker’s force need be considered:W = Fdcosθθθθ = Fd.h/d = Fh



Example 7-2: Work on a backpack.

(a) Determine the work a hiker must do on a 15.0 kg backpack to carry it up a hill of height h = 10.0 m, as shown.

W = FHh = mgh = 15.0××××9.80××××10.0 = 1470 J

(b) the work done by gravity on the backpack.

W = -mgh = -15.0××××9.80××××10.0 = -1470 J

(gravitational force in opposite direction to

motion)

(c) the net work done on the backpack. For simplicity.

Wnet = Whiker + Wgravity = 1470 – 1470 = 0

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Conceptual Example 7-3: Does the Earth do work on the Moon?

The Moon revolves around the Earth in a nearly circular orbit, with approximately constant tangential speed, kept there by the gravitational force exerted by the Earth. Does gravity do

(a) positive work,

(b) negative work,

(c) no work at all on the Moon?

Since the only force is at right angles to the motion (cos90°°°° = 0) no work is done by gravity on the moon


7-2 Scalar Product of Two Vectors

Definition of the scalar, or dot, product:

Therefore, we can write:

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7-2 Scalar Product of Two Vectors

Example 7-4: Using the dot product.

The force shown has magnitude FP = 20 N and makes an angle of 30°to the ground. Calculate the work done by this force, using the dot product, when the wagon is dragged 100 m along the ground.

J 170030cos10020cos =°××==•= θdFWP

dFP

rr


7-3 Work Done by a Varying Force

Particle acted on by a varying force. Clearly, ·d is not constant!F

r

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For a force that varies, the work can be approximated by dividing the distance up into small pieces, finding the work done during each, and adding them up.



In the limit that the pieces become infinitesimally narrow, the work is the area under the curve:

Or:

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Work done by a spring force:

The force exerted by a spring is given by Hooke’s:

xFS

rrk−=

Note that the force exerted by the spring is always in the opposite direction to the displacement.



Plot of F vs. x. Work done is equal to the shaded area.

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Example 7-5: Work done on a spring.

(a) A person pulls on a spring, stretching it 3.0 cm, which requires a maximum force of 75 N. How much work does the person do?

(b) If, instead, the person compresses the spring 3.0 cm, how much work does the person do?

( )J 1.1

2

03.02500kx W:Work

N/m 250003.0

75

x

FkkxF :Constant Spring

2

2

21 =

×==

===→=

Work is the same as in (a)



Example 7-6: Force as a function of x.

where F0 = 2.0 N, x0 = 0.0070 m, and x is the position of the end of the arm. If the arm moves from x1 = 0.010 m to x2 = 0.050 m, how much work did the motor do?

A robot arm that controls the position of a video camera in an automated surveillance system is manipulated by a motor that exerts a force on the arm. The force is given by

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F0 = 2.0 Nx0 = 0.0070 mx1 = 0.010 mx2 = 0.050 m

( )

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−

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dxx

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x

x

x

x


7-4 Kinetic Energy and the Work-Energy Principle

Energy was traditionally defined as the ability to do work. We now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition.

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This means that the work done is equal to the change in the kinetic energy:

• If the net work is positive, the kinetic energy increases.

• If the net work is negative, the kinetic energy decreases.

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Because work and kinetic energy can be equated, they must have the same units: kinetic energy is measured in joules. Energy can be considered as the ability to do work:

A moving hammer strikes a nail and comes to rest. The hammer exerts a force F on the nail; the nail exerts a force -F on the hammer (Newton’s third law).

The work done on the nail by the hammer is positive (Wn = Fd >0). The work done on the hammer by the nail is negative (Wh = -Fd)



Example 7-7: Kinetic energy and work done on a baseball.

A 145 g baseball is thrown so that it acquires a speed of 25 m/s.

(a) What is its kinetic energy?

(b) What was the net work done on the ball to make it reach this speed, if it started from rest?

J 4525145.0 2

212

21 =××== mvK

As net work is equal to the change in kinetic energy, the net work also equals 45 J

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Example 7-8: Work on a car, to increase its kinetic energy.

How much net work is required to accelerate a 1000 kg car from 20 m/s to 30 m/s?

( )( ) J 102.520301000 522

21

2

1

2

2212

1212

221

12

×=−×=

−=−=−= vvmmvmvKKWnet


Example 7-9: Work to stop a car.

A car traveling 60 km/h can brake to a stop within a distance d of 20 m. If the car is going twice as fast, 120 km/h, what is its stopping distance? Assume the maximum braking force is approximately independent of speed.

If the car’s velocity is doubled then the kinetic energy is increased by a factor of 4 (K ∝∝∝∝ v2).If the force is constant the stopping distance will

be 4××××20 = 80 m

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Example 7-10: A compressed spring.

A horizontal spring has spring constant k = 360 N/m.

(a) How much work is required to compress it from its uncompressed

length (x = 0) to x = 11.0 cm?

( )J 2.18

2

110.03602

2

21 === kxW



A horizontal spring has spring constant k = 360 N/m. Spring is

compressed 11.0 cm = 0.110 m →→→→ (a) W = 2.18 J

(b) If a 1.85 kg block is placed against the spring and the spring is

released, what will be the speed of the block when it separates

from the spring at x = 0? Ignore friction.

m/s 1.5485.1

18.222

2

21

=×

==

==

m

Kv

mvKW

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A horizontal spring has spring constant k = 360 N/m.

(a) How much work is required to compress it from its uncompressed

length (x = 0) to x = 11.0 cm? (2.18 J)

(b) If a 1.85 kg block is placed against the spring and the spring is

released, what will be the speed of the block when it separates

from the spring at x = 0? Ignore friction. (1.54 m/s)

(c) Repeat part (b) but assume that the block is moving on a table and

that some kind of constant drag force FD = 7.0 N is acting to slow it

down, such as friction.

m/s 1.2385.1

41.122

J 1.41770182

J 0.770.1107.0

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Summary of Chapter 7

• Work:

• Work done by a variable force:

• Kinetic energy is energy of motion:

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Summary of Chapter 7

• Work-energy principle: The net work done on an object equals the change in its kinetic energy.

chapter 7 work and energy - wordpress.com · 7-4 kinetic energy and the work-energy principle...

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