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Chapter 4

Work & Energy

Dr. Ali

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CHAPTER OUTLINE

Work Power Energy and Its Forms Kinetic Energy Potential Energy Conservation of Energy

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WORK

Who is doing more work?

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WORK

Work is defined as the product of the net force acting on a body and its displacement (in the direction of the force).

Work = force x parallel distance

W = F x d

Unit of Energy: 1 Joule = 1 J = 1 Nm = 1 kgm2/s2

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WORK

Only the component of the applied force, F, in the direction of the motion of the lawn mower, FHorizontal, is used to do work on the lawn mower.

For work to be done, force and motion must be in the same direction

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Example 1:

An object is moved with a force of 15 N across a horizontal surface. How much work is done if the object is moved 50 m?

F = 15 N

d = 50 m

W = ???W = 750 J

W = F x d

W = (15 N)(50 m)

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Example 2:

650 J of work is done in moving a desk a horizontal distance of 5 m. How much force is used to move the desk?

WF =

dF = ???

d = 5 m

W = 650 J

650 J=

5 m

F = 130 N

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Example 3:

How much work is done in lifting a 10 kg box 1.5 m off the floor?

m = 10 kg

d = 1.5 m

F = ???

W = ???W = 150 J

F = w

W = F x d

= m x g

F = 10 kg x 10 m/s2 = 100 N

= 100 N x 1.5 m

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Example 4:

How much work is done while walking 5.0 m holding an object with mass of 3.0 kg?

No work is being done, since force and motion are not in the same direction.

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POWER

Power is the rate at which work is done.

workPower=

time

SI Unit for Power: 1 watt = 1 J/s

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Example 1:

A force of 150 N is used to push a motorcycle 10 m along a road in 20 s. Calculate the power in watts.

F = 150 N

d = 10 m

t = 20 s

P = ???P = 75 watts

150 N x 10 m=

20 s

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Example 2:An 80 kg man runs up a flight of stairs 5.0 m high in 10 seconds. What is the man’s power output in watts?

W F x dP = =

t t

m = 80 kg

d = 5.0 m

t = 10 s

P = ???P = 400 watts

F = w

F = 80 kg x 10 m/s2 = 800 N

800 N x 5.0 m=

10 s

= m x g

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Example 3:

A pump lifts 30 kg of water a vertical distance of 20 m each second. What is the power output?

W F x dP = =

t t

m = 30 kg

d = 20 m

t = 1 s

P = ???P = 6000 watts

F = w = m x g

F = 30 kg x 10 m/s2 = 300 N

300 N x 20 m=

1 s

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Example 4:

A crane uses 750 kwatts of power to lift a car 0.5 m in 12 seconds. How much work is done?

P = 750 kwatts

d = 0.5 m

t = 12 s

W =??? W = 9000 kJ

W = P x t

= 750 kwatts x 12 s

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Example 4:

A crane uses 750 kwatts of power to lift a car 0.5 m in 12 seconds. How much work is done?

WF =

dW = 9000 kJ

d = 0.5 m

F =??? F = 18000 kN

What force is used by the crane?

90000 kJ=

0.5 m

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Example 5:

Which person does more work, 1 or 2?

Which person has greater power, 1 or 2?

1 2

F1 = F2 d1 = d2

W1 = W2

t1 > t2

P1 < P2

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ENERGY

Energy is the ability to do work

The SI units for energy is joules (J).

The brick and the hammer possess

energy and thus can do work on the nail

1J = 1 joule = 1 kgm2/s2

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ENERGY

There are two types of energy here:

KineticPotential

Energy of motion

Stored energy

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ENERGY

Energy can be converted from one type to another.

Kinetic E

Potential E

Kinetic & potential

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KINETICENERGY

Kinetic energy is energy of motion.

21KE = m v

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KE is a scalar quantity

(Note: Energy is never a vector)

Velocity has greater effect on KE than mass

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Example 1:

What is the kinetic energy of a 60 kg girl on skis traveling at 20 m/s?

21KE = mv

2m = 60 kg

v = 20 m/s

KE = ???

KE = 12000 J

21= (60 kg)(20 m/s)

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Example 2:

A sports car is moving at 4.0 m/s. If the mass of the car is 800 kg, how much KE does it have?

21KE = mv

2m = 800 kg

v = 4.0 m/s

KE = ???

KE = 6400 J

21= (800 kg)(4.0 m/s)

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Example 3:

Two identical cars are moving, one with twice the velocity of the other. How much more kinetic energy does the faster car possess?

21KE = mv

2v2 = 2 v1

Faster car has 4 times greater KE

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POTENTIALENERGY

Potential energy is stored energy.

The compressed spring has potential

energy because when released it can do

work on the mass, m.

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POTENTIALENERGY

Gravitational potential energy is energy of position.

Potential energy = Weight x height

PE = m g h

Low PE

High PE

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Example 1:

A mass of 100 kg is lifted a distance of 50 m. How much PE does it possess?

m = 100 kg

h = 50 m

PE = ??? PE = 50000 J

PE = m g h

= (100 kg) (10 m/s2) (50 m)

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Example 2:

A 70-kg diver standing on a diving platform possesses 35000 J of PE. How high is the platform?

PEh =

mgm = 70 kg

h = ???

PE = 35000 J h = 50 m

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35000 J=

(70 kg)(10 m/s )

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CONSERVATIONOF ENERGY

The sum of KE and PE in a system is constant, in the absence of friction.

KE + PE = constant

KE + PE = ETotal

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CONSERVATIONOF ENERGY

Energy cannot be created or destroyed

It may be transformed from one form to another

PE

KE

ETotal in a system remains constant.

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Example 1:

A 10-kg boulder rests at the edge of a 100-m cliff. How much PE does the rock possess?

m = 10 kg

h = 100 m

PE = ???PE = 10000 J

PE = m g h

= (10 kg) (10 m/s2) (100 m)

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Example 1:The rock rolls off the cliff and falls to the bottom. How much KE does the rock possess at the bottom of the cliff?

KEbottom = PEtop = 10000 J

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Example 1:

What speed does the rock have just before hitting the ground?

m = 10 kg

v = ???

KE = 10000 J

Thus, v = 45 m/s

21KE= mv

2

v2 = 2000 m2/s2

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Example 2:

A 60-kg boy and his sled (neglect the mass of the sled) are at a 10 m high slope. How much PE do the boy and sled possess?

m = 60 kg

h = 10 m

PE = ???PE = 6000 J

PE = mgh

= (60 kg)(10 m/s2)(10 m)

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Example 2:

What kinds of energy, and how much of each, do they possess halfway down the slope?

PE = KE = 3000 J

PE = KE

@ halfway

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THE END