Work and Energy

Work: The word looks the same, it spells the same

but has different meaning in physics from the way it

is normally used in the everyday language

Definition of work W done by a constant force F

exerted on an object through distance d

work = force along distance × the distance moved

Only the component that acts in the same direction as the motion is

doing work on the box.

Vertical component is just trying to lift the object up.

d

F

θFd

F

θFd

dW = F d = Fd cosθ

work = force × distance moved × cos of the angle between them

The SI unit for work is the newton–metre and is called

the joule named after the 19th Century physicist

James Prescott Joule.

1 J (Joule) = 1N x 1 m

Work is a scalar (add like ordinary numbers)

Units

According to the physics definition, you

are NOT doing work if you are just

holding the weight above your head

(no distance moved)

Work - like studying very hard, trying to lift up the car and get

completely exausted, holding weights above head for half an hour is

no work worth mentioning in physics.

you are doing work only while you are

lifting the weight above your head

(force in the direction of distance moved)

A force is applied. Question: Is the work done by that force?

Who’s doing the work around here?

NO WORK WORK

If I carry a box across the room I do not do work on it because the force is not in the direction of the motion (cos 900 = 0)

θ = 00 00< θ <900 θ = 900 900< θ <1800 θ = 1800

cos θ = 1 cos θ = + cos θ = 0 cos θ = – cos θ = –1

Work done by a force F is zero if:

F

motion

normalforce

F

dd

F

force is exerted but no motion is involved:

no distance moved, no work

force is perpendicular to the direction of motion (cos 900 = 0)

gravitationalforce

tension in the string

when the force and direction of motion are generally in the same directions

(the force helps the motion – can increase kinetic energy by increasing speed)

when the force and direction of motion are generally in the opposite directions

(force opposes the motion – decreases kinetic energy by decreasing speed)

00< θ < 900 → cos θ = + cos 00 = 1

Work done by force F is:

W = Fd

W = - Fd

(the work done by friction

force is always negative)

positive

negative

900< θ < 1800 → cos θ = – cos 1800 = –1

Mike is cutting the grass using a human-powered lawn mower.

He pushes the mower with a force of 45 N directed at an angle

of 41° below the horizontal direction. Calculate the work that

Mike does on the mower in pushing it 9.1 m across the yard.

F

410

d

F = 45 N

d = 9.1 m

θ = 410

W = Fd cos θ = 310 J

Forward force is 200 N. Friction force is 200 N.

The distance moved is 200 km. Find

a. the work done by forward force F on the car.

b. the work done by friction force Ffr on the car.

c. the net work done on the car.

F = 200 N

Ffr = 200 N

d = 2x105 m

a. WF = Fd cos 00 = 4x107 J

b. Wfr = Ffr d cos 1800 = - 4x107 J

c. the net work done on the car means the

work done by net force on the car.

It can be found as:

W = WF + Wfr = 0

or

W = Fnet d cos θ = 0 (Fnet = 0)

Work done by a varying force - graphically

W = Fd cos θ applies only when the force is constant.

Force can vary in magnitude or direction during the action.

Examples: 1) rocket moving away from the Earth – force of gravity

decreases 2.) varying force of the golf club on a golf ball

Work done is determined graphically.

The lady from the first slide is pulling the car for 2 m with force of 160 N

at the angle of 60o , then she gets tired and lowers her arms behind her at

an angle of 45o pulling it now with 170 N for next 2 m. Finally seeing the

end of the journey she pulls it horizontally with the force of 40 N for 1 m.

Work done by her on the car is:

W = (160 N)(cos 60o)(2m) +(170 N)(cos 45o)(2m) + (40 N)(cos 0o)(1m)

W = 80x2 + 120x2 + 40x1 = pink area + green area + blue area = 440 J

http://www.kcvs.ca/map/java/applets/workE

nergy/applethelp/lesson/lesson.html#1

In general:

The area under a Force - distance graph equals

the work done by that force

Spring/Tension force – Hooke’s law

Holding one end and pulling the other produces a tension

(spring) force in the spring.

You’ll notice as you pull the spring, that the further you extend

the spring, then the greater the force that you have to exert in

order to extend it even further.

As long as the spring is not streched beyond a certain

extension, called elastic limit, the force is directly proportional

to the extension.

Beyond this point the proportionality is lost.

If you stretch it more, the spring can become permanently

deformed in such a way that when you stop pulling, the spring

will not go back to its original length.

Unstretchedspring

L

Stretchedspring

xp

ulli

ng

forc

e

sp

rin

gfo

rce

F

Compressedspring

x

pu

sh

ing

forc

esp

rin

gfo

rce

F

In the region of proportionality

we can write

F = kx

k is a constant whose value depends on the particular spring.

For this reason k is called the spring constant.

It measures stiffness of the spring in Newtons per metre.

k = 20 N/m

L = 10 cm

How much force do you have to exert

if you want to extend the spring for

a. 1cm

b. 2 cm

c. 3 cm

Example:

a. F = kx = (20 N/m)(0.01 m) = 0.2 N

b. F = kx = (20 N/m)(0.02 m) = 0.4 N

c. F = kx = (20 N/m)(0.03 m) = 0.6 N

proportionality

Note: The spring force has the same value, but

it is in the opposite direction !!!!!

Spring force – Hooke’s law: F = – kx

Note: The spring force has the same value, but it is in the opposite direction !!!!!

The spring force is the force exerted by a compressed or stretched spring upon any object that is attached to it.

x is the displacement (extension/compression) of the spring's end from its equilibrium position

F is the spring force exerted by the spring

k is a constant called spring constant

(in SI units: N/m).

The work done by a non-constant applied force on a Hooke’s

spring is found from the area under the graph F vs. x

Work done by a force F = kx when extending

a spring from extension x1 to x2 is:

2 2 1 1

1 1( ) ( )

2 2W kx x kx x

2 2

2 1

1( )

2W k x x

–

extensionx2x1

F

kx2

kx1

Work done by a force F = kx when extending

a spring from extension 0 to x is:

W = ½ kx2

Energy, E

In physics energy and work are very closed linked; in some senses

they are the same thing. If an object has energy it can do a work. On

the other hand the work done on an object is converted into energy.

work done = change in energy

W = ∆ E

Elastic potential energy, EPE

If some force stretches a spring by extension x, the work done by

that force is ½ kx2. Since work is the transfer of energy, we say

that the energy was transferred into the spring, and that work is

now stored in stretched spring as elastic potential energy.

EPE = ½ kx2

A spring can be stretched or compressed. The same

mathematics holds for stretching as for compressing springs.

Just imagine how much

energy is stored in the

springs of this scale.

Gravitational potential energy, PE = mg h

Imagine a mass m lifted a vertical distance h following two different paths.

1. straight up:

– minimum force needed is F1 = mg.

– the work done against gravitational force is:

W = F1 h cos 00

W = mg h

2. along a ramp a distance d (no friction)

– minimum force needed is F2 = mg sinθ.

– the work done against gravitational force is:

W = F2 d cos 00 = mg sinθ d

W = mg h

hF1d

mg

F2

θ

the work W = mg h is stored

as potential energy of the object.

PE = mg h

It depends on VERTICAL distance

from initial position not on the path taken.

A ramp can reduce the force – can make life easier

1. W = F h = mg h

2. W = F d = mg sinθ h/sinθ = mg h

θ

h

hd

W = F h or F d (along the ramp)

Kinetic energy, KE = ½ m v2

A moving object possesses the capacity to do work. A hammer by virtue

of its motion can be used to do work in driving a nail into a piece of wood.

How much work a moving object is capable of doing? Let’s find out.

Imagine an object of mass m accelerated from rest by a resultant

force, F. After traveling a distance d the mass has velocity v.

The work done is: W = Fd cos 00 = Fd = mad

From v2 = u2 + 2ad → ad = v2/2 and W = ½ mv2

The work done (Fd) by the force F in moving the object a

distance d is now expressed in terms of the properties of the body

and its motion. The quantity ½ mv2 is called the kinetic energy

(KE) of the body and is the energy that a body possesses by

virtue of its motion. The kinetic energy of a body essentially tells

us how much work the body is capable of doing.

Useful relationship between momentum and kinetic energy:

22 21 1 p

KE = mv = (mv) =2 2m 2m

KE depends on the reference frame in which it is measured. When

you are sleeping, you have zero KE realtive to your bed.

But relative to the sun, you have KE = ½ (60 kg) (30000 m/s)2

Summary:

Work done on an object by applied force against gravitational force

(when the net force is zero, so there is no acceleration) is stored

as gravitational potential energy.

PE depends on the reference frame in which it is measured, so we

keep in mind some reference level, like desk, lower reservoir of

water, ground,…. PE = mg ∆h

Work done on a spring by applied force against spring force

(when the net force is zero, so there is no acceleration) is stored

as elastic potential energy… EPE = ½ kx2

If the net force acting on a body is not zero, then:

The work done by net force on a body is equal (results)

to the change in the kinetic energy of the body:

W = ∆KE = KEf – KEi

Units

W = Fd (W) = (1N)(1m) = 1 kg m2 s-2 = 1 J

PE = mgh (PE) = (1kg)(1 m s-2 )(1m) = 1 kg m2 s-2 = 1 J

EPE = ½ kx2 (EPE) = (1N/1m)(1m2) = 1 kg m2 s-2 = 1 J

Even in units we see that the work and energy are equivalent.

KE = ½ mv2 (KE) = (1kg)(1m/s)2 = 1 kg m2 s-2 = 1 J

(1) A father pushes his child on a sled on level ice, a distance 5 m

from rest, giving a final speed of 2 m/s. If the mass of the child

and sled is 30 kg, how much work did he do?

W = ∆ KE = ½ m v2 – 0 = ½ (30 kg)(2)2 = 60 J

other way: W = Fd cosθ = Fd

F = ma

F = 12 N W = 60 J

(2) What is the average force he exerted on the child?

W = Fd = 60 J, and d = 5 m, so F = 60/5 = 12 N

2 22v -u

a = = 0.4 m/s2d

(3)A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the

road does work on the car through a backward-directed friction force.

How much work must this friction force do in order to stop the car?

W = ∆ KE = 0 – ½ m u2 = – ½ (1000 kg) (45 x1000 m/3600 s)2

W = – 78125 J = – 78 kJ

(the – sign just means the work leads to a decrease in KE)

d

u

Ffr

(work done

by friction

force)

v = 0

(4) Instead of slamming on the brakes the work required to stop the car isprovided by a tree!!! What average force is required to stop a 1000-kg car going at 45 km/h if the car collapses one foot (0.3 m) upon impact?

corresponds to 29 tons hitting you OUCH

Do you see why the cars should not be rigid. Smaller collapse distance,

gretaer force, greater acceleration. For half the distance force would

double!!!!

OUCH, OUCH

The net force acting on the car is F

W = – 78125 J

W = – F d = – ½ m u2

F = 260 x 103 N

a = (v2 - u2 )/2d = - 520 m/s2 HUGE!!!!

- W 78125F = =

d 0.3

The principle of energy conservation !!!!!!

An object of mass 4.0 kg slides from rest without friction down an

inclined plane. The plane makes an angle of 30° with the horizontal

and the object starts from a vertical height of 0.5 m. Determine the

speed of the object when it reaches the bottom of the plane.

we shall see in nuclear physics that this should

actually be the principle of mass-energy conservation

To demonstrate the so-called principle of energy conservation

we will solve a dynamics problem in two different ways, one

using Newton’s laws and the kinematics equations and

the other using the principle of energy conservation.

point A: h = 0.5 m, KE = 0 (u = 0)

point B: h = 0 , object has gained KE

h=0

N

300B

A

h=0.5mv=?

u=0

Method 1: Newton’s law Method 2: Energy conservation

As the object slides down the plane its

PE becomes transformed into KE.

If we assume that no energy is ‘lost’ we

can write

F mgsinθa = = = gsinθ

m m

v2 = u2 + 2ad = 2gsinθ

hd =

sinθ

d

h

sinθ

v = 2gh v = 3.2 m/s

PEA = KEB

mgh = ½ mv2

v = 2gh v = 3.2 m/s

Note that the mass of the object does not come into the question, nor does the

distance travelled down the plane, only vertical height.

The second solution involves making the assumption that potential

energy is transformed into kinetic energy and that no energy is converted

into other forms of energy (heat, sound,…).

This is ‘Law of conservation of energy’, this means the energy is

conserved. Energy cannot be created or destroyed; it may be

transformed from one form into another, but the total amount of energy

never changes.

When using the energy principle we are only concerned with the initial

and final conditions and not with what happens in between.

No time included in energy principle.

Very powerful tool of enormous importance.

Clearly in this example it is much quicker to use the energy principle.

This is often the case with many problems and in fact with some

problems the solution can only be achieved using energy considerations.

If it is transformed into thermal energy, we say it is dissipated. The object can’t keep that energy. It is shared with environment.

and now problems and examples. Some of them would be very, very hard to solve using kinematic equations and Newton’s laws.

Dropping down from a pole. As he dives,

PE becomes KE. Total energy is always constant,

equal to initial energy.

In presence of air, some energy gets transformed to

heat (which is random motion of the air molecules).

Total energy at any height would be PE + KE + heat,

so at a given height, the KE would be less than in

vacuum.

What happens when he hits the ground?

Just before he hits ground, he has large KE (large

speed). This gets transformed into heat energy of

his hands and the ground on impact, sound energy,

and energy associated with deformation .

If accounted for air resistance, then how

would the numbers change?

Amusement park physics

the roller coaster is an excellent

example of the conversion of

energy from one form into another

work must first be done in lifting the

cars to the top of the first hill.

the work is stored as gravitational

potential energy

you are then on your way!

PE is being transformed into

KE and vice versa

Up and down the track

PE

PE + KE

PE

If friction is negligible the ball will get

up to the same height on the right side.

Three balls are thrown from the top of the cliff along paths A, B, and C with

the same initial speed (air resistance is negligible). Which ball strikes the

ground below with the greatest speed?

a. A b. B c. C d. All strike with the same speed

h

The initial PE + KE of each ball is: mgh + ½ mu2.

This amount of energy becomes KE before impact ½ mv2.

mgh + ½ mu2 = ½ mv2 m cancels

The speed of impact for each ball is the same:

It depends ONLY on HEIGHT and initial SPEED, not mass, not path !!!!!

2v = 2gh + u

A child of mass m is released from rest at the top of a water slide, at

height h = 8.5 m above the bottom of the slide. Assuming that the slide

is frictionless because of the water on it, find the child’s speed at the

bottom of the slide.

m cancels

on both

sides

Do you think you could use kinematics

equations and Newton’s laws? Try it. Good luck !

v = 2gh = 13 m/s

mgh = ½ mv2

a baby, an elephant and you would reach

the bottom at the same speed !!!!!

And, by the way, it is the same speed as you were to fall straight down

from the same height.

v2 = u2 + 2gh = 2gh

Work is a way of transferring energy from one form to another, but

itself is not a form of energy. All types of energies and work have the

same units. Everything can be transformed into each other.

When work is done on an object, energy is

transferred to that object.

Example: A spring at the bottom of a slide is

compressed by an external force. A parent

releases the spring when a child sits on it.

EPE is transferred to the child by spring

force doing the work which is transformed

into KE of the child. On the way up, KE is

being transformed into PE.

This energy is what enables that child to

then do work. How? First that PE has to be

transformed back into KE – child slides

down a slide – KE of the child can do work

on the waiting parent – it can knock him

over.

Parent does the work on the ground.

A car (toy car – no engine) is at the top of a hill on a

frictionless track. What must the car’s speed be at the top of

the first hill if it can just make it to the top of the second hill?

m cancels out ; v2 = 0

40m

v1mgh1 + ½ mv1

2 = mgh2 + ½ mv22

gh1 + ½ v12 = gh2

v1 = 22 m/s

v2 = 0PE1 + KE1 = PE2 + KE2

20m v12 = 2g( h2 – h1)

A simple pendulum consists of a 2.0 kg mass attached

to a string. It is released from rest at A

as shown. Its speed at the lowest point B is:

2 2

A A B B

1 1mv + mgh = mv + mgh

2 2

2

B Av = 2gh = 2×9.80m/s ×1.85m = 6m/s

1.85 m

A

B

PEA + KEA = PEB + KEB

2

A B

10 + gh = mv + 0

2

● Conservation of energy law AGAIN – WITH FRICTION INCLUDED

All previous examples were neglecting friction force, air resistance …

What happens if the friction force acts?

Can we still use conservation of energy principle?

YES, WE CAN

● Example: An object of mass 4.0 kg slides 1.0 m down

an inclined plane starting from rest. Determine the

speed of the object when it reaches the bottom of

the plane if

a. friction is neglected b. constant friction force of 16 N acts

on the object as it slides down.

all PE was transformed into KE Friction converts part of KE of the object

into heat energy. This energy equals to

the work done by the friction. We say that

the frictional force has dissipated energy.

initial energy = final energy initial energy – Ffr d = final energy

Wfr = – Ffr d decreases KE of the object

PEA = KEB PEA – Ffr d = KEB

mgh = ½ mv2 mgh – Ffr d = ½ mv2

20 – 16 = 2.5 v2

v = 1.4 m/sv = 2gh = 3.2 m/s

Types of energy and energy transformations

We usually classify energy as the following forms:

1 Kinetic energy

2 Gravitational potential energy

3 Elastic potential energy

4 Thermal or heat energy

5 Light energy

6 Sound energy

7 Chemical energy

8 Electrical energy

9 Magnetic energy

10 Nuclear energy

There are energy changes happening all around us all of

the time. Here are a few examples to think about.

In any given transformation, some of the energy

is almost inevitably changed to heat.

• A bouncing ball

The gravitational potential energy is changed to kinetic

energy and back again with each bounce losing energy

due to work done by air friction and nonelastic contact

with the ground . Eventually the ball comes to rest and

all the energy has become low grade heat.

• An aeroplane taking off

As fuel is burned in the engines, chemical energy is converted

to heat, light and sound. The plane accelerates down the

runway and its kinetic energy increases. There will be heat

energy due to friction between the tyres and the runway. As

the plane takes off and climbs into the sky, its gravitational

potential energy increases.

• A laptop computer

The electrical energy form the mains or chemtcal energy from

the battery is changed to light on the screen and sound through

the speakers. There is kinetic energy in the fan, magnetic energy

in the motors, and plenty of heat is generated.

• A human body

Most of the chemical energy from our food is changed to heat

to keep us alive. Some is changed to kinetic energy as we

move, or gravitational potential energy if we climb stairs. There

is also elastic potential energy in our muscles, sound energy

when we talk and electrical energy in our nerves and brain.

• A nuclear power station

Nuclear energy from the uranium fuel changes to thermal

energy that is used to boil water. The kinetic energy of the

steam molecules drives turbines, and as the kinetic energy of

the turbines increases, it interacts with magnetic energy to

give electrical energy.

Power

Power is the work done in unit time or energy converted in unit time

measures how fast work is done or how quickly energy is converted.

Power is a scalar quantity.

◘

Units: ◘1J(joule)

1 W(Watt) = 1s

A 100 W light bulb converts electrical energy to heat and light at

the rate of 100 J every second.

Sometimes you’ll see power given in kW or even MW.

orW E

P = P = t t

Calculate the power of a worker in a supermarket who

stacks shelves 1.5 m high with cartons of orange juice,

each of mass 6.0 kg, at the rate of 30 cartons per minute.

0W Fdcos0 (30×60N)×1.5mP = = =

t t 60sP = 45 W

There is another way to calculate power

0W Fdcos0 dP = = = F

t t tP = Fv

Power is equal to force times velocity, providing that both

force and velocity are constant and in the same direction.

Constant velocity with a force applied ??????

That’s because we are interested in one force only. Net force

is obviously zero. Like power of the engine of the car.

Efficiency

Efficiency is the ratio of how much work, energy or power we get

out of a system compared to how much is put in.

◘

useful outputefficiency =

total input

No units◘

Efficiency can be expressed as percentage by multiplying by 100%.◘

out out out

in in in

W E Peff = = =

W E P

No real machine or sysem can ever be 100% efficient,

because there will always be some energy changed to

heat due to friction, air resistance or other causes.

◘

A car engine has an efficiency of 20 % and produces an

average of 25 kJ of useful work per second.

How much energy is converted into heat per second.

out

in

Eeff =

E

in

25000J0.2 =

EEin = 125000 J

heat = 125 kJ – 25 kJ = 100 kJ