Work & Kinetic Energy

Kinetic Energy: We define a new physical parameter to

describe the state of motion of an object of mass m and

speed v

We define its kinetic energy K as:

2

2

mvK =

Work:(symbol W)

If a force F is applied to an object of mass m it can accelerate it

and increase its speed v and kinetic energy K. Similarly F can

decelerate m and decrease its kinetic energy.

We account for these changes in K by saying that F has

transferred energy W to or from the object. If energy ittransferred energy W to or from the object. If energy it

transferred to m (its K increases) we say that work was done by

F on the object (W > 0). If on the other hand energy its

transferred from the object (its K decreases) we say that work

was done by m (W < 0)

m m

Consider a bead of mass that can move

without friction along a straight wire along

the x-axis. A constant force applied at an

angle

Finding an expression for Work

to the wire is acting on th b

:

e

m

F

φ

�

ead

We apply Newton's second law: We assume that the bead had an initial

velocity and after it has travelled a distance its velocity is . We apply the

x x

o

F ma

v d v

=�

� �

2

velocity and after it has travelled a distance its velocity is . We apply the

third equation of kinematics:

ov d v

v v− 2

2 2 2

2

2 We multiply both sides by / 2

2 2 cos 2 2 2 2 2

The change in kinetic energy cos 2

Thus the work done the force the beby on

o x

xo x x i o

f f i

a d m

Fm m m m mv v a d d F d F d K v

m

mK v K K Fd

W

φ

ϕ

= →

− = = = = =

= → − =

ad is given by: cos

xW F d Fd ϕ= =

cosW Fd ϕ= W F d= ⋅��

Scalar Product of Two Vectors

Definition of the scalar, or dot, product:

Therefore, we can write:

The unit of is the same as that of i.e.

The expressions for work we have developed apply when is constant

We have made the implicit assumption that the m

Note 1:

Note oving objec t

jo

i

ule

s p2 -

s

: oint

W K

F

like

0 if 0 90 , 0 if 90 180

If we have several forces acting on a body (say three as in the picture)

there are two methods that can be used to calculate the

Note 3:

Net Wor :

n

k

et

W Wφ φ> < < ° < ° < < °

work

First calculate the work done by each force: by force ,Method 1:

net

A A

W

W F

= + +

�

� �

by force , and by force . Then determine

CMethod 2: alculate first ;

nB B C C

net B C

C

A

et A BWW F W F

F F F

W W

F

W

=

=

+

+

+

+� �

� � � �

Then determine netW F d= ⋅��

AF�

BF�

CF�

In the SI system, the units of work are joules:

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 init. The force he exerts has no component in

the direction of motion.

The person does no work on the bag of groceries

since FP is perpendicular to the displacement d.

Conceptual : 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)gravity do (a) positive work, (b)

negative work, or (c) no work at all on

the Moon?

Gravity does no work on the Moon, as the force is always perpendicular

to the displacement (assuming a circular orbit).

A

B

C o n sid er a to m ato o f m ass th a t is th ro w n u p w ard s a t p o in t A

W o rk D o n e b y th e G rav ita tio n a l F o rce :

m

w ith in itia l sp eed . A s th e to m ato rises , it s lo w s d o w n b y th e

g rav ita tio n a l fo rce so th a t a t p o

o

g

v

F

( )

( )

( )

in t B its h as a sm alle r sp eed .

T h e w o rk A B d o n e b y th e g rav ita tio n a l fo rce o n th e

to m ato as it trave ls fro m p o in t A to p o in t B is :

A B co s 1 8 0

T h e w o rk B A d o n e b y th e g rav ita tio n a l fo rc

g

g

g

v

W

W m g d m g d

W

→

→ = ° = −

→

( )

e o n th e

to m ato as it trave ls fro m p o in t B to p o in t A is :

B A co s 0gW m g d m g d→ = ° =

A

B

m

.Consider an object of mass m that is lifted by a force F form

point A to point B. The object starts from rest at A a

Work done by a force in Lifti

nd arrives

at B with zero spee

ng an o

d. The

bject:

force F is not necessarily constant

during the trip.

The work-kinetic energy theorem states that:

We also have that 0 0 There are two forces

f i net

i f net

K K K W

K K K W

∆ = − =

= → ∆ = → =

acting on the object: The gravitational force and the applied force

t

gF F

( ) ( )

( ) ( )

( ) ( )

( )

Work done by

hat lifts the

a force in Lo

object. A B A B

wering an obje

0

A B A B

A B cos180 - A B

In this case the object moves from B to A

B A cos 0

ct:

net a g

a g

g a

g

W W W

W W

W mgd mgd W mgd

W mgd mgd

= → + → = →

→ = − →

→ = ° = → → =

→ = ° = ( ) ( ) B A B A = a gW W mgd→ = − → −

A force that is not constant but instead varies as function of

is shown in fig

Work done by a variable

.a. We wish to calcula

force (

te

) act

the work that does

on an obje

ing along the

c

-axis:

F x

W F

F x x

( )

,

t it moves from position to position .

We partition the interval , into "elements" of length

each as is shown in fig.b. The work done by in the - th

interval is: Where

i f

i f

j j avg

x x

x x N

x F j

W F x F

∆

∆ = ∆ , is the average value of F

over the -th element. We then take the limit of

j avg

N

j W F x= ∆∑ ,

1

,

1

over the -th element. We then take the limit of

the sum as 0 , (or equivalently )

lim ( ) Geometrically, is the area

b

f

i

j avg

j

xN

j avg

j x

j W F x

x N

W F x F x dx W

=

=

= ∆

∆ → → ∞

= ∆ =

∑

∑ ∫

etween ( ) curve and the -axis, between and

(shaded blue in fig.d)

i fF x x x x

( )

f

i

x

x

W F x dx= ∫

Work Done by a Varying Force

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

constant!

F�

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.

Work done by a force F is (a) approximately equal to the sum of the

areas of the rectangles.

In the limit that the pieces become infinitesimally narrow, the work is

the area under the curve:

Or:

Work done by a force F is exactly equal to the area under the curve of F cos θ vs. l.

Fig.a shows a spring in its relaxed state.

In fig.b we pull one end of the spring and

stretch it by an amount . The spring

resits by exerting a force on our hand

T

he Spring

in

the opp

o

Forc

si di

e:

te

d

F

rection.

In fig.c we push one end of the spring and

compress it by an amount . Again the dcompress it by an amount . Again the

spring resists by exerting a force on our

hand in the opposite direction

d

F

The force exerted by the spring on whatever agent (in the picture our hand)

is trying to change its natural length either by extending or by compressing it

is given by the equation: Here x

F

F kx= − is the amount by which the spring

has been extended or compressed. This equation is known as "Hookes law"

k is known as "spring constant" F kx= −

Work done by a spring force:

The force exerted by a spring is given by:

(a) Spring in normal (unstretched) position. (b) Spring is stretched by a

person exerting a force FP to the right (positive direction). The spring

pulls back with a force FS where FS = -kx. (c) Person compresses the

spring (x < 0) and the spring pushes back with a force FS = kx where FS >

0 because x < 0.

Plot of F vs. x. Work done is equal to

the shaded area.

Work done to stretch a spring a distance x equals the triangular area under

the curve F = kx. The area of a triangle is ½ x base x altitude, so W =

½(x)(kx) = ½ kx2.

O (b)

xi

x

O (c)

xf

x

O (a)

x

i f

Consider the relaxed spring of spring constant k shown in (a)

By applying an external force we change the spring's

length from x (see b) to x (s

Work Done by

ee c). We wi

a

ll

Spring F

calcul

orce

ate the work done by the spring on the external agent

(in this case our hand) that changed the spring length. We

assume that the spring is massless and that it obeys Hooke's law

sW

x x x

222

We will use the expression: ( )

Quite often we start we a relaxed 2 2 2

spring ( 0) and we either stretch or compress the spring by

f f f

i i i

f

i

x x x

s

x x x

x

fis

x

i

W F x dx kxdx k xdx

kxkxxW k

x

= = − = −

= − = −

=

∫ ∫ ∫

2

an

amount ( ). In this case 2

sf

kx Wx x

x= −= ±

The spring constant: k = 2.5 x 103 N/m. Then W = 1.1 J, whether

stretching or compressing.

Example : Force as a function of x.

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

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?

Do the integral. W = 0.36 J.

In the general case the force acts in three dimensional space and moves an object

on a three dimensional path from an initial point A to a fin

Three dimensi

al point B

on

Th

al An

e for

aly

c

sis:

e has t

F�

( ) ( ) ( )

( ) ( )

ˆˆ ˆhe form: , , , , , ,

Po int s A and B have coordinates , , and , , , respectively

f f f

x y z

i i i f f f

x y z

x y zB

F F x y z i F x y z j F x y z k

x y z x y z

dW F dr F dx F dy F dz

W dW F dx F dy F dz

= + +

= ⋅ = + +

= = + +∫ ∫ ∫ ∫

�

�

�

i i i

x y z

A x y z

W dW F dx F dy F dz= = + +∫ ∫ ∫ ∫

f f f

i i i

x y z

x y z

x y z

W F dx F dy F dz= + +∫ ∫ ∫ Ox

y

z

A

B

path

Kinetic Energy and the Work-Energy Principle

If we write the acceleration in terms of the velocity and the distance, we find that the work

done here is

We define the kinetic energy as:

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.

We define the kinetic energy as:

A constant net force Fnet accelerates a car from speed v1 to speed v2 over a displacement

d. The net work done is Wnet = Fnetd.

Work-Kinetic Energy Theorem

W e h a v e s e e n e a r lie r th a t : .

W e d e f in e th e c h a n g e in k in e t ic e n e rg y a s :

. T h e e q u a t io n a b o v e b e c o m e s

th w o rk -k in e t ic e n e rg y te h e o re m

f i n e t

f i

K K W

K K K

− =

∆ = −

f i netK K K W∆ = − =

Change in the kinetic net work done on

energy of a pareticle the particle

=

The work-kinetic energy theorem holds for both positive and negative values of

If 0 0

If 0 0

net

net f i f i

net f i f i

W

W K K K K

W K K K K

> → − > → >

< → − < → <

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 positive, the kinetic energy increases.

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

Example : 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.

Example : 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? (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. (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 (or perhaps your finger).7.0 N is acting to slow it down, such as friction (or perhaps your finger).

Conside a variable force F(x) which moves an object of mass m from point A( )

to point B( ). We apply Newton's second law: We th

Work-Kinetic Energy Theorem with a Variable Force:

i

f

x x

dvx x F ma m

dt

=

= = =

f f

i i

x x

x x

en

multiply both sides of the last equation with and get:

We integrate both sides over from to : i f

dvdx Fdx m dx

dt

dvdx x x Fdx m dx

dt

=

=∫ ∫

Thus the integral becodv dv dx dv dv dx

dx dx vdvdt dx dt dt dx dt

= → = =

2 22

mes:

2 2 2

Note: The work-kinetic energy theorem has exactly the same form as in the case

when is constant!

f

f

i

i

xx f i

f ixx

mv mvmW m vdv v K K K

F

= = = − = − = ∆ ∫

We define "power" as the rate at which work is done by a force .

If does work in a time interval then we define as the as: average pow

Power

er

P F

F W t∆

avg

WP

t=

∆

The instantaneous po is definedwer as: dW

Pdt

=

The SI unit of power is the watt. It is defined as the power Unit of :P

POWER

The SI unit of power is the watt. It is defined as the power

of an engine that does work = 1 J in a time = 1 second

A commonly used non-SI power unit is the horsepower

Unit of

(hp) de i

:

f ne

t

P

W

6

d as:

1 hp = 746 W

The kilowatt-hour (kWh) is a unit of work. It is defined

as the work performed by an engine of power = 1000 W in a time = 1 hour

1000 3600 3.60 10

The

J

kilowatt-hour

P t

W Pt= = × = × The kWh is used by electrical utility

companies (check your latest electric bill)

Consider a force acting on a particle at an angle to the motion. The rate

cosat which does work is given by: cos cos

F

dW F dx dxF P F Fv

dt dt dt

φ

φφ φ= = = =

cosP Fv F vφ= = ⋅�

�

v