Work, Energy, Power and Conservation Laws. In this week we will introduce the following concepts: o Kinetic energy of a moving object o Work done by a.

Work, Energy, Power and Conservation Laws.

In this week we will introduce the following concepts:

o Kinetic energy of a moving object

o Work done by a force

o Power

o Potential Energy

o Conservative and non-conservative forces

o Mechanical Energy

o Conservation of Mechanical Energy

The conservation of energy theorem will be used to solve a variety of problems

(7-1)

o In addition we will develop the work-kinetic energy theorem and apply it to solve a variety of problems

o This approach uses scalars such as work and kinetic energy rather than vectors such as velocity and acceleration. Therefore it simpler to apply.

o It cannot be used to solve all problems, particularly those which demand an answer involving position as a function of time. But it is best to try to use it first.

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

We can use the equation above to define the SI unit for work (the joule, symbol: J ). An object of mass m = 1kg that moves with speed v = 1 m/s has a kinetic energy K = 1J

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 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. If on the other hand energy its transferred from the object (its K decreases) we say that work was done by m (W < 0) (7-2)

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

ead

2

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

third equation of kinematics:

x x

o

F ma

v 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

mm

K v K K Fd

W

ad is given by: cos

xW F d Fd

cosW Fd W F d

(7-3)

m m

AF

BF

CF m m

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 W

work

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

by force , and by force . Then determine

C

Method 1:

Method 2: alculate first ;

n

net

A A

B B C C

net B C

C

A

et A BW

W

W F

W F W F

F F F

W W

F

W

Then determine netW F d

cosW Fd

W F d

(7-4)

m m

We have seen earlier that: .

We define the change in kinetic energy as:

. The equation above becomes

th work-kinetic energy te heorem

f i net

f i

K K W

K K K

Work-Kinetic Energy Theorem

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

(7-5)

A

B

Consider a tomato of mass that is thrown upwards at point A

with initial speed . As the tomato rises, it slows down by the

Work Done by the Gravitational

gravitational f

For

orce so that at po

ce:

o

g

m

v

F

int B its has a smaller speed .

The work A B done by the gravitational force on the

tomato as it travels from point A to point B is:

A B cos180

The work B A done by the gravitational forc

g

g

g

v

W

W mgd mgd

W

e on the

tomato as it travels from point B to point A is:

B A cos 0gW mgd mgd

(7-6)

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

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

t

f i net

i f net

g

K K K W

K K K W

F 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 cos0

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

(7-7)

A

B

m

.

Work Done against Friction

W F d

So work done is

W = Force x distance = R d

Where is the coefficient of dynamic friction R = mg is the force down (due to gravity) d is the distance pushed

10kg

F

d

Push a weight at constant speed against friction over a surface

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

,

,1

,1

is the average value of F

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

N

j avgj

xN

j avgj 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 (7-8)

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

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 (7-9)

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

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 2sf

kx Wx x

x

(7-10)

O (b)

xi

x

O (c)

xf

x

O (a)

x

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: , , , , , ,

Point s A and B have coordinates , , and , , , respectively

f f f

i i i

x y z

i i i f f f

x y z

x y zB

x y z

A x y z

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

Oxy

z

A

B

path

f f f

i i i

x y z

x y z

x y z

W F dx F dy F dz

(7-11)

. .O x-axis

xdx

F(x)A Bm

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 :

Thus the integral beco

i f

dvdx Fdx m dx

dt

dvdx x x Fdx m dx

dt

dv dv dx dv dv dxdx dx vdv

dt 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!

ff

ii

xx f i

f ixx

mv mvmW m vdv v K K K

F

f iW K K K (7-12)

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:

dWP

dt

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)(7-13)

(7-14)

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

cosP Fv F v

v