10 Work 021

CHAPTER 7

WORK ,KINETIC ENERGY & POWER

Objectives

At the end of the lesson, you should be able to:

State and explain work, kinetic energy, and the relationship between the two Solve problems involving work, kinetic energy, and the work-energy theorem Calculate the work done by a varying force like that of a spring Solve problems involving average and instantaneous power

OVERVIEW

I. Work and Kinetic Energy A. Work B. Kinetic Energy C. Work Done by a Varying Force D. Power

II. Potential Energy and Energy Conservation E. Gravitational Potential Energy F. Conservation of Mechanical Energy G. Elastic Potential Energy H. Conservative and Nonconservative Forces I. Law of Conservation of Energy

Kinetic Energy

KINETIC ENERGY (KE) - physical parameter to describe the state of motion

of an object of mass m and speed v

2

21 mvKE

J joule 1 2

2

1:s

mkgunitIS

Work

WORK (W) -energy transferred to or from the object that account for changes in KE. SIGN CONVENTION: • W (positive )>> work W is done on/to the object by force F >> Energy is transferred TO the object >> KE increases

• W (negative) >> work W is done by the object ; >> Energy is transferred FROM the object >> KE decreases

Work Done by a Constant Force

m m

Ex. Work done BY the force ON the bead

cosW Fd

dF

W

Worthy of NOTE:

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

AF

BF

CF

Some important assumptions: 1. Applicable when F is constant 2. W (+) if : 0< ϴ<90ᵒ , W(-) if : 90°< ϴ<180ᵒ 3. For several forces , get Fnet first or individual W to

get Wnet

dFWnet

WWWWnet

net

CBA

Work Done by a Constant Force

cosW Fd

Work Done by a Constant Force

d

Work Done by a Constant Force

1180cos

090cos

10cos

d

d

cosW Fd

2.17 x 103 J

Work Done by a Constant Force

Example 1. Pulling a Suitcase-on-Wheels. Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m.

d

d

F=5.4 N

Work Done by a Constant Force

Example 2. A particle moving in the xy plane undergoes a displacement d = (2.0i + 3.0j) m as a constant force F = (5.0i + 2.0j) N acts on the particle. (a) Calculate the magnitude of the displacement and that of the

force. (b) Calculate the work done by F. (c) Calculate the angle between F and d.

d=3.6 m

W=16J

Ө=35°

W O R K ! W O R K! W O R K!

Work done by a Force in Lifting an Object Work done by a Force in Lowering an Object

Work Done by a Force

mgdmgdFdW

mgdmgdFdW

F

g

0cos0cos

180cos180cos

d

d mgdmgdFdW

mgdmgdFdW

F

g

180cos180cos

0cos0cos

Work by

GRAVITATIONAL

FORCE

Work by

GRAVITATIONAL

FORCE

Work by

APPLIED FORCE

Work by

APPLIED FORCE

Work Done by a Force

Example 3. Bench-Pressing. The weight lifter is bench-pressing a barbell whose weight is 710 N. He raises the barbell a distance of 0.65 m above his chest, and he lowers it the same distance. The weight is raised and lowered at a constant velocity. Determine the work done on the barbell by the weight lifter during (a) the lifting phase and (b) the lowering phase.

4.6 x 102 J

-4.6 x 102 J

Work Done by a Variable Force

Work done by a Variable Force

f

i

x

x

dxxF )(W

The Spring Force:

Hookes Law where : k –spring constant

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"

Work Done by a Variable Force

kxs F

Work Done by the Spring Force where : k –spring constant

O (b)

xo

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

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

Work Done by a Variable Force

fx

x

s

s

kxW

dx)x(FW

0

2

2

Work in 3D analysis

Work Done by a Variable Force

O x

y

z

A

B

path

f

i

f

i

f

i

z

z

z

y

y

z

x

x

x dzzFdyyFdxxF )()()(W

Work Done by a Force

Example 5. A force acting on a particle varies with x. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m

W=25J

W O R K ! W O R K! W O R K!

The Work-Kinetic Energy Theorem

THE WORK-ENERGY THEOREM When a net external force does work on and object, the kinetic energy of the object changes according to

2

212

f21

of KEKE omvmvW

d

of

ofxx

ofxxx

KEKEFd

vvmxamdF

vvmamF

cos

)(

22

21

22

21 xavv xof 222

22

21

ofxx vva

Example 6. A 6.0-kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N. a. Find the speed of the block after it has moved 3.0 m (using

Work KE Theorem) b. Find the acceleration of the block and determine its final speed, (using the kinematics equation)

a. v = 3.5m/s

b. a=2.0 m/s2; v = 3.5m/s

The Work-Kinetic Energy Theorem

Example 8. A block of mass 1.6 kg is attached to a horizontal spring that has a force constant of 1.0 x 103 N/m. The spring is compressed 2.0 cm and is then released from rest. (a) Calculate the speed of the block as it passes through the equilibrium position x =0 if the surface is frictionless. (b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 4.0 N retards its motion from the moment it is released.

vf =0.50 m/s

vf = 0.39 m/s

The Work-Kinetic Energy Theorem

O (b)

x

O (c)

xf

x

O

x xo

Example 9. Downhill Skiing. A 58-kg skier is coasting down a 25° slope. Near the top of the slope, her speed is 3.6 m/s. She accelerates down the slope because of the gravitational force, even though a kinetic frictional force of magnitude 71 N opposes her motion. Ignoring air resistance, determine the speed at a point that is displaced 57 m downhill.

v = 19 m/s

d

The Work-Kinetic Energy Theorem