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King Fahd University of Petroleum& Minerals

Mechanical Engineering

Dynamics ME 201

BY

Dr. Meyassar N. Al-HaddadLecture # 15

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Chapter 14

Kinetics of a Particle ; Work Energy

Objectives To develop the principle of work and energy and

apply it to solve problems that involve force, velocity, and displacement.

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Work of a force

Force F does work on a particle onlywhen the particle undergoes a

displacement in the direction of the force

F

S

U = F . S

F

U = (F cos q) . S

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The Work of a Force

The unit of the work: Joule : 1 joule of work is done whena force of 1 Newton moves 1 meter along its line of action1 J = 1 N. m

English Unit Ib.ft

negativeisU

zeroisU

positiveisU

90

90

90

o

o

o

q

F

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Work of a Constant Force Moving Along a Straight Line

2

1

cos21s

sc dsFU

)(cos 1221 ss

FU c

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Work of a Variable Force

dsFdUs

s 2

1

cos212

1

r

r

rF

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Work of a Weight

)(

)()(

12

21

2

1

yyWWdy

dzdydxWdU

y

y

2

1

r

rkjijrF

)(21 yWU

::

::ntdisplacemevertical:

W

W

U

Uy

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Work of a Spring Force

)2

1

2

1(

21

2221 ksksU

)sskdsskdsFUS

S

S

S s-

2

1

2

221 (2

12

1

2

1

Positive work

Negative work

ksFs

spring takes workeither compressed orStretched. Therefore,Negative sign is placed

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Constant towing force T do positive work,

W displacement vector negative work ??

Normal force no displacement no work !

Positive and negative work ?

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J505198908692 ..UUUWFP s

J869230cos24001 .)(NdsPUP )

J90)5052(302

1)(

2

12 22

2

1

2

2 ..sskU) sF

J198)30sin2(198)(3 ..ymgU) W

ntdisplacemethetolarperpendicualwaysisitsinceworknodoesNB

?

5.0

1030

30400

TotalU

msretchedoriginal

kgm

N/mkNP

Example 14.1

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14.2 Principle of Work and Energy

dvmvdsF

dsdvmvF

ds

vdvadvvdsa

maF

v

v

s

s t

t

tt

tt

2

1

2

1

equationKinetictheapplying

2

1

2

22121

21 mvmvU

2211 TUT

Useful equationto obtain final speedor

displacement

EnergyKinatic21 2mvT

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14.3 Principle of Work and Energy for a System of

Particles

2211 TUT

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Work and Energy relation

2211 TUT

)2

1

(])(2

1

[)2

1

(

2

2

22

1 mvmv NsyWksFs k Initial K.E + Work Done = Final K.E

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2

2

2

1

2211

2

1)sin(

2

1

Energy&WorkofprincipletheUsing

mvsFsWmv

TUT

A

Example 14.2

w= 3500 Ibv1= 20 ft/sV2=0k=0.5s= ?

x

y

Ib4.17238.3446*5.0cos q WNFA

ft519

0)(4.172310sin)(3500)20)(2.32

3500(2

1 2

.S

SS

Ib8.344610cos3500cosN0 qWFy

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Review

Example 14-3

Example 14-4

Example 14-6