presentation chapter6

7/27/2019 Presentation CHAPTER6

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CHAPTER 6

WORK and KINETIC ENERGY

Work Done by a Constant Force

Work Done by a Variable Force

Work- KineticEnergy Theorem

Center of Mass Work


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The energy is one of the most important concept that helps us describe

many processes in the world around us.

Falling water releases stored gravitational potential energy turninginto a kinetic energy of motion. This mechanical energy can be used

to spin turbines and alternators doing work to generate electrical

energy.

Human beings transform the stored chemical energy of food intovarious forms necessary for the maintenance of the functions of the

various organ system, tissues and cells in the body.Burning gasoline in car engines converts chemical energy stored in

the atomic bonds of the constituent atoms of gasoline into heat that

then drives a piston of combustion engine.Stretching or compressing a spring stores elastic potential energy that

can be released as kinetic energy.

The process of vision begins with stored atomic energy released as

electromagnetic radiation (light) that is detected by exciting atoms in theeye.

In physics Work it is transfer energy by force.


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WORK DONE BY CONSTANT FORCE

In physics Work it is transfer energy by force.

The Work is scalar value and can be positive negative and zero.

The work done by constant force is equals to the component force in the

direction of the displacement times the magnitude of the displacement.

Or another definition Work done by constant force is scalar (or dot)product of force vector and displacement vector.

coslFlFW

Assuming risinx direction.

xFxFW x cos


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The SI unit of work is thejoule (J), which is equals

The product of Newton and meter.

1J = 1N*m

In the U.S. customary system, the unit of work is

foot-pound.

1ft*lb = 1.356J

Unit of work that use in nuclear and quantum physics.

is electron volt (eV).

1 eV = 1.62*10^-19J

For the multiple forces, that act on object we will have.

.....332211 xFxFxFWtotal


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When force Fnetapplied to object and object move under influence

of this force with acceleration afrom A to B along the x axis.

The physical value Kmv 22 is scalar represented energy associated

with motion and call Kinetic Energy and

Work Kinetic Energy Theorem Kmvmv

W if

22

22

Completed work is equal change in kinetic energy of object

222

122

2 AB

v

v

v

v

B

A

AB

net

B

A

netAB

mvmvmvdvvmdtv

dt

dvmW

dt

dxv

dt

dvmmaFdxFW

B

A

B

A

AB

dlFldFdW cos

ABAB KKW


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WORK IN SCALAR (or DOT) PRODUCT NOTATION

The workdWdone by forceFfor displacement dlis.

dlFldFdW cos

2

1

ldFW

Definition of Work

cosABBA

Definition for scalar or dot product of two vectors.zzyyxxzyxzyx BABABAkBjBiBkAjAiABA

dzFdyFdxFkdzjdyi dxkFjFi FldFdW zyxzyx

2222

21

22

21

22

21

22

21

22

2

1

2

1

2

1

2

1

2

1

2

1

vvmvvmvvmvvm

dvmvdvmvdvmvdzFdyFdxFW

zzyyxx

zzyyxxzyx


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kBjBiBkAjAiABA zyxzyx

02cos

10cos

jijiiiiizzyyxx BABABABA

Component of vector in specific direction

of unit vector:

xzyx AikAjAiAiA )(

Same rules is applied to sum of vectors

xxx

CBAiCiBACBA )(

dt

BdAB

dt

AdBA

dt

d

The rules for differentiating

Dot Product of Two Vectors


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Negative work means

A the kinetic energy of the object increases.

B the applied force is variable.

C the applied force is perpendicular to the displacement.

D the applied force is opposite to the displacement.

E nothing; there is no such thing as negative work.


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Two athletes numbers 1 and 2 run a race in beginning they have same

kinetic energy, but athlete #2 is running faster than the 1st. When athlete

#1 increase speed by 25 percent, he is running at the same speed as theathlete #2. If mass athlete #1 is 85 kg, what is mass athlete #2?

22221

21121 vmvm

Using the definition of kinetic energy2

2

112

v

vmm

Express the condition on speed that enables to run athlete #1 at the

same speed as the athlete #2: v2 = 1.25v1

Substitute forv2 and simplify to obtain:2

1

2

1

112

25.1

1

25.1

m

v

vmm

kgkg 5425.1

185

2

2

m

Conceptual problem:


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Work Done by Variable Force

Straight Line Motion

Work Done by Variable Force StraightLine Motion

2

1

x

x

xdxFW

This is area under curveFversusx

i

ixix

xFlimWi

0

xFW x


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A 3.0-kg object moving along thex axis has a velocity of +2.0 m/s as

it passes through the origin. It is subjected to a single force,

that varies with position. How much work is done by

the force as the object moves fromx = 0.0 m tox = 4.0 m? What isthe speed of the object when it is atx = 4.0 m?

xkNFx sin3

mk 3

The force and displacement are parallel, the work done is the area

under the curve.

JkxkN

dxxkNdxFW x 3.4cos3

sin3

4

0

4

0

4

0

According to the work-kinetic energy theorem:

if KKW if KWK

smvm

Wv

mvWmv

if

if

/.622

22

2

22

What work is done by the force if object move 6m?


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Work Done by Spring

Hookes Law

kxFx

The force varies withx.

Work by spring force is.

22)(

21

22

2

1

2

1

xxkdxkxdxFW

x

x

x

x x

22

22

21

fikxkx

AAW

We can use geometry to calculate area

under the curve

22

22fi

kxkxW


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One end of a light spring (force constant k) is attached to the ceiling,

the other end is attached to an object of mass m. The spring initially

is vertical and unstressed. You then ease the object down to an

equilibrium position a distance h below its initial position. Next, yourepeat this experiment, but instead of easing the object down, you

release it, with the result that it falls a distanceHbelow the initial

position before momentarily stopping. Find h and H.

Applyyy amF

to the object in its equilibrium position:0gs FF

khF s ,g mgF 0mgkh

k

mgh

Apply the work-kinetic energy theorem to the spring-object

system to obtain: KW extnet, or, because the object begins and ends at rest, 0extnet, W

0springbygravityby WW hH 2021 2

kHmgH k

mg

H 2

1


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A block with mass m is pulled up a slope with constant velocity.

Calculate the amount of work done by the force after the block has

moved to a height h.

sin

0sin

mgF

mgFFFmaFgx

This force acts over a distance d. The value of d is fixed by the angle

and the height h: sindh

The work done by the force on the block is given by

mghh

mgdFWF

sin

sin

Notice that work by force on block das not

depend from angle and length of incline

and only function of height h.


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The work done on the block by the gravitational force is given by

mghmgddFW GG sin

The work done on the crate by the normal force N is zerosince N is perpendicular to d.

We conclude that the total work done on the block is given by

00 mghmghWWtotal

gF


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A 6.0-kg block slides 1.5 m down a frictionless incline that makes an

angle of 60 with the horizontal. (a) Find the work done by each force

when the block slides 1.5 m (measured along the incline). (b) What is

the total work done on the block? (c) What is the speed of the block afterit has slid 1.5 m, if it starts from rest? (d) What is its speed after 1.5 m, if

it starts with an initial speed of 2.0 m/s?

The work done by gF

cosggg sFsFWF J7mgsWgF 630cos

The work done bynF cosnnn sFsFWF

90 0WF n

(c) The total work done on the block is the sum of the work done by

JWWWng FFtot

76

(a)


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(c) Apply the work-kinetic energy theorem to the block to obtain:

ifgKKKW

F or, because vi= 0

fg KWF 2f21cos mvmgs

smgsv /5cos2f (d) IfKi 0, equation (1) becomes:

(1)

2

i

2

fif2

1

2

1g

mvmvKKWF

Solving forvf and simplifying yields:

smvgsv /4.5cos22

if


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You have two icy slops.

Assume that the surfaces of objects 1 and 2 of same mass are frictionless.

Which object will have greater final speed 1 or 2?

h Fnmg

1

ghv

mghmvmvmv

W

mghdymgW

mgdyjdyidxjmgdlgmdW

dlFldFdW

WWW

f

fif

total

hg

g

nnn

gntotal

2

222

0)2/cos(

222

0

2


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True or false:

(a) The scalar product cannot have units.

(b) If the scalar product of two nonzero vectors is zero, then they areparallel.

(c) If the scalar product of two nonzero vectors is equal to the

product of their magnitudes, then the two vectors are parallel.

(d) As an object slides up an incline, the sign of the scalar product ofthe force of gravity on it and its displacement is negative.


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(a) False. Work is the scalar product of force and displacement.

(b) False. Because ,cosABBA where is the angle betweenif the scalar product of the vectors is zero, then must be 90

(c) True. If the scalar product of the vectors is equal to the product

of their magnitudes, then must be 0 and the vectors are parallel.

(d) True. Because the angle between the gravitational force and the

displacement of the object is greater than 90, its cosine is negative


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A particle undergoes a displacement . During

this displacement a constant force acts on

the particle. Find (a) the work done by the force, and (b) thecomponent of the force in the direction of the displacement.

jNiNF 43

jmiml 52

Nll

W

l

lFFlFlF

FFlFlF

lll

JmNmNW

lFlFlFW

yx

ll

llFlF

yx

yyxx

6.2

coscos

145423

22

22

Solution:

Problem

a)

b)

lF

C t l bl


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Find the unit vector that is in the same direction as the

kjiA 0.10.10.2

Find the component of the vector kjiA

0.1

0.1

0.2

in the direction of the vector jiB 0.40.3

By definition, the unit vector that is in the same direction as and:A

kji

kjiuA

41.041.082.00.10.10.2

0.10.10.2

222

BBuAA

ofdirection

4.05

4

5

30.10.10.2

ofdirection

jikjiAB

Conceptual problem;

A

Au

A

jiji

B

BuB

5

4

5

3

0.40.3

0.40.3

22


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POWER

In physics, the time rate at which a force does work is called

the Power P.

Or another definition, the Power Pis work performed per unit timeby force.

dtldvwheredtvFldFdW

dt

dWP

where

vFdt

dWP

The SI unit of power is one joule per second (J/s) and called a watt (w).

Power define by formula

The modern pulsing lasers can output more than 1.0 GW of power. A

typical large modern electric generation plant typically produces 1.0 GW

of electrical power. Does this mean the laser outputs a huge amount of

energy? No, the power of laser may only last for a short time interval.


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You lift a package vertically upward a distanceL in time t. You

then lift a second package that has twice the mass of the first

package vertically upward the same distance while providing the

same power as required for the first package. How much timedoes lifting the second package take?

L


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Express the power you exert in lifting the package one meter

in tseconds:

t

W

t

WP

1

1

11

Express the power you develop in lifting a package of twice the mass

one meter in tseconds:

2

1

2

2

2

2

t

W

t

WP

Because you exert the same power in lifting both packages:

2

11

2 t

W

t

W tt 2 2


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A object of mass m moves from rest at t = 0 under the influence

of a single constant force . Find the power delivered by

the force at any time t.

F

Express the rate at which this force does work

The velocity of the object, in terms of its acceleration

tav

:

Using Newtons 2nd law, substitute for a

tm

Ft

m

FFt

m

FFP

2

vFP

Substitute forvin equation (1) and simplify to obtain:

(1)

tm

Fv


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A 7.5-kg box is being lifted by a light rope that is threaded through a

single, light, frictionless pulley that is attached to the ceiling. (a) If

the box is being lifted at a constant speedof 2.0 m/s, what is the

power delivered by the person pulling on the rope? (b) If the box is

lifted, at constant acceleration, from rest on the floor to a height of

1.5 m above the floor in 0.42 s, what average power is delivered by

the person pulling on the rope?

(a) The power exerted by the person pulling on the rope is given by:cosTvvTP T vand are in the same direction, TvP

ymaFT gSecond Newtowns law ay= 0,because 0mgT

kWmgvP 15.0

(b) The average power exerted by the person pulling the rope is given by:

t

yF

t

WP

av 2

21

0 tatvy yy


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or because the box starts from rest, 221 tay

y

2

2

t

yay

2

2

t

ymmaF

y

t

yF

t

W

P

av

kWt

ym

t

yt

ym

P 46.0

2

2

3

22

av


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Let show that the power delivered by the net force acting on

a particle equals the rate at which the kinetic energy of the

particle is changing.

dtdKvFP

mv

dt

dv

dt

dmP

vadt

vd

vvdt

vd

vvdt

d

vdt

d

vamvFP

2)(

2

22

2

2

Rate at which kinetic energy change.


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CENTER OF MASS WORK

In Chapter 5 we learn that for system of objects net force define

Center Mass WorkTranslation Kinetic Energy Relation

transK Translation Kinetic Energy

transcmnetcm

transcmcmcmcmnet

cmi

netinet

KldFW

dt

dKM v

dt

dvaMvF

MaFF

2

1

2

2

presentation chapter6

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