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WORK AND ENERGY Scalars are back

W ORK AND E NERGY Scalars are back. R EVIEW Equations for Motion Along One Dimension

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WORK AND ENERGYScalars are back

REVIEW

Equations for Motion Along One Dimension

dt

dx

t

xv

t

xv

t

ave

0lim

dt

dv

t

va

t

va

t

ave

0lim

REVIEW

Motion Equations for Constant Acceleration

•1.

•2.

•3.

•4.

atvv 0

221

00 attvxx

20vv

vave

xavv 220

2

REVIEW

3 Laws of Motion If in Equilibrium

If not in equilibrium Change in Motion is Due to Force

Force causes a change in acceleration

0F

maF

SPRINGS AND OTHER PROBLEMS

Force exerted by a spring is dependent on amount of deformity of the spring

Amount of force applied changes continuously over time

What is the velocity of an object launched from the spring?

WORK

Work done on an object by all forces is equal to the change in kinetic energy of the object.

This definition is valid even if the force is not constant

WORK – CONSTANT FORCE

When a force, F, is doing work on an object, the object will move and be displaced.

The work done, by the force, F, is defined as

Where d is the objects displacement

FdW

WORK – CONSTANT FORCE

We are only interested in the component of the force that is parallel to the direction of motion dFW ||

WORK – CONSTANT FORCE

We are only interested in the component of the force that is parallel to the direction of motion

or

cosFdW

dFW

JOULE

Work done by 1N of force to move an object 1 meter in the same direction

cosFdW

JJouleW

mNW

mNW

11

1

11

JAMES PRESCOTT JOULE

December 24, 1818- October 11, 1889

The mechanical equivalent of heat 838 ft.lbf of work to

raise temperature of 1 lb of water by 1 degree farenheit

Led to the theory of conservation of energy

Helped Lord Kelvin develop the absolute scale of temperature

WORK – ZERO, NEGATIVE, POSITIVE

When defining work done, its always important to specify which force is acting on what object Work done by man Work done by

gravity Work done by

barbell

TOTAL WORK

Compute work done by forces individually

Then just add to get total work done on the object

Note: work is scalar

...21 WWWWtot

EXAMPLE Farmer hitches a tractor

with firewood and pulls it a distance 20m on level ground. Total weight of the sled and wood is 14700N and the tractor pulls with a constant force of 5000N at an angle 36.9o above the horizontal. There is a 3500N friction force opposing the motion. Find the work done by each of the forces and the total work done by all forces.

EXAMPLE

JW

W

dFW

JW

W

W

dFW

friction

friction

frictionfriction

tractor

tractor

tractor

Tractortractor

70000

)180cos()20)(3500(

cos

80000

79968

)9.36cos()20)(5000(

cos

WORK DONE BY NON-CONSTANT FORCE

Requires the use of integrals

ENERGY

Energy is a hard to define concept

Simplified definition The ability of a

physical system to do work on another physical system

Many types of energy- these are much easier to define

KINETIC ENERGY

Energy of motion When work is done

to an object the object moves

It also affects an objects speed W>0 – object speeds

up W<0 – object slows

down W=0 – no effect

KINETIC ENERGY

Newton’s 2nd LawFdW

madW

KINETIC ENERGY

20

2

20

2

20

2

20

2

20

2

20

2

2

1

2

1

2

2

2

2

2

mvmvW

vvmW

dd

vvmW

d

vva

advv

advv

madW

KINETIC ENERGY

Work done is the change in kinetic energy of an object

This is translational kinetic energy

20

2

2

1

2

1mvmvW

2

2

1mvK

WORK – ENERGY THEOREM

Assuming mass is constant

Unit of work is Joules Unit of energy is also

Joules

Note: Energy is also scalar

KW

mvW

2

2

1

EXAMPLE Farmer hitches a tractor

with firewood and pulls it a distance 20m on level ground. Total weight of the sled and wood is 14700N and the tractor pulls with a constant force of 5000N at an angle 36.9o above the horizontal. There is a 3500N friction force opposing the motion. Suppose it’s initial speed is 2.0 m/s, what is its final speed after travelling 20m.

EXAMPLE

sm

total

total

total

v

v

v

v

vvmW

KEW

JW

4

1633.4

2333.13

)2)(1500(2

110000

)(2

1

10000

22

22

20

2

EXAMPLE

A 15kg block is placed on a 40o incline and allowed to slide for 5m. What is it’s final speed? kg15

POTENTIAL ENERGY

Energy due to a body’s configuration or surroundings.

Many different types Springs Electrical Gravitational

GRAVITATIONAL POTENTIAL

An object held in the air has the “potential” to do work once released.

Assume object at some height

After travelling some distance y

00 v

gyv

ygv

advv

2

))((2

2

2

2

20

2

GRAVITATIONAL POTENTIAL

An object held in the air has the “potential” to do work once released.

KE after travelling some distance y

mgyK

mgymvK

gyv

22

1

2

1

2

2

2

GRAVITATIONAL POTENTIAL

An object held in the air has the “potential” to do work once released.

Amount of potential work

mgyKW

mgyK

K

00

GRAVITATIONAL POTENTIAL

An object held in the air has the “potential” to do work once released.

Note: choose your origin and be consistent

mgyUPEgrav

EXAMPLE- GIANCOLI 6-28

By how much does the gravitational potential energy of a 64-kg pole vaulter change if his center of mass rises 4.0m?

EXAMPLE- GIANCOLI 6-28

By how much does the gravitational potential energy of a 64-kg pole vaulter change if his center of mass rises 4.0m?

JU

JU

mgU

mgU

mgyU

2500

2500

)4)(8.9)(64()4(

0)0(0

WORK DONE EXAMPLE

What is work done to lift a block by 5 m?

If a 40o was used?

kg15

CONSERVATIVE AND NON-CONSERVATIVE FORCE

Conservative Force Work Done is

independent of the path taken Gravity Elastic Electric

You can “store” energy in these types of systems by doing work on the system

Non Conservative Force

Work done depends on the path taken Friction Air resistance Tension Push-Pull from a

person Cannot define

potential energy for these types of forces

CONSERVATION OF MECHANICAL ENERGY

If only gravity is acting on the object

Valid for all conservative forces

If only conservative forces are acting, the total mechanical energy of a system neither increase nor decrease in any process. It stays constant- it is conserved.00

00 )(

UKUK

UUKK

UK

UWW

KW

grav

CONSERVATION OF MECHANICAL ENERGY

If a non-conservative force is acting on the object

Most common non-conservative energy is friction

NC

NC

NC

grav

NCgrav

NCgrav

WUKUK

UKW

UWK

UW

KWWW

WWW

00

fdWNC

EXAMPLE – FROM OUR 2ND LECTURE

A motorcycle stuntman rides over a cliff. Just at the cliff edge his velocity is completely horizontal with magnitude 9.0 m/s. Find the motorcycles speed after 0.50s.

LIST THE GIVEN

Origin is cliff edge a=-g=-9.80m/s2

At time t=0s

At time t=0.50s

0v

v00 x

00 y

?d?v

smv 0.90

SPLIT INTO COMPONENTS

0v

v

yD

xD

DDD

y

x

yx

yx vvv

sm

xv 0.90

00 yv

CALCULATE COMPONENTS INDEPENDENTLY

0v

vxv

yv

sm

xx vv 0.90

sm

y

y

yy

v

gtv

gtvv

9.4

)5.0)(8.9(

0

CALCULATE VELOCITY

0v

vxv

yv

sm

xv 0.9s

myv 9.4

sm

sm

yx

xv

v

vvv

100.125.10

)9.4()0.9( 22

22

NOT NEEDED

29o below the horizontal

0v

v

xv

yv

sm

xv 0.9s

myv 9.4

smxv 100.1

544.09

9.4tan

x

y

v

v

2956.28

smxv 100.1

ALTERNATE SOLUTION

0v

vxv

yv

0

0.9

0

202

10

0

00

U

mvK

v

UKUK

sm

ALTERNATE SOLUTION

0v

vxv

yv

my

gty

gttvyy y

225.1

)5.0)(8.9( 2212

21

221

00

0

0.9

0

202

10

0

00

U

mvK

v

UKUK

sm

ALTERNATE SOLUTION

0v

vxv

yv

247.10

1.105

)225.1)(8.9(29

0

0

2

22

20

2

202

1221

00

0

202

10

00

v

v

v

gyvv

mgymvmv

UUKK

mgyU

U

mvK

UKUK

PROBLEM – YOUNG AND FREEDMAN 7.14

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swung so that it makes a maximum angle of 45o with the vertical. (a) What is the speed of the rock when it passes the vertical position? (b) What is the tension in the string when it makes an angle 45o with the vertical? (c) What is the tension in the string when it passes through the vertical?

PROBLEM – SERWAY 7.33

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100N parallel to the incline, which makes an angle of 20o with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system due to friction. (c) How much work is done by the 100N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5m?

OTHER TYPES OF POTENTIAL ENERGY

Elastic Potential For Ideal Springs If a spring is to be

stretched a certain distance x

Where k is the spring constant (the spring’s stiffness)

It’s me again

kxF

POTENTIAL ENERGY OF SPRINGS

Restoring Force

Hooke’s Law – valid for small x

kxFs

POTENTIAL ENERGY OF SPRINGS Work done ON the

spring (from equilibrium)

NO Force is not constant We can still use

average force Luckily F varies linearly

with x

))(( xkxW

FdW

kxF

POTENTIAL ENERGY OF SPRINGS Work done ON the spring

(from equilibrium)

Where U is the elastic potential

221

221

21

021

0

))((

)(

0)0(

kxU

kxW

xkxW

dFW

FFF

kxF

kF

ave

ave

CONSERVATION OF MECHANICAL ENERGY EXPANDED

Conservative

With Non conservative

springgravspringgrav UUKUUK 000

NCspringgravspringgrav WUUKUUK 000

YOUNG AND FREEDMAN 7.20

A 1.20kg piece of cheese was placed on a vertical spring of negligible mass and force constant k=1800 N/m that is compressed 15.0 cm. When the spring is released how high does the cheese rise from its original position?

POWER

Rate at which work is done

SI unit is called the Watt = 1J/s

Horsepower = 550ftlb/s = 746W

Time

WorkPAve

POWER

Rate at which work is done

Efficiency

Time

WorkPAve

aveAve Fvt

FdP

in

out

P

Pe

EXAMPLE GIANCOLI 6-58

How long will it take a 1750W motor to lift a 315 kg piano to a sixth story window 16.0m above?

EXAMPLE GIANCOLI 6-58

How long will it take a 1750W motor to lift a 315 kg piano to a sixth story window 16.0m above?

sP

Wt

JFdWTime

WorkPAve

2.28

4939216)8.9(315

1750

PROBLEM SERWAY 7.40

A 650 kg elevator starts from rest. It moves upward for 3s with constant acceleration until it reaches its cruising speed of 1.75m/s. (a) What is the average power of the elevator motor during this period? (b) How does this compare when the elevator moves at cruising speed?

YOUNG AND FREEDMAN – 7.42

A 2.00 kg block is pushed against a spring with negligible mass and force constant k= 400 N/m, compressing it 0.220m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0o. (a) what is the speed of the block on the horizontal surface after leaving the spring? (b) How far up the slope does the block travel before starting to slide back down?

GIANCOLI 6-56

A 280 g wood block is firmly attached to the end of a horizontal spring. The block can slide along the table with a coefficient of friction of 0.30. A force of 22 N compresses the string 18 cm. if the spring is released, how far from the equilibrium position will it stretch at its first maximum extension.

GROUP WORK

A 1500 kg rocket is to be launched with an initial upward speed of 50.0 m/s. In order to assist the engines, the engineers will start it from rest on a ramp that rises 53o above the horizontal. The engines provide a constant forward thrust of 2000N and the coefficient of kinetic friction with the ramp is 0.05. At what height should the rocket start?