Chapter 6: Conservation of Energy

• CQ: 1, 3. Problems: 1, 3, 5, 11, 17, 25, 29, 32, 49, 51, 53, 69, 71.

• Law of Conservation of Energy

• Work done by a constant force

• Kinetic energy

• Gravitational potential energy

• Work by variable force, Hooke’s law

• Elastic potential energy1

2

Energy & Work

• Energy is the capacity to do work, • Energy is position & speed dependent• Unit: joule = newton·meter (J = N·m)• Work = force x distance (Fd) when force is

in direction of motion (or opposite to motion)

• Ex. 50N pushes distance of 4 meters.• W = (50N)(4m) = 200 J• /

2

Work & Force• Work is energy transferred by part of force

in line of motion

• Ex. Force 60° above path of motion

dFW )cos(

FddFW 21)60cos(

dFW //3

4

Machines

• change an applied force by increasing it, decreasing it, or changing its direction.

• Types:

• inclined plane, screw, wedge

• pulley, wheel

• lever

4

levers• Work input Fd = Work output Fd

• Ex. Your hand moves 100m, causes car to rise 0.10m. The force amplification factor is,

FF

dd

=__ __

10001.0

100 mm

5

6

inclined plane

• Weight x height change

• = Force x distance along plane

• Force along ramp less than Weight

• Ramp distance greater than height change

• ADA Standards: Ramp must be at least 12x longer than vertical rise

• Ex. A 1ft vertical rise requires 12ft of ramp.

6

ADA Ramp

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Energy of Motion

• Called Kinetic Energy (KE)

• KE = ½(mass)(velocity)2 = ½mv2.

• Ex. 2000kg car moving at 2m/s.

• KE = ½ (2000)(2)2 = 4000J.

• Position Dependent Energies are called Potential Energies “PE” or “U”

• /88

Gravitational Potential Energy

• U = weight x height (mgh)

• 1kg at 1m height:

• U = (1kg)(9.8N/kg)(1m) = 9.8J

• Energy released in falling

• /

9

Hooke’s Law

• The restoring force an object exerts is proportional to the amount it has been deformed (F = -kx)

10

Elastic Potential Energy

• PE-elastic = average force x distance

• 10N compresses a spring 1m. U = (avg. force, 5 N)(1m) = 5 J

• k = spring constant = force/distance

• U = (½kx)x = ½kx2

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Work Energy Theorem

xFW xnetnet

Let direction of motion be +x

xmax2

22ovv

m

KWnet 12

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Power• Power is the rate work is performed• Power = work/time = Force x velocity• Unit: watt = joule/second = J/s

• Other Unit: horsepower• 1 horsepower = 746 watts• /

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Energy & Power

• Energy = power x time

• Ex. A toy car has 1000 J of energy at full charge.

• How long can it run at 100 watts? At 10 watts?

• Time = Energy/power

• = 1000J/100watts = 10 seconds

• = 1000J/10watts = 100 seconds/1414

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Conservation of Energy

• E = K + U = constant• Ex. Falling: Kinetic ↑ as Potential ↓

• /

UW UK

0 UK

0 UK

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Energy Summary

• PE-gravitational = mgy

• PE-elastic = ½kx2

• KE = ½mv2

• Mechanical Energy ME = KEs + PEs

• E = ME + Thermal Energy

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Energy E1 E2 E3

Kinetic 0 ½mv22 0

U-grav 0 0 mgh

U-elastic ½kx2 0 0

U-therm 0 0 0

Totals ½kx2 ½mv22 mgh

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Energy h y

Kinetic 0 ½mv2

U-grav mgh mgy

U-elast 0 0

U-therm 0 0

Totals mgh ½mv2 + mgy

Energy and speed are same at height y

Accelerations are not same

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Energy Ei Ef

Kinetic ½mvi2 0

U-grav 0 0

U-elastic 0 0

U-therm 0 fkd

Totals ½mvi2 fks

Ex. Sled slides to a stop

d

dfmv ki 221

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A 2.00kg ball is dropped from rest from a height of 1.0m above the floor. The ball rebounds to a height of 0.500m. A movie-frame type diagram of the motion is shown below.

Type E1 E2 E3 E4 E5

gravita-tional

mg(1) 0 0 0 mg(1/2)

kinetic 0 ½ m(v2)2 0 ½ m(v4)2 0

elastic 0 0 U-el 0 0

thermal 0 0 U-therm U-therm U-therm

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Terminology

• E: total energy of a system

• E-mech = total energy minus the thermal energy

• E-mech = E – Utherm.

• Mech. Energy conserved in a frictionless system

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Power: The time rate of doing work.

SI Unit: watt, W = J/stime

workPavg

Example: How much average power is needed to accelerate a 2000kg car from rest to 20m/s in 5.0s?

work = KE 2212

21

if mvmv 2

212

21 )/0)(2000()/20)(2000( smkgsmkg

J000,400

s

J

t

workPavg 0.5

000,400 watts000,80

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avgavg vFt

sF

t

sF

t

WP )(cos)(cos

)(cos

Another equation for Power:

Ex: A car drives at 20m/s and experiences air-drag of 400N. The engine must use (400N)(20m/s) = 8,000 watts of engine power to overcome this force. 8,000 watts = 10.7 hp.

Work Example

mg

mgsin

dmgh

(mgsin)d = mg(dsin) = mgh

Moving down an inclined plane

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Summary

• Energy: Kinetic + Potential + Thermal, is conserved.

• Mech. Energy: Kinetic + Potential, conserved in frictionless systems

• Work is energy transfer (+ or -)

• Power is rate of energy transfer

Vehicle Efficiency

• 1 gallon gasoline has 138,000,000 J

• Engines only get a fraction of this:

• Ex. A 25% efficient car gets (0.25)(138,000,000 J) = 34,500,000J out of 1 gallon.

• A 20% efficient car gets 27,600,000J.

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Vehicle Frictional Work

• = Total Frictional Force x distance

• Ex. 400N friction for 1600 meters (1 mile)

• Work = (400N)(1600m) = 640,000J for one mile traveled

• /

28

Mpg

)( Mileper Work Frictional

)( Efficiency

J/mile

J/gal

gal

mile

J/mile

J/gal

)(

)(

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Ex. Mpg 20% Efficiency, f = 400N

• Engine gets 27,600,000 J/gal

• Frictional Work/Mile = 640,000J/mile

• = 43 mpg (at constant speed)

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)( 640,000

)( 27,600,000

J/mile

J/gal

30

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hpwatt

hpwatts107

746

1000,80

Horsepower: 1 hp = 746 watts

For the previous example:

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What size electric motor is needed to raise 2000lbs = 9000N of bricks at 10cm/s?

Minimum Power:

Pavg = Fvavg = (9000N)(0.1m/s)

P = 900 W = 1.2 hp

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35

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Similar to 5-45.

• 3gram bullet moves at 200m/s and goes 10cm into a tree. What is the average force on the bullet? Tree?

• Wnet on bullet = -Fd = change in K

• Change in K = 0 – ½ (0.003kg)(200)(200)

• -F(0.1m) = - 60Nm

• F = 600N

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By energy conservation, the sum of all energies in each column is the same, = E1 = mg(1) = 19.6J

Calculate v2: (use 1st and 2nd columns)mg(1) = ½ m(v2)2.

g = ½ (v2)2.v2 = 4.43m/s

Calculate PE-thermal: (use 1st and 5th columns)mg(1) = mg(1/2) + PE-thermal

mg(1/2) = PE-thermalPE-thermal = 9.8J

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Calculate PE-elastic: (use 1st and 3rd columns)PE-elastic + PE-thermal = mg(1)

PE-elastic + 9.8 = 19.6PE-elastic = 9.8J

Calculate v4: (use 1st and 4th columns)½ m(v4)2 + PE-thermal = mg(1)

½ m(v4)2 + 9.8 = 19.6½ m(v4)2 = 9.8 (v4)2 = 2(9.8)/2

v4 = 3.13m/s

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Energy A B C

Kinetic ½m22 ½mvB2 ½mvC

2

U-grav mg7 mg3 mg0

U-elastic 0 0 0

U-therm 0 1000 1200

Totals39