Chapter 8B - Work and Chapter 8B - Work and EnergyEnergy

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

The Ninja, a roller coaster at Six Flags over Georgia, has a height of 122 ft and a speed of 52 mi/h. The potential energy due to its height changes into kinetic energy of motion.

Objectives: After completing Objectives: After completing this module, you should be this module, you should be able to:able to:

• Define Define kinetic energykinetic energy and and potential energypotential energy, , along with the appropriate units in each along with the appropriate units in each system.system.

• Describe the relationship between work Describe the relationship between work and kinetic energy, and apply the and kinetic energy, and apply the WORK-WORK-ENERGY THEOREM.ENERGY THEOREM.

• Define and apply the concept of Define and apply the concept of POWERPOWER,, along with the appropriate units.along with the appropriate units.

EnergyEnergyEnergyEnergy is anything that can be con-is anything that can be con-verted into work; i.e., anything that verted into work; i.e., anything that can exert a force through a can exert a force through a distancedistance.

Energy is the capability for doing Energy is the capability for doing work.work.

Potential EnergyPotential Energy

Potential Energy:Potential Energy: Ability to do work Ability to do work by virtue of position or conditionby virtue of position or condition.

A stretched bowA stretched bowA suspended weightA suspended weight

Example Problem:Example Problem: What is the What is the potential energy of a 50-kg person potential energy of a 50-kg person in a skyscraper if he is 480 m above in a skyscraper if he is 480 m above the street below?the street below?

Gravitational Potential Gravitational Potential EnergyEnergy

What is the P.E. of a 50-kg person at a height of 480 m?

U = mgh = (50 kg)(9.8 m/s2)(480 m)

U = 235 kJU = 235 kJ

Kinetic EnergyKinetic EnergyKinetic Energy:Kinetic Energy: Ability to do work Ability to do work by virtue of motion. (Mass with by virtue of motion. (Mass with velocity)velocity)A speeding car A speeding car

or a space or a space rocketrocket

Examples of Kinetic Examples of Kinetic EnergyEnergy

What is the kinetic energy of a 5-g What is the kinetic energy of a 5-g bullet traveling at 200 m/s?bullet traveling at 200 m/s?

What is the kinetic energy of a What is the kinetic energy of a 1000-kg car traveling at 14.1 m/s? 1000-kg car traveling at 14.1 m/s?

5 g5 g

200 200 m/sm/s

K = 100 JK = 100 J

K = 99.4 JK = 99.4 J

2 21 12 2 (0.005 kg)(200 m/s)K mv

2 21 12 2 (1000 kg)(14.1 m/s)K mv

Work and Kinetic EnergyWork and Kinetic EnergyA resultant force changes the velocity A resultant force changes the velocity

of an object and does work on that of an object and does work on that object.object.

m

vo

m

vfx

F F

( ) ;Work Fx ma x 2 2

0

2fv v

ax

2 21 102 2fWork mv mv

The Work-Energy The Work-Energy TheoremTheorem

Work is Work is equal to the equal to the change inchange in ½mv½mv22

If we define If we define kinetic energykinetic energy as as ½mv½mv22 then we can state a very important then we can state a very important

physical principle:physical principle:

The Work-Energy Theorem:The Work-Energy Theorem: The work The work done by a resultant force is equal to done by a resultant force is equal to the change in kinetic energy that it the change in kinetic energy that it produces.produces.

2 21 102 2fWork mv mv

Example 1:Example 1: A A 20-g20-g projectile strikes a projectile strikes a mud bank, penetrating a distance of mud bank, penetrating a distance of 6 6 cmcm before stopping. Find the stopping before stopping. Find the stopping force force FF if the entrance velocity is if the entrance velocity is 80 80 m/sm/s..

x

F = ?F = ?

80 80 m/sm/s

6 cm6 cmWork = ½Work = ½ mvmvff

2 2 - ½- ½ mvmvoo

22

0

F x = - F x = - ½½ mvmvoo22

FF (0.06 m) cos 180 (0.06 m) cos 18000 = - = - ½½ (0.02 kg)(80 (0.02 kg)(80 m/s)m/s)22

FF (0.06 m)(-1) = -64 J (0.06 m)(-1) = -64 J F = 1067 NF = 1067 N

Work to stop bullet = change in K.E. for Work to stop bullet = change in K.E. for bulletbullet

Example 2:Example 2: A bus slams on brakes to A bus slams on brakes to avoid an accident. The tread marks of avoid an accident. The tread marks of the tires are the tires are 80 m80 m long. If long. If kk = 0.7 = 0.7, , what was the speed before applying what was the speed before applying brakes?brakes?

2525 mm fff = f = k.k.n = n = k k mgmg

Work =Work = F(F(cos cos ) ) xx

Work = - Work = - k k mg mg xx

0K =K = ½½ mvmvff

2 2 - ½- ½ mvmvoo22

-½-½ mvmvoo22 = - = -k k mgmg xx vvoo = = 22kkgxgx

vo = 2(0.7)(9.8 m/s2)(25 m) vo = 59.9 ft/s

vo = 59.9 ft/s

Work = K

K = Work

Example 3:Example 3: A A 4-kg4-kg block slides from block slides from rest at top to bottom of the rest at top to bottom of the 303000 inclined inclined plane. Find velocity at bottom. plane. Find velocity at bottom. ((hh = 20 = 20 mm and and kk = 0.2 = 0.2))

hh3030

00

nnff

mgmg

xx

Plan: Plan: We must calculate We must calculate both the resultant work both the resultant work and the net displacement and the net displacement xx. Then the velocity can . Then the velocity can be found from the fact be found from the fact that that Work = Work = K.K.

Resultant work = (Resultant force down Resultant work = (Resultant force down the plane) the plane) xx (the (the displacement down the displacement down the plane)plane)

Example 3 (Cont.):Example 3 (Cont.): We first find the We first find the net displacement net displacement xx down the plane: down the plane:

h

300

nf

mg

x

From trig, we know that the Sin From trig, we know that the Sin 303000 = = h/x h/x and:and:

0

20 m40 m

sin 30x 0sin 30

h

x

hx

303000

Example 3(Cont.):Example 3(Cont.): Next we find the Next we find the resultant work on resultant work on 4-kg4-kg block. ( block. (x x = 40 m= 40 m and and kk = 0.2 = 0.2))

WWyy = = (4 kg)(9.8 m/s(4 kg)(9.8 m/s22)(cos 30)(cos 3000) = 33.9 N

hh

303000

nnff

mgmg

x = x = 4040

mm

Draw free-body diagram to find the resultant Draw free-body diagram to find the resultant force:force:

nnff

mgmg303000

x

y

mgmg cos 30 cos 3000 mgmg sin 30 sin 3000

WWxx = = (4 kg)(9.8 m/s(4 kg)(9.8 m/s22)(sin 30)(sin 3000) = 19.6 N

Example 3(Cont.):Example 3(Cont.): Find the resultant Find the resultant force on force on 4-kg4-kg block. ( block. (x x = 40 m= 40 m and and kk = = 0.20.2))

nnff

mgmg303000

x

y

33.9 N33.9 N19.6 N19.6 N

Resultant force down Resultant force down plane: plane: 19.6 N - 19.6 N - ffRecall that Recall that ffkk = = k k

nnFFyy = 0 or = 0 or nn = 33.9 = 33.9 NN

Resultant Force = 19.6 N – kn ; and k = 0.2Resultant Force = 19.6 N – (0.2)(33.9 N) = 12.8 N

Resultant Force Down Plane = 12.8 N

Example 3 (Cont.):Example 3 (Cont.): The resultant The resultant work on work on 4-kg4-kg block. ( block. (xx = 40 m = 40 m and and FFRR = 12.8 N = 12.8 N))

(Work)(Work)RR = = FFRRxx

FFRR

303000

xxNet Work = (12.8 N)(40 m)Net Work = 512 J

Finally, we are able to apply the work-energy theorem to find the final velocity:

2 21 102 2fWork mv mv

0

Example 3 (Cont.):Example 3 (Cont.): A A 4-kg4-kg block slides block slides from rest at top to bottom of the from rest at top to bottom of the 303000 plane. Find velocity at bottom. plane. Find velocity at bottom. ((hh = 20 = 20 mm and and kk = 0.2 = 0.2))

hh3030

00

nnff

mgmg

xx Resultant Work = Resultant Work = 512 J512 JWork done on block Work done on block

equals the change in K. equals the change in K. E. of block.E. of block.

½½ mvmvff2 2 - ½- ½ mvmvoo

22 = = WorkWork

0 ½½ mvmvff

22 = = 512 J512 J

½½(4 kg)(4 kg)vvff22 = 512 = 512

JJ vf = 16 m/svf = 16 m/s

PowerPowerPower is defined as the rate at

which work is done: (P = dW/dt )

10 kg

20 mh

m

mg

t

4 s

F

Power of 1 W is work done at rate of 1 J/s

Power of 1 W is work done at rate of 1 J/s

Work F rPower

time t

Work F rPower

time t

2(10kg)(9.8m/s )(20m)

4 s

mgrP

t

490J/s or 490 watts (W)P 490J/s or 490 watts (W)P

Units of PowerUnits of Power

1 W = 1 J/s and 1 kW = 1000 W

One watt (W) is work done at the rate of one joule per second.

One ft lb/s is an older (USCS) unit of power.One horsepower is work done at the rate of 550 ft lb/s. ( 1 hp = 550 ft lb/s )

Example of PowerExample of Power

Power Consumed: P = 2220 W

Power Consumed: P = 2220 W

What power is consumed in lifting a 70-kg robber 1.6 m in

0.50 s?Fh mghP

t t

2(70 kg)(9.8 m/s )(1.6 m)

0.50 sP

Example 4:Example 4: A 100-kg cheetah moves A 100-kg cheetah moves from rest to 30 m/s in 4 s. What is the from rest to 30 m/s in 4 s. What is the power?power?

Recognize that work is equal Recognize that work is equal to the change in kinetic to the change in kinetic energy:energy: Work

Pt

2 21 102 2fWork mv mv

2 21 12 2 (100 kg)(30 m/s)

4 sfmv

Pt

m = 100 kg

Power Consumed: P = 1.22 kW

Power Consumed: P = 1.22 kW

Power and VelocityPower and VelocityRecall that average or constant

velocity is distance covered per unit of time v = x/t.

P = = Fx

t

F x

tIf power varies with time, then calculus is needed to integrate over time. (Optional)

Since P = dW/dt: ( )Work P t dt ( )Work P t dt

P FvP Fv

Example 5:Example 5: What What power is required to lift power is required to lift a 900-kg elevator at a a 900-kg elevator at a constant speed of 4 constant speed of 4 m/s?m/s?

v = 4 m/s

P = (900 kg)(9.8 m/s2)(4 m/s)

P = F v = mg v

P = 35.3 kW

P = 35.3 kW

Example 6:Example 6: What work is done by a What work is done by a 4-hp4-hp mower in mower in one hourone hour? The ? The conversion factor is needed: conversion factor is needed: 1 hp 1 hp = 550 ft lb/s= 550 ft lb/s..

Work = 132,000 ft lb

Work = 132,000 ft lb

550ft lb/s4hp 2200ft lb/s

1hp

(2200ft lb/s)(60s)Work

; Work

P Work Ptt

The Work-Energy Theorem:The Work-Energy Theorem: The work The work done by a resultant force is equal to the done by a resultant force is equal to the change in kinetic energy that it produces.change in kinetic energy that it produces.

The Work-Energy Theorem:The Work-Energy Theorem: The work The work done by a resultant force is equal to the done by a resultant force is equal to the change in kinetic energy that it produces.change in kinetic energy that it produces.

SummarySummaryPotential Energy:Potential Energy: Ability to do Ability to do work by virtue of position or work by virtue of position or conditioncondition.

U mgh

Kinetic Energy:Kinetic Energy: Ability to do work Ability to do work by virtue of motion. (Mass with by virtue of motion. (Mass with velocity)velocity)

212K mv

Work = ½ mvf2 - ½

mvo2

Work = ½ mvf2 - ½

mvo2

Summary (Cont.)Summary (Cont.)

Power is defined as the rate at which work is done: (P = dW/dt )

WorkP

t

Work F rPower

time t

Work F rPower

time t

Power of 1 W is work done at rate of 1 J/s

Power of 1 W is work done at rate of 1 J/s

P= F v

CONCLUSION: Chapter 8BCONCLUSION: Chapter 8BWork and EnergyWork and Energy