Chapter 8B Chapter 8B -- Work and EnergyWork and EnergyA PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 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 this Objectives: After completing this module, you should be able to:module, you should be 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 and Describe the relationship between work and kinetic energy, and apply the kinetic energy, and apply the WORKWORK-- 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 conis anything that can be con-- vertedverted into work; i.e., anything that can into work; i.e., anything that can exert a force through a distanceexert a force through a distance.

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

Potential EnergyPotential Energy

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

A stretched bowA stretched bowA suspended weightA suspended weight

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

Gravitational Potential EnergyGravitational Potential Energy

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 Energy

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

A speeding car A speeding car or a space rocketor a space rocket

Examples of Kinetic EnergyExamples of Kinetic EnergyWhat is the kinetic energy of a 5What is the kinetic energy of a 5--g bullet g bullet

traveling at 200 m/s?traveling at 200 m/s?

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

5 g5 g

200 m/s200 m/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 of an A resultant force changes the velocity of an

object and does work on that object.object and does work on that object.

m

vo

m

vfxF F

( ) ;Work Fx ma x 2 2

0

2fv v

ax

2 21 102 2fWork mv mv

The WorkThe Work--Energy TheoremEnergy TheoremWork is equal Work is equal to the change to the change inin ½½mvmv22

If we define If we define kinetic energykinetic energy as as ½½mvmv22 then we then we can state a very important physical principle:can state a very important physical principle:

The WorkThe Work--Energy Theorem: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.

2 21 102 2fWork mv mv

Example 1:Example 1: A A 2020--gg projectile strikes a mud projectile strikes a mud bank, penetrating a distance of bank, penetrating a distance of 6 cm6 cm before before stopping. Find the stopping force stopping. Find the stopping force FF if the if the entrance velocity is entrance velocity is 80 m/s80 m/s..

x

F = ?F = ?

80 m/s80 m/s 6 cm6 cm

Work = Work = ½½ mvmv ff 2 2 -- ½½ mvmv oo 22

0

F x = F x = -- ½½ mvmv oo 22

FF (0.06 m) (0.06 m) coscos 18018000 = = -- ½½ (0.02 kg)(80 m/s)(0.02 kg)(80 m/s)22

FF (0.06 m)((0.06 m)(--1) = 1) = --64 J64 J F = 1067 NF = 1067 N

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

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

2525 mm fff = f = k.k. nn = = kk mgmgWork =Work = F(F(coscos ) x) x

Work = Work = -- kk mg xmg x0

K =K = ½½ mvmv ff 2 2 -- ½½ mvmv oo 22

--½½ mvmv oo 22 = = --kk mgmg xx vv oo = = 22kk gxgx

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

Work = K

K = Work

Example 3:Example 3: A A 44--kgkg block slides from rest at block slides from rest at top to bottom of the top to bottom of the 303000 inclined plane. Find inclined plane. Find velocity at bottom. velocity at bottom. ((hh = 20 m= 20 m and and kk = 0.2= 0.2))

hh303000

nnff

mgmg

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

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

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

h300

nf

mg

x

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

0

20 m 40 msin 30

x 0sin 30 hx

hx

303000

Example 3(Cont.):Example 3(Cont.): Next we find the resultant Next we find the resultant work on work on 44--kgkg block. (block. (x x = 40 m= 40 m and and kk = 0.2= 0.2))

WW yy = = (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 freeDraw free--body diagram to find the resultant force:body diagram to find the resultant force:

nnff

mgmg303000

x

y

mgmg coscos 303000 mgmg sin 30sin 3000

WW xx = = (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 force on Find the resultant force on 44--kgkg block. (block. (x x = 40 m= 40 m and and kk = 0.2= 0.2))

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 -- ff

Recall that Recall that ff kk = = kk nnFF yy = 0 or = 0 or nn = 33.9 N= 33.9 N

Resultant Force = 19.6 N – k n ; and k = 0.2

Resultant 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 work on The resultant work on 44--kgkg block. (block. (xx = 40 m= 40 m and and FFRR = 12.8 N= 12.8 N))

((Work)Work) RR = = FF RR xx

FF RR303000

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 44--kgkg block slides from block slides from rest at top to bottom of the rest at top to bottom of the 303000 plane. Find plane. Find velocity at bottom. velocity at bottom. ((hh = 20 m= 20 m and and kk = 0.2= 0.2))

hh303000

nnff

mgmg

xx Resultant Work = Resultant Work = 512 J512 J

Work done on block equals Work done on block equals the change in K. E. of block.the change in K. E. of block.

½½ mvmv ff 2 2 -- ½½ mvmv oo 22 = Work= Work

0½½ mvmv ff 22 = = 512 J512 J

½½(4 kg)(4 kg)vv ff 22 = 512 J= 512 J 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/sPower of 1 W is work done at rate of 1 J/s

Work F rPowertime t

2(10kg)(9.8m/s )(20m)4 s

mgrPt

490 J/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 WPower Consumed: P = 2220 W

What power is consumed in lifting a 70-kg robber 1.6 m in 0.50 s?

Fh mghPt t

2(70 kg)(9.8 m/s )(1.6 m)0.50 s

P

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

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

WorkPt

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 kWPower 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

P Fv

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

v = 4 m/s

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

P = F v = mg v

P = 35.3 kWP = 35.3 kW

Example 6:Example 6: What work is done by a What work is done by a 44--hphp mower in mower in one hourone hour? The conversion ? The conversion factor is needed: factor is needed: 1 hp = 550 ft lb/s1 hp = 550 ft lb/s..

Work = 132,000 ft lbWork = 132,000 ft lb

550ft lb/s4hp 2200ft lb/s1hp

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

; WorkP Work Ptt

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

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

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

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

212K mv

Work = ½ mvf 2 - ½ mvo

2Work = ½ mvf2 - ½ mvo

2

Summary (Cont.)Summary (Cont.)

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

WorkPt

Work F rPowertime t

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

P= F v

CONCLUSION: Chapter 8BCONCLUSION: Chapter 8B Work and EnergyWork and Energy