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ChapterChapter 8B8B -- WorkWork andand EnergyEnergy AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

© 2007 and a of 122 ft

, a roller coaster at Six Flags over

. The potential due to its

The Ninja Georgia, has a height of 52 mi/h height changes into of . Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to:

•• DefineDefine kinetickinetic energyenergy andand potentialpotential energyenergy,, alongalong withwith thethe appropriateappropriate unitsunits inin eacheach system.system. •• DescribeDescribe thethe relationshiprelationship betweenbetween workwork andand kinetickinetic energy,energy, andand applyapply thethe WORKWORK-- ENERGYENERGY THEOREM.THEOREM. •• DefineDefine andand applyapply thethe conceptconcept ofof POWERPOWER,, alongalong withwith thethe appropriateappropriate units.units. EnergyEnergy EnergyEnergy isis anythinganything thatthat cancan bebe concon-- vertedverted intointo ;work; i.e.,i.e., anythinganything thatthat cancan exertexert aa forceforce throughthrough aa .distance

EnergyEnergy isis thethe capabilitycapability forfor doingdoing work.work. PotentialPotential EnergyEnergy

PotentialPotential Energy:Energy: AbilityAbility toto dodo workwork byby virtuevirtue ofof positionposition oror condition.condition

AA suspendedsuspended weightweight AA stretchedstretched bowbow ExampleExample Problem:Problem: WhatWhat isis thethe potentialpotential energyenergy ofof aa 5050--kgkg personperson inin aa skyscraperskyscraper ifif hehe isis 480480 mm aboveabove thethe streetstreet below?below? GravitationalGravitational PotentialPotential 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)

UU == 235235 kJkJ KineticKinetic EnergyEnergy

KineticKinetic Energy:Energy: AbilityAbility toto dodo workwork byby virtuevirtue ofof motion.motion. ((Mass withwith )velocity)

AA speedingspeeding carcar oror aa spacespace rocketrocket ExamplesExamples ofof KineticKinetic EnergyEnergy WhatWhat isis thethe kinetickinetic energyenergy ofof aa 55--gg bulletbullet travelingtraveling atat 200200 m/s?m/s? 5 g 1122 Kmv22(0.005 kg)(200 m/s)

200 m/s KK == 100100 JJ WhatWhat isis thethe kinetickinetic energyenergy ofof aa 10001000--kgkg carcar travelingtraveling atat 14.114.1 m/s?m/s? 1122 Kmv22(1000 kg)(14.1 m/s)

KK == 99.499.4 JJ WorkWork andand KineticKinetic EnergyEnergy AA resultantresultant forceforce changeschanges thethe velocityvelocity ofof anan objectobject andand doesdoes workwork onon thatthat object.object.

x vf vo F m F m

vv22 Work Fx(); ma x a  f 0 2x

1122 Work 22mv f mv0 TheThe WorkWork--EnergyEnergy TheoremTheorem

WorkWork isis equalequal 1122 toto thethe changechange Work 22mv f mv0 inin ½½mvmv2 IfIf wewe definedefine kinetickinetic energyenergy asas ½½mvmv2 thenthen wewe cancan statestate aa veryvery importantimportant physicalphysical principle:principle:

TheThe WorkWork--EnergyEnergy Theorem:Theorem: TheThe workwork donedone byby aa resultantresultant forceforce isis equalequal toto thethe changechange inin kinetickinetic energyenergy thatthat itit produces.produces. ExampleExample 1:1: AA 2020--gg projectileprojectile strikesstrikes aa mudmud bank,bank, penetratingpenetrating aa distancedistance ofof 66 cmcm beforebefore stopping.stopping. FindFind thethe stoppingstopping forceforce FF ifif thethe entrance velocity is 80 m/s. entrance velocity is 80 m/s. 6 cm 0 80 m/s 2 2 x WorkWork == ½½ mvmvf -- ½½ mvmvo

2 F = ? FF xx == -- ½½ mvmvo FF (0.06(0.06 m)m) coscos 1801800 == -- ½½ (0.02(0.02 kg)(80kg)(80 m/s)m/s)2

FF (0.06(0.06 m)(m)(--1)1) == --6464 JJ FF == 10671067 NN

WorkWork toto stopstop bulletbullet == changechange inin K.E.K.E. forfor bulletbullet ExampleExample 2:2: AA busbus slamsslams onon brakesbrakes toto avoidavoid anan accident.accident. TheThe treadtread marksmarks ofof thethe tirestires areare 8080 mm long.long. IfIf k == 0.70.7,, whatwhat waswas thethe speedspeed beforebefore applyingapplying brakes?brakes? Work = K

WorkWork == F(F(coscos )) xx 2525 mm ff ff == k.n n == k mgmg 0 2 2 WorkWork == -- k mgmg xx KK == ½½ mvmvf -- ½½ mvmvo

2 --½½mvmvK =o Work== --k mgmg xx vv o == 22kgx gx

2 v = 59.9 ft/s vo = 2(0.7)(9.8 m/s )(25 m) voo = 59.9 ft/s ExampleExample 3:3: AA 44--kgkg blockblock slidesslides fromfrom restrest atat toptop toto bottombottom ofof thethe 30300 inclinedinclined plane.plane. FindFind velocityvelocity atat bottom.bottom. ((hh == 2020 mm andand k == 0.20.2)) Plan:Plan: WeWe mustmust calculatecalculate bothboth ff nn xx thethe resultantresultant workwork andand thethe hh netnet displacementdisplacement xx.. ThenThen thethe velocityvelocity cancan bebe foundfound fromfrom mgmg 0 30 thethe factfact thatthat WorkWork == K.K.

ResultantResultant workwork == (Resultant(Resultant forceforce downdown thethe plane)plane) xx (the(the displacementdisplacement downdown thethe plane)plane) ExampleExample 33 (Cont.):(Cont.): WeWe firstfirst findfind thethe netnet displacementdisplacement xx downdown thethe plane:plane:

f n x x h h 300 mg 300

FromFrom trig,trig, wewe knowknow thatthat thethe SinSin 30300 == h/xh/x and:and: h 20 m sin 300  x 40 m x sin 300 ExampleExample 3(Cont.):3(Cont.): NextNext wewe findfind thethe resultantresultant workwork onon 44--kgkg block.block. ((xx == 4040 mm andand k == 0.20.2)) DrawDraw freefree--bodybody diagramdiagram toto findfind thethe resultantresultant :force:

y x = 40 m ff ff nn nn mg sin 300 mg cos 300 mg sin 30 hh x mgmg 300 300 mgmg

2 0 WW x == (4(4 kg)(9.8kg)(9.8 m/sm/s )(sin)(sin 3030 ) = 19.6 N 2 0 WW y == (4(4 kg)(9.8kg)(9.8 m/sm/s )(cos)(cos 3030 ) = 33.9 N ExampleExample 3(Cont.):3(Cont.): FindFind thethe resultantresultant forceforce onon 44--kgkg block.block. ((xx == 4040 mm andand k == 0.20.2)) y Resultant force down ff Resultant force down nn plane:plane: 19.619.6 NN -- ff 19.6 N 33.9 N RecallRecall thatthat ff k == k nn x 300 mg 30 FF y == 00 oror nn == 33.933.9 NN

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 ExampleExample 33 (Cont.):(Cont.): TheThe resultantresultant workwork onon 44--kgkg block.block. ((xx == 4040 mm andand FFR == 12.812.8 NN))

((Work)Work) R == FF Rx x xx Net Work = (12.8 N)(40 m) FF R 300 Net Work = 512 J

Finally, we are able to apply the work-energy theorem to find the final velocity: 0

1122 Work 22mv f mv0 ExampleExample 33 (Cont.):(Cont.): AA 44--kgkg blockblock slidesslides fromfrom restrest atat toptop toto bottombottom ofof thethe 30300 plane.plane. FindFind velocityvelocity atat bottom.bottom. ((hh == 2020 mm andand k == 0.20.2))

ff nn xx ResultantResultant WorkWork == 512512 JJ hh WorkWork donedone onon blockblock equalsequals the change in K. E. of block. mgmg 300 the change in K. E. of block. 0 2 2 2 ½½ mvmvf -- ½½ mvmvo == WorkWork ½½ mvmvf == 512512 JJ

½½(4(4 kg)kg)vv 2 == 512512 JJ v = 16 m/s f vff = 16 m/s PowerPower is defined as the at which work is done: (P = dW/dt ) t F Work F r Power  m t 4 s 10 kg mgr (10kg)(9.8m/s2 )(20m) h mg P  20 m t 4 s P  490J/s or 490 (W)

PowerPower ofof 11 WW isis workwork donedone atat raterate ofof 1J/s1J/s UnitsUnits ofof PowerPower

One (W) is work done at the rate of one per . 1 W = 1 J/s and 1 kW = 1000 W

One ft lb/s is an older (USCS) unit of power.

One is work done at the rate of 550 ft lb/s. ( 1 hp = 550 ft lb/s ) ExampleExample ofof PowerPower

What power is consumed in lifting a 70-kg robber 1.6 m in 0.50 s? Fh mgh P  tt (70 kg)(9.8 m/s2 )(1.6 m) P  0.50 s

PowerPower Consumed:Consumed: PP== 22202220 WW ExampleExample 4:4: AA 100100--kgkg cheetahcheetah movesmoves fromfrom restrest toto 3030 m/sm/s inin 44 s.s. WhatWhat isis thethe power?power?

RecognizeRecognize thatthat workwork isis equalequal toto thethe changechange inin kinetickinetic energy:energy:

1122Work Work mvf mv0 P  22 t m = 100 kg 1 mv2 1 (100 kg)(30 m/s)2 P 2 f 2 t 4 s

PowerPower Consumed:Consumed: PP== 1.221.22 kWkW PowerPower andand VelocityVelocity

Recall that average or constant velocity is distance covered per unit of time v = x/t.

F x x P = = F P  Fv t t If power varies with time, then is needed to integrate over time. (Optional)

Since P = dW/dt: Work  P() t dt ExampleExample 5:5: WhatWhat powerpower is required to lift a 900-kg is required to lift a 900-kg v = 4 m/s elevatorelevator atat aa constantconstant speedspeed ofof 44 m/s?m/s? P = F v = mg v P = (900 kg)(9.8 m/s2)(4 m/s)

PP == 35.335.3 kWkW ExampleExample 6:6: WhatWhat workwork isis donedone byby aa 44--hphp mowermower inin oneone hourhour?? TheThe conversionconversion factorfactor isis needed:needed: 11 hphp == 550550 ftft lb/slb/s.. 550ft lb/s 4hp 2200ft lb/s 1hp Work P ; Work Pt t Work (2200ft lb/s)(60s)

WorkWork == 132,000132,000 ftft lblb 2 mv 2 1   K Umgh

2 2 o o v

The work done The work done

-½mv -½m 2 2 f f v

Ability to do work Ability to do work

Summary Ability to do work by Summary Ability to do work by

Work = ½ mv Energy Theorem: Energy Theorem: Work = ½ m - -

The Work-Energy Theorem: The work done by a resultant force is equal to the change in kinetic energy that it produces. : Potential Energy: by virtue of or condition. by virtue of position or condition The Work The Work 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. Kinetic Energy: Kinetic Energy: virtue of motion. (Mass with velocity) virtue of motion. (Mass with velocity) SummarySummary (Cont.)(Cont.)

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

Work F r Power  P= F v time t

PowerPower ofof 11 WWisis workwork donedone atat raterate ofof 11 J/sJ/s CONCLUSION:CONCLUSION: ChapterChapter 8B8B WorkWork andand EnergyEnergy