Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

Physics1120:Work&EnergySolutions

Energy

1. Inthediagrambelow,thespringhasaforceconstantof5000N/m,theblockhasamassof6.20kg,andtheheighthofthehillis5.25m.Determinethecompressionofthespringsuchthattheblockjustmakesittothetopofthehill.Assumethattherearenononconservativeforcesinvolved.

Sincetheprobleminvolvesachangeisheightandhasaspring,wemakeuseoftheGeneralizedWorkEnergyTheorem.Sincetheinitialandfinalspeedsarezero,

WNC=E=UgravfUspringi=mghKx2.

TherearenononconservativeforcessoWNC=0.Gettingxbyitselfyields

x=[2mgh/K]=0.357m.

2. Asolidball,acylinder,andahollowballallhavethesamemassmandradiusR.TheyareallowedtorolldownahillofheightHwithoutslipping.Howfastwilleachbemovingonthelevelground?

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Theonlyexternalforceontheobject,excludinggravitywhichistakenintoaccountthroughgravitationalpotentialenergy,isstaticfriction.Forobjectswhichrollacrossastationarysurface,staticfrictiondoesnowork,sohere0=E,whereEistheobservableormechanicalenergy.

Astheobjectdropsdownthehill,itlosesgravitationalpotentialenergywhileitgainsbothlinearandrotationalkineticenergy.Thuswehave

0=mgH+mv2+I2

Nowsincetheobjectrollswithoutslipping,theangularvelocityisrelatedtothelinearvelocityvby=v/R.Soourequationbecomes

0=mgH+mv2+I(v/R)2

Isolatingv2terms,wefindv2[m+I/R2]=2mgH.Solvingforvweget

Toproceedfurther,weneedtoknowthemomentofinertiaofeachobjectaboutitsCM.Weconsultatableofmomentsofinertiaandfind

Shape ICM v

solidsphere 2/5MR2

solidcylinder MR2

hollowsphere 2/3MR2

3. AcylinderofmassMandradiusR,onaninclineofangle,isattachedtoaspringofconstantK.Thespringisnotstretched.FindthespeedofthecylinderwhenithasrolledadistanceLdowntheincline.

Theonlyexternalforceonthesystem,excludinggravitywhichistakenintoaccountthroughgravitationalpotentialenergy,isstaticfriction.Forobjectswhichrollacrossastationarysurface,staticfrictiondoesnowork,sohere0=E,whereEistheobservableormechanicalenergy.

AstheobjectrollsdistanceLdowntheincline,itlosesgravitationalpotentialenergywhileitgainsbothlinearandrotationalkineticenergy.Moreover,thespringgainsSpringPotentialEnergy.Thuswehave

0=mgH+mv2+I2+KL2

TherelationshipbetweenHandtheLisH=Lsin.Sincetheobjectrollswithoutslipping,theangularvelocityisrelatedtothelinearvelocityvby=v/R.Furthermore,consultingatableofvalues,themomentofinertiaofacylinderabouttheperpendicularaxisthroughtheCM,isMR2.Ourequationbecomes

0=mgLsin+mv2+(MR2)(v/R)2+KL2

Isolatingv2terms,wefindv2[m+m]=mgLsinKL2.Solvingforvweget

4. AblockofmassmisconnectedbyastringofnegligiblemasstoaspringwithspringconstantKwhichisinturnfixedtoawall.Thespringishorizontalandthestringishungoverapulleysuchthatthemasshangsvertically.ThepulleyisasoliddiskofmassMandradiusR.Asshowninthediagrambelow,thespringisinitiallyinitsequilibriumpositionandthesystemisnotmoving.(a)Useenergymethods,todeterminethespeedvoftheblockafterithasfallenadistanceh.Expressyouranswerintermsofg,m,K,M,andh.(b)Theblockwilloscillatebetweenitsinitialheightanditslowestpoint.Atitslowestpoint,itturnsaround.Useyouranswertopart(a)tofindwhereitturnsaround.(c)Useyouranswertopart(a)andcalculustofindtheheightatwhichthespeedisamaximum.

(a)Theprobleminvolvesachangeinheightandspeedandhasaspring,sowewouldapplythegeneralizedWorkEnergyTheoremevenifnotdirectedtodo,

WNC=E=(KfKi)+(UfUi),(1)

whereKisthesumofallthelinearandrotationalkineticenergiesofeachobject,andUisthesumofthespringandgravitationalpotentialenergies.Sincethereisnokineticfrictionactingonthesystem,WNC=0.

Examiningtheproblemobjectbyobjectweseethatthespringstretches,sothereisanincreaseinspringpotentialenergy.Thepulleystartsrotating,sothereisanincreaseinitsrotationalkineticenergy.Theblockdrops,sothereisadecreaseinitsgravitationalpotentialenergy.Aswell,astheblockdrop,itincreasesitskineticenergy.Equation(1)forthisproblemisthus

0=Kx2+I2mgh+mv2.

Sincethespringisconnectedtotheblock,thespringstretchesasmuchastheblockdrops,sox=h.Wearetoldthatthepulleyisasoliddisk,soI=MR2.Sincetheropedoesnotslipthetangentialspeedofthepulleyisthesameastheropeandthus=v/R.Substitutingtheserelationsbackintoourequationyields,

0=Kh2+Mv2mgh+mv2.

Collectingthetermswithv,andsolvingforvyields

v=[(2mghkh2)/(m+M/2)].(2)

(b)Recallfromourdiscussionsonkinematicsthatanobjectturnsaroundwhenitsvelocityiszero.Settingequation(2)tozero

[(2mghkh2)/(m+M/2)]=0,

weseethatthenumeratoriszerowhen

2mghkh2=0.

Solvingthisforhrevealsthattheobjectturnsaroundwhenh=2mg/k.

(c)Tofindthemaximumvelocity,weneedtofinddv/dh=0.Takingthederivativeofequation(2)yields

dv/dh={1/[(2mghkh2)/(m+M/2)]}[2mg2kh]/[m+M/2]=0.

Weseethatthenumeratortozerowhen

2mg2kh=0.

Solvingthisforhrevealsthattheobjectturnsaroundwhenh=mg/k,halfwaybetweenthestartingpositionandtheturnaroundpoint.

5. Asystemofweightsandpulleysisassembledasshownbelow.Thepulleysareallfixed.ThepulleysonthesidesaredisksofmassmdandradiusRdthecentralpulleyisahoopofmassmhandradiusRh.Amasslessidealropepassesaroundthepulleysandjoinstwoweights.Theropedoesnotslip.Thetwoweightshavemassesm1andm2respectively.Useenergymethodstofindthevelocityoftheweightsasafunctionofdisplacement,x.ThemomentofinertiaofadiskisIdisk=MR2andofahoopisIhoop=MR2.

Theprobleminvolveschangesinheight,speed,androtation,sowewouldapplythegeneralizedWorkEnergyTheoremevenifnotdirectedtodo,

WNC=E=(KfKi)+(UfUi),(1)

whereKisthesumofallthelinearandrotationalkineticenergiesofeachobject,andUisthesumofthegravitationalpotentialenergies.Sincethereisnokineticfrictionactingonthesystem,WNC=0.

Let'sassumeblock2movesdown.Thusblock1movesupanidenticalamounth.Sinceblock2drops,thereisadecreaseinitsgravitationalpotentialenergy.Block1willincreaseitspotentialenergy.Aswell,asbothblocksmove,theyincreasesitskineticenergy.Thepulleysstartrotating,sothereisanincreaseintherotationalkineticenergyofeach.Equation(1)forthisproblemisthus

0=m2gh+m1gh+m2v2+m1v2+2ID(D)2+IH(H)2.

WearetoldthatIdisk=MD(RD)2andIhoop=MH(RH)2.Sincetheropedoesnotslipthetangentialspeedofthepulleysisthesameastheropeandthus=v/R.Substitutingthisrelationsbackintoourequationyields,

0=(m2m1)gh+m2v2+m1v2+MD(RD)2(v/RD)2+MH(RH)2(v/RH)2.

Collectingthetermswithv,andsolvingforvyields

v=[2(m2m1)gh/(m1+m2+MD+MH)].

6. Inthediagrambelow,amovingskierontopofacircularhillofradiusRh=62.0mfeelsthathis"weight"isonly(3/8)mg.Whatwouldbetheskier'sapparentweight(inmultiplesofmg)atthebottomofthecircularvalleywhichhasaradiusRv=43.0m?Neglectfrictionandairresistance.

Theprobleminvolvesachangeinheightandspeed,soweapplythegeneralizedWorkEnergyTheorem.

WNC=E=(KfKi)+(UfUi)=[m(vv)22mgRv][m(vh)2+2mgRh].

Thereisnofrictionorairresistance,WNC=0.Ourequationisthus

0=[m(vv)22mgRv][m(vh)2+2mgRh].(1)

Wetoldwhattheskierfeelshisweighttobe.ThesensationofweightisN,thenormalactingontheperson.TofindaforceweneedtodrawaFBDandapplyNewton'sSecondLaw.Wemustdothisbothatthetopofthehillandatthebottomofthevalley.Sincebotharecircularpaths,wearedealingwithcentripetalacceleration.

hill valley

Fy=may Fy=may

Nhmg=m(vh)2/Rh Nvmg=m(vv)2/Rv

WearetoldthatNh=(3/8)mg,sothefirstforceequationbecomes(5/8)mg=m(vh)2/Rh.Solvingforvh,wefindvh=[(5/8)gRh].Thiscanbesubstitutedintoequation(1)

0=[(vv)22gRv][{(5/8)gRh}+2gRh].

Rearrangingtoisolatevv,wefind

vv=[g{(37/8)Rh+Rv}]=67.08m/s.

Thisresult,alongwiththesecondforceequationlet'sfindtheapparentweightinthevalley,

Nv=mg+m(vv)2/Rv=mg[1+{(37/8)Rh+Rv}/Rv]=mg[2+(37/8)(Rh/Rv)]=(11.7)mg.

Theskierfeels11.7timesheavieratbottomofvalleywhichisnotverylikely!

7. AtpointAinthefigureshownbelow,aspring(springconstantk=1000N/m)iscompressed50.0cmbya2.00kgblock.WhenreleasedtheblocktravelsoverthefrictionlesstrackuntilitislaunchedintotheairatpointB.ItlandsatpointC.Theinclinedpartofthetrackmakesanangleof=55.0withthehorizontalandpointBisaheighth=4.50mabovetheground.HowfarhorizontallyispointCfrompointB?

Theprobleminvolvesachangeinheightandspeedandhasaspring,soweapplythegeneralizedWorkEnergyTheorem.,WNC=E.

Thereisnofrictionorairresistance,soWNC=0.Thespringiscompressedinitially,soitlosesspringpotentialenergy.Theblockincreaseskineticenergyandgainsgravitationalpotentialenergy.Ourequationisthus

0=kx2+mv2+mgh.

Wecanusethistofindthespeedoftheblockatlaunch

v=[kx2/m2gh]=[(1000)(0.5)2/22(9.81)(4.5)]=6.0589m/s.

Nowtheblockisaprojectile.Tosolveaprojectileproblemwebreakthemotionintoitsxandycomponentsandapplyourkinematicsequations.

i jv0x=vcos(55)=3.47523m/s v0y=vsin(55)=4.96314m/s

x=? y=4.50max=0 ay=9.81m/s2

t=? t=?

Wehaveenoughinformationintheycolumntofindtusingy=v0yt+at2,

4.50=4.96314t4.905t2.

Usingthequadraticequation,thesolutionsaret=0.5773sandt=1.5892s.Wewantthepositive,or

forwardintime,solution.Hencethehorizontaldistancetraveledbytheblockis

x=v0xt=3.475231.5892=5.52m.

PointCistherefore5.52mfromB.

8. Asmallsolidsphereofradiusr=1.00cmandmassm=0.100kgatpointAispressedagainstaspringandisreleasedfromrestwiththespringcompressed20.0cmfromitsnaturallength.Thespringhasaforceconstantk=20.0N/m.ThesphererollswithoutslippingalongahorizontalsurfacetopointBwhereitsmoothlycontinuesontoacirculartrackofradiusR=2.00m.TheballfinallyleavesthesurfaceofthetrackatpointC.Findtheanglewheretheballleavesthetrack.Assumethatfrictiondoesnowork.HintfindanexpressionforthespeedofthesphereatpointC.ThemomentofinertiaofasphereisI=2/5mr2.

Theprobleminvolvesachangeinheightandspeedandhasaspring,soweapplythegeneralizedWorkEnergyTheorem.,WNC=E.

WearetoldtoassumeWNC=0.Thespringiscompressedinitially,soitlosesspringpotentialenergy.Theballincreasesbothlinearandrotationalkineticenergy.Theballlosesgravitationalpotentialenergy.Ourequationisthus

0=kx2+mv2+I2mgh.(1)

Wearetoldthemomentofinertia,I,andwearetoldthattheballdoesnotslipso=v/r.Usingsometrigonometry,thedistancedroppedish=(R+r)(R+r)cos.Usingtheseresultswithequation(1)yields

0=kx2+(7/10)mv2mg(R+r)[1cos].(2)

Wehavetwounknowns,vand.Toproceedfurtherwenotethattheballlosescontactwiththesurface.RecallthatlosingcontactimpliesthatN=0.ThenormalisaforceandtofindforceswedrawaFBDandapplyNewton'sSecondLaw.Sincetheballmovesinacircle,wearedealingwithcentripetalacceleration.

j

Fy=may

Nmgcos=mv2/(R+r)

SinceN=0,andaftersomerearranging,theforceequationyields

v2=(R+r)gcos.

Substitutingthisintoequation(2),weget

0=kx2+(7/10)mg(R+r)cosmg(R+r)[1cos].

Collectingtermswithcosonthelefthandsideyields

(17/10)mg(R+r)cos=kx2+mg(R+r).

Solvingfor,

=cos1{[kx2+mg(R+r)]/(17/10)mg(R+r)}=44.9.

Theballcomesoffthesurfacewhentheangleis44.9.

9. Thepotentialenergyofasystemofparticlesinonedimensionisgivenby:

,

wherethepotentialenergyisinJoules.Whatistheworkdoneinmovingaparticleinthispotentialfromx=1mtox=2m?Whatistheforceonaparticleinthispotentialatx=1mandatx=2m?Wherearethepointsofstableandunstableequilibrium(peaksandtroughs)?

Assumingtheparticlestartsandendsatrest,theworkdoneis

WNC=UfUi=[52+3(2)22(2)3][51+3(1)22(1)3]=2J5J=7J.

Forceisthenegativederivativeofthepotential,

F=dU/dx=16x6x2.

Thustheforceatx=1mis

F=16(1)6(1)2=1N,

andatx=2mis

F=16(2)6(2)2=13N.

TheequilibriumpointsoccurattheminimaandmaximaofU(x).WefindthembysettingdU/dx=0,or

1+6x6x2=0.

Thisequationhasrootsatx=0.211mandx=0.789m.

Lookingatthesecondderivative,

d2U/d2x=612x.

Weseethatwhenx=0.211m,d2U/d2x=612(0.211)>0,sox=0.211misaminimum.Wealsoseethatwhenx=0.789m,d2U/d2x=612(0.789)

Examiningthegraph,atx=1.5m,thepotentialenergyapproximatelyU=75J.SothetotalenergyisEtotal=K+U=145J.

(b)Zerovelocityimplieszerokineticenergy.Thusalltheenergywouldhavetobepotential.Examiningthegraph,weseethatEtotal=150Jinterceptsthepotentialenergycurveatx=1.1mandatx=3.4m.

Themaximumkineticenergywouldoccurwherethepotentialenergyisaminimum.Examiningthegraphtheminimumpotentialisatx=2.2m,andis40J.ThekineticenergyisK=EtotalU=150J40J=110J.

SinceK=mv2

v=[2K/m]=[2(110J)/2.5kg)]=9.4m/s.

(c)Examiningthegraph,aparticlewithanenergyofabout210Jcouldescapethepotentialwell.

WorkandEnergy

11. Inthediagrambelow,calculatetheworkdoneif:(a)F=15.0N,=15,andx=2.50m,(b)F=25.0N,=75,andx=12.0m,(c)F=10.0N,=135,andx=5.50m,

Forconstantforces,workisdefinedbyW=Fxcos().(a)W=36.2J(b)W=77.6J(c)W=38.9J

12. Inthediagrambelow,aropewithtensionT=150Npullsa15.0kgblock3.0mupanincline(=25.0).Thecoefficientofkineticfrictionisk=0.20.Findtheworkdonebyeachforceactingontheblock.

Tofindtheworkdonebyaforce,weneedtoknowthemagnitudeoftheforceandtheangleitmakeswiththedisplacement.Tofindforces,wedrawaFBDanduseNewton'sSecondLaw.

i j

Fx=max Fy=may

Tfkmgsin=ma Nmgcos=0

ThesecondequationinformsusthatN=mgcos.Weknowfk=kN=kmgcos.

Force Force(N) W=Fxcos(J)Tension 150 0 450Weight 147.15 +/2 187Normal 133.36 /2 0Friction 26.67 80

13. Awinchliftsa150kgcrate3.0mupwardswithanaccelerationof0.50m/s2.Howmuchworkisdonebythewinch?Howmuchworkisdonebygravity?

Tofindtheworkdonebyaforce,weneedtoknowthemagnitudeoftheforceandtheangleitmakeswiththedisplacement.Tofindforces,wedrawaFBDanduseNewton'sSecondLaw.

j

Fy=may

Tmg=ma

Theworkdonebythewinchistheworkdonebytension.Theworkdonebygravityistheworkdonebytheobjectsweight.Sinceweknowmandg,wefindT=mg+ma=1546.5N.TheworkdonebytensionisWtension=Tycos(0)=4.64103J.TheworkdonebygravityisWgravity=mgycos()=4.41103J.

14. Whatworkdoesabaseballbatdoonabaseballofmass0.325kgwhichhasaforwardspedof36m/sandafinalspeedof27m/sbackwards.Assumemotionishorizontal.

Sinceweareaskedfortheworkdoneandhaveachangeinspeed,wemakeuseofthegeneralizedWorkEnergyTheorem.Sincetheheightoftheballdoesnotchange,thereisonlyachangeinkineticenergy.

WNC=E=KfKi=m[(vf)2(v0)2]=(0.325kg)[(27m/s)2(36m/s)2]=92.1J.

15. Whatistheworkdonebyfrictioninslowinga10.5kgblocktravelingat5.85m/stoacompletestopinadistanceof9.65m?Whatisthekineticcoefficientoffriction?

Sinceweareaskedfortheworkdoneandhaveachangeinspeed,wemakeuseofthegeneralizedWorkEnergyTheorem.Sincetheheightoftheblockdoesnotchange,thereisonlyachangeinkineticenergy.

Wfriction=E=KfKi=m[(vf)2(v0)2]=(10.5kg)[(0)2(5.85m/s)2]=179.67J.

Tofind,weneedtoknowtheforceandtheangleitmakeswiththedisplacement.Tofindforces,wedrawaFBDanduseNewton'sSecondLaw.

i j

Fx=max Fy=may

fk=ma Nmg=0

ThesecondequationgivesN=mgandweknowfk=kN,sofk=kmg.Therefore,theworkdonebyfrictionisWfriction=fkx=kmgx.Rearrangingthisyields

k=Wfriction/mgx=0.18.

16. A50.0Nforceisappliedhorizontallytoa12.0kgblockwhichisinitiallyatrest.Aftertraveling6.45m,thespeedoftheblockis5.90m/s.Whatisthecoefficientofkineticfriction?

Sincetheprobleminvolvesachangeisspeed,wemakeuseoftheGeneralizedWorkEnergyTheorem

WNC=E=KfKi=m[(vf)2(v0)2]=m(vf)2.

Therearetwononconservativeforcesinthisproblem,frictionandtheappliedforce.TheworkdonebyfrictionisgivenbyWfriction=fkx.TheworkdonebytheappliedforceisWF=Fx.

Fxfkx=m(vf)2.(1)

Tofindoutmoreaboutfk,wedrawaFBDanduseNewton'sSecondLaw.

i j

Fx=max Fy=may

Ffk=ma Nmg=0

ThesecondequationgivesN=mgandweknowfk=kN,sofk=kmg.ThusWfriction=kmgx.Combiningthisresultwithequation(1),weget

(Fkmg)x=m(vf)2.

Rearrangingyieldsanexpressionfork

k=[Fxm(vf)2]/mgx.

Usingthegivenvalues,wefindk=0.15.

17. Aropeiswrappedaboundacylindricaldrumasshownbelow.Itispulledwithaconstanttensionof100Nforsixrevolutionsofthedrum.Thedrumhasaradiusof0.500m.Abrakeisalsoapplyingaforcetothedrum.Thebrakepushesinwardsonthedrumwithaforceof200N.Thepressurepointis0.350mfromthecentreofthedrum.Thecoefficientofkineticfrictionbetweenthebrakeandthedrumis0.50.Determinetheworkdonebyeachtorque.

WorkisdefinedbytheformulaW=inrotationalcases.Sincetheropedoesnotslipasitispulled,theobjectrotates6timesclockwiseor=12.Weknowfk=kN.Inthisproblem,Nequalshowhardthebrakeispressed.Notethatthetensionandthefrictionaretangentialtothedrum.

(a)WT=(RT)=(0.5m)(100N)(12)=1.88103J.

(b)Wf=(rfk)=(0.35m)(0.50200N)(12)=1.32103J.

18. Determinetheworkdonebythefollowing.Determinetheanglesbetweentheforcesandthedisplacements.TheforcesareinNewtonsandthedisplacementsareinmetres:

(a)F=(1,2,3)andr=(4,5,6)(b)F=(1,2,3)andr=(4,5,6)(c)F=(4,2,4)andr=(2,8,2)

In3D,workisdefinedW=Fr,whichmeansW=Fxx+Fyy+Fzz.ItisalsodefinedbyW=Frcos,whereFandRarethemagnitudesofthevectors,Fandr.Usingthe3DversionofPythagoras'Theorem,F=[(Fx)2+(Fy)2+(Fz)2]andr=[(x)2+(y)2+(z)2].Ifwefindtheworkusingthefirstform,thencanbefoundfromthesecondby=cos1(W/Fr).

(a)W=14+25+36=32J.

SinceF=[(1)2+(2)2+(3)2]=3.7417Nandr=[(4)2+(5)2+(6)2]=8.7750m,then=cos1(W/Fr)=12.9.

(b)W=14+25+3(6)=4J.

SinceF=[(1)2+(2)2+(3)2]=3.7417Nandr=[(4)2+(5)2+(6)2]=8.7750m,then=cos1(W/Fr)=97.0.

(c)W=42+2(8)+42=0J.

SinceF=[(4)2+(2)2+(4)2]=6.1644Nandr=[(2)2+(8)2+(2)2]=8.4853m,then=cos1(W/Fr)=90.

19. Inthediagrambelow,a5.00kgblockslidesfromrestataheightofh1=1.75mdowntoahorizontalsurfacewhereitpassesovera2.00mroughpatch.Theroughpatchhasacoefficientofkineticfrictionk=0.25.Whatheight,h2,doestheblockreachonthe=30.0incline?

Sincetheprobleminvolvesachangeofheightandspeed,wemakeuseoftheGeneralizedWorkEnergyTheorem.Sincetheblock'sinitialandfinalspeedsarezero,wehave

WNC=E=UfUi=mgh2mgh1.(1)

Thenonconservativeforceinthisproblemisfriction.Tofindtheworkdonebyfriction,weneedtoknowthefriction.Tofindfriction,aforce,wedrawaFBDattheroughsurfaceanduseNewton'sSecondLaw.

i j

Fx=max Fy=may

fk=ma Nmg=0

ThesecondequationgivesN=mgandweknowfk=kN,sofk=kmg.Therefore,theworkdonebyfrictionisWfriction=fkx=kmgx.Puttingthisintoequation(1)yields

kmgx=mgh2mgh1.

Solvingforh2,wefind

h2=h1kx=1.25m.

20. Inthediagrambelow,a5.00kgblockslidesfromrestataheightofh1=1.75mdowntoasmoothhorizontalsurfaceuntilitencountersaroughincline.Theinclinehasacoefficientofkineticfrictionk=0.25.Whatheight,h2,doestheblockreachonthe=30.0incline?

Sincetheprobleminvolvesachangeofheightandspeed,wemakeuseoftheGeneralizedWorkEnergyTheorem.Sincetheblock'sinitialandfinalspeedsarezero,wehave

WNC=E=UfUi=mgh2mgh1.(1)

Thenonconservativeforceinthisproblemisfriction.Tofindtheworkdonebyfriction,weneedtoknowthefriction.Tofindfriction,aforce,wedrawaFBDattheroughsurfaceanduseNewton'sSecondLaw.

i j

Fx=max Fy=may

fkmgsin=ma Nmgcos=0

ThesecondequationgivesN=mgcosandweknowfk=kN,sofk=kmgcos.Therefore,theworkdonebyfrictionisWfriction=fkx=kmgcosx.Puttingthisintoequation(1)yields

kmgcosx=mgh2mgh1.

Alittletrigonometryshowsthatxisrelatedtoh2byx=h2/sin.Puttingthisintotheaboveequationyields

kcos[h2/sin]=h2h1.

Solvingforh2,wefind

h2=h1/[1+k/tan]=1.22m.

21. Aropeiswrappedexactlythreetimesaroundacylinderwithafixedaxisofrotationatitscentre.Thecylinderhasamassof250kgandadiameterof34.0cm.Theropeispulledwithaconstanttensionof12.6N.ThemomentofinertiaofacylinderaboutitscentreisI=MR2.(a)Whatistheworkdownbytheropeasitispulledoffthecylinder.Notethattheropedoesnotslip.(b)Ifthecylinderwasinitiallyatrest,whatisitsfinalangularvelocity?Notethatropesarealwaystangentialtothesurfacesthattheyarewrappedaround.Notethattheworkdonebythetensionisnonconservative.

Sincetheprobleminvolvesachangeisspeed,wemakeuseoftheGeneralizedWorkEnergyTheorem.Sincethereisonlyachangeinrotationalspeed,

WNC=E=KfKi=I[(f)2(0)2]=I(f)2.

Thenonconservativeforceinthisproblemistension,WNC=WT.Sowehave

WT=I(f)2.(1)

(a)ThedefinitionofworkinrotationalsituationsisW=.TensionisalwaystangentialtocylinderssoT=RT.Sincetheropedoesnotslip,thecylinderrotatesclockwisethreetimesso=6.Wecanthusfindtheworkdonebythetension

WT=(RT)(6)=(0.34m)(12.6N)6=40.4J.

(b)Thenwefindthefinalvelocityfromequation(1),

f=[2WT/I].

AccordingtothetableofMomentsofInertia,I=MR2forasolidcylinder.So

f=[4WT/MR2]=6.10rad/s.

22. Supposethatthereisfrictioninproblem12andthatthecompressionmustinfactbe0.425mfortheblocktojustreachthetopofthehill.Whatworkisdonebythefrictionalforce?

FrictionisanonconservativeforcessoWNC=Wfriction0.Thus

Wfriction=mghKx2=132J.

Frictiondoes132Joulesofwork.

23. AlargecylinderofmassM=150kgandradiusR=0.350m.Theaxleonwhichthecylinderrotatesis

NOTfrictionless.Aropeiswrappedaroundthecylinderexactlytentimes.Fromrest,theropeispulledwithaconstanttensionof25.0N.Theropedoesnotslipandwhentheropecomesfree,thecylinderhasaforwardangularvelocityf=10.5rad/s.ThemomentofinertiaofacylinderisI=MR2.(a)Whatanglewasthecylinderrotatedthrough?(b)Whatisthefrictionaltorqueoftheaxle?(c)Howlongwillittakethefrictionaltorquetobringthecylindertoastop?(d)Howmanyrevolutionswillithaveturned?

(a)Sincetheropedoesnotslip,thecylinderrotatestentimesso=20,wheretherotationisassumedtobeclockwise.

(b)Sincetheprobleminvolvesforcesandachangeisrotationalspeed,wemakeuseoftheGeneralizedWorkEnergyTheorem.Sincethereisonlyachangeinrotationalkineticenergy,

WNC=E=KfKi=I[(f)2(0)2]=I(f)2.

Thenonconservativeforcesinthisproblemarethetensionandtheaxlefriction,WNC=WT+Wf.Sowehave

WT+Wf=I(f)2.(1)

ThedefinitionofworkinrotationalsituationsisW=.TensionisalwaystangentialtocylinderssoT=RT,againassumingtheropepullsthecylinderclockwise.Thustheworkdonebythetensionis

WT=(RT)(20)=(0.35m)(25N)(20)=549.78J.

Usingthisresultandequation(1),wecanfindtheworkdonebyfriction

Wf=I(f)2WT=M(Rf)2WT=506.46J549.78J=43.3178J.

SinceWf=f,wefindthefrictionaltorquetobe

f=Wf/=Wf/20=+0.6894Nm

thesignindicatingthatitiscounterclockwise.

(c)Aftertheropecomesoffthecylinder,theonlyforceactingisfrictionsothegeneralizedWorkEnergyTheorembecomes

Wf=E=KfKi=I[(f)2(0)2]=I(0)2,

where0hereisffromthefirstpartoftheproblemandthenewf=0sincethedrumcomestoastop.AgainWf=f,wherefistheresultfrompart(b).Therefore

f=I(0)2=(506.46J)/0.6894Nm=734.6rad=117revclockwise.

(d)Tofindthetimeittakestoslowdown,notethatwehavetheinitialandfinalangularvelocitiesandtheangulardisplacement.Referringtoourkinematicsequations,wefind

=(f+0)t.

Rearrangingfortyields,

t=2/(f+0)=140s.

24. Apersonwithanaxetogrindisusingawhetstone.Thewhetstoneisconnectedtoamotorwhichkeepsitrotatingat85rev/min.Thewhetstoneisasolidcylinderofmadeofaspecialtypeofstoneofmass45kgandradius20cm.Theheavyaxebladeisbeingpressedontothewhetstonewithasteadyforceof25Ndirectedintothecentreofthewhetstone.Suddenly,thepowertothemotoriscutbutthepersonmaintainstheforceoftheaxeonthewhetstone.Thecoefficientofkineticfrictionbetweentheaxeandthestoneis0.60.Themomentofinertiaofacylinderaboutit'scentreofmassisI=MR2.(a)Calculatethetorqueofthefrictionalforceonthewheel.(b)Calculatethetotalanglethroughwhichthewheelturnsfromthetimethepowergoesofftothetimethewhetstonestopsrotating.(c)Howlongdoesittakeforthewheeltostopafterthepoweroutage?

(a)Thefrictionalforceistangentialtothesurfaceofthewhetstone.Thefrictiontorqueisf=Rfk.Thefrictionalforceisgivenbyfk=N.AccordingtoNewton'sThirdLaw,thenormalisequaltotheforcepressingtheaxeintothewhetstone.Hence

f=RF=(0.2m)(0.6)(25N)=3.0Nm.

(b)Sincetheobjectisrotatingandwearegivenspeeds,weapplythegeneralizedWorkEnergyTheorem.Notethatweonlyhaverotationalkineticenergy,so

WNC=E=KfKi=I[(f)2(0)2]=I(0)2=M(R0)2,

whereWNCistheworkdonebyfrictioninslowingthewhetstoneandweusedthegivenI.Theangularvelocityisgiveninrev/minbutinSIunitsitis

0=85rev/min(2rad)/revmin/(60s)=8.9012rad/s.

ThedefinitionofworkisWNC=f.Ifweusethisrelation,wefindthat

=M(R0)2/f=(45kg)(0.208.9012rad/s)2/f=11.9rad=1.89rev.

(c)Tofindthetimeittakestoslowdown,notethatwehavetheinitialandfinalangularvelocitiesandtheangulardisplacement.Referringtoourkinematicsequations,wefind

=(f+0)t.

Rearrangingfortyields,

t=2/0=2(11.88rad)/(8.9012rad/s)=2.67s.

25. Inthefigurebelow,ablockofmass5.0kgstartsatpointAwithaspeedof15.0m/sonaflatfrictionlesssurface.AtpointB,itencountersaninclinewithcoefficientofkineticfrictionk=0.15.Theblockmakesituptheinclinetoasecondflatfrictionlesssurface.Whatistheworkdonebyfriction?WhatisthevelocityoftheblockatpointC?Theinclineis2.2mlongatanangle=15.

Theprobleminvolvesachangeinheightandspeed,soweapplythegeneralizedWorkEnergyTheorem.

WNC=E=(KfKi)+(UfUi)=m(vC)2m(vA)2+mgh.(1)

Herethenonconservativeforceisfriction,soWNC=Wf.Tofindfriction,aforce,wedrawaFBDanduseNewton'sSecondLaw.

i j

Fx=max Fy=may

fkmgsin=ma Nmgcos=0

ThesecondequationgivesN=mgcosandweknowfk=kN,sofk=kmgcos.Therefore,theworkdonebyfrictionis

Wf=fkx=kmgcosx=(.15)(5kg)(9.81m/s2)cos(15)(2.2m)=15.635J.

Notefromthediagram,thattheheighthisrelatedtothelengthoftheinclinebyh=xsin.Puttingboth

resultsintoequation(1)yields

Wf=m(vC)2m(vA)2+mg[xsin].

SolvingforvCyields

vC=[2Wf/m2gxsin+(vA)2]=14.4m/s.

26. InthediagramblockM1isconnectedtoM2byaverylightstringrunningoverthreeidenticalpulleys.ThepulleysarediskswithmassMpandtheropedoesnotslip.ThecoefficientofkineticfrictionforthehorizontalsurfacethatM1isonisk.FindanexpressionforthespeedvofblockM2intermsofthedistanceLthatblockM1movestotheright.YouranswershouldbeexpressedintermsofL,M1,M2,Mp,k,andgonly.Hint:workandenergymethodsprovidethequickestsolution.

Theprobleminvolveschangesinheight,speed,androtation,sowewouldapplythegeneralizedWorkEnergyTheoremevenifnotdirectedtodoso,

WNC=E=(KfKi)+(UfUi),(1)

whereKisthesumofallthelinearandrotationalkineticenergiesofeachobject,andUisthesumofthegravitationalpotentialenergies.Sincethereiskineticfrictionactingonthesystem,WNC=Wf.TofindWf,weneedtothefrictionalforce.Tofindaforce,wedrawaFBDanduseNewton'sSecondLaw.

i j

Fx=max Fy=may

Tfk=M1a NM1g=0

ThesecondequationgivesN=M1gandweknowfk=kN,sofk=kM1g.Therefore,theworkdonebyfrictionisWf=fkx=kM1gx.

Nextconsiderthechangeinenergyofeachobject.M1increasesitslinearkineticenergy.M2alsoincreasesitslinearkineticenergybutlosesgravitationalpotentialenergy.Thethreeidenticalpulleysincreasetheirrotationalkineticenergies.Thusequation(1)becomes

kM1gx=M1v2+M2v2M2gh+3Idisk2.

SinceM1andM2areconnectedbythesamerope,theyhavethesamespeedandmovethesamedistancesothatx=h.ConsultingatableofMomentsofInertia,wefindIdisk=MPR2.Sincetheropedoesnotslip,thetangentialspeedofthepulleysisthesameastheropeandthus=v/R.Substitutingthisrelationsbackintoourequationyields,

kM1gh=M1v2+M2v2M2gh+3[MPR2](v/R)2.

Collectingthetermswithvyields

[M1+M2+(3/2)MP]v2=[M2kM1]gh.

Solvingforvyields,

v=[2(M2kM1)gh/(M1+M2+(3/2)MP)].

27. Tarzan,LordofApes,isswingingthroughthejungle.Inthediagrambelow,TarzanisstandingatpointAonatreebranchh1=22.0mabovethefloorofthejungle.Tarzanisholdingoneendofavinewhichisattachedtoabranchonasecondtree.ThevineisL=21.0mlong.WhenTarzanswingsonthevine,hispathisinanarcofacircle.AtthebottomofhisswingheisatpointB,13.0mabovetheground.IgnoreTarzan'sheight.Tarzanhasamassof90.0kg.Thevinedoesnotstretchandhasnegligiblemass.

(a)Whydoesthetensioninthevinedonowork?(b)Whatwillbehisspeedatthispoint?(c)Whatwillbethetensionintherope?

Theprobleminvolvesachangeinheightandspeed,soweapplythegeneralizedWorkEnergyTheorem.

WNC=E=(KfKi)+(UfUi)=m(vB)2mgh.(1)

(a)Heretheonlypossiblenonconservativeforceisfriction,soWNC=WT.ThedefinitionofworkisW=Fxcos,butinthisproblemthetensionisalongaradiusandisthusalwaysat90tothedisplacement.Asaresult,WT=0.Thuswehave

0=m(vB)2mgh.

(b)TofindthespeedatpointB,weneedtoknowh,thedistanceTarzandropped.Examiningthequestion,weseethath=h1h2=22.0m13.0m=9.0m.Rearrangingourequation,wefind

vB=[2gh]=[2(9.81m/s2)(9m)]=13.29m/s.

(c)Tensionisaforce.TofindaforceweneedtodrawaFBDandapplyNewton'sSecondLaw.SinceTarzanisswinginginacircle,wearedealingwithcentripetalacceleration.

j

Fy=may

Tmg=mv2/L

SolvingforT,

T=mg+mv2/L=(90.0kg)[9.81m/s2+(13.288m/s)2/(21.0m)]=1.64103N.

28. AblockofmassMonaflattableisconnectedbyastringofnegligiblemasstoaverticalspringwithspringconstantKwhichisfixedtothefloor.ThestringgoesoverapulleythatisasoliddiskofmassM

andradiusR.Asshowninthediagrambelow,thespringisinitiallyinitsequilibriumpositionandthesystemisnotmoving.ApersonpullstheblockwithforceFthroughadistanceL.DeterminethespeedvoftheblockafterithasmoveddistanceL.Thetabletopisfrictionless.

Theprobleminvolveschangesinheight,speed,androtation,sowewouldapplythegeneralizedWorkEnergyTheoremevenifnotdirectedtodoso,

Wext=E,(1)

whereEisthesumofallthemechanicalenergiesofeachobject.Ifthesystemconsistsofthespring,string,pulley,blockandtheearth,thenFisanexternalforceactingonthesystemandWext=FL.

Nextconsiderthechangeinenergyofeachobject.Thespringstretchesassoincreasesitspotentialenergy.Thepulleyturnsfromrestsoincreasesitsrotationalkineticenergy.Theblockmovesfromrestsoitincreasesitslinearkineticenergy.Thusequation(1)becomes

FL=Kx2+Idisk2+Mv2.

Sincetheblockandspringareconnectedbythesamestring,thespringhasstretchedx=L.ConsultingatableofMomentsofInertia,wefindIdisk=MR2.Sincethestringdoesnotslip,thetangentialspeedofthepulleysisthesameastheropeandthus=v/R.Substitutingthisrelationsback

FL=KL2+[MR2](v/R)2+Mv2.

Collectingthetermswithvyields

3/4Mv2=FLKL2.

Solvingforvyields,

v=[(4M2KL2)/3M].

29. Whatpowerisrequiredtopulla5.0kgblockatasteadyspeedof1.25m/s?Thecoefficientoffrictionis0.30.

ThepowerrequiredtomovetheblockatconstantspeedisP=Fv.Wearegivenv,thespeedoftheblock.TogetF,aforce,wedrawaFBDandapplyNewton'sSecondLaw,

i j

Fx=max Fy=may

Ffk=0 Nmg=0

ThesecondequationgivesN=mgandweknowfk=kN,sofk=kmg.Therefore,theappliedforceisF=kmg.Thusthepoweris

P=kmgv=(0.3)(5kg)(9.81m/s2)(1.25m/s)=18.4Watts.

30. A7500Wengineispropellingaboatat12km/h.Whatforceistheengineexertingontheboat?Whatforceandhowmuchpoweriswaterresistanceexertingonthespeedboat?

FirstweconvertthevelocitytoSIunits,

12km/h(1000m)/km(1h)/(3600s)=3.333m/s.

WeknowP=Fv,so

F=P/v=7500W/3.333m/s=2250N.

ByNewton'sThirdLaw,thewaterisexerting2250Ninthereversedirection.Itisalsoremoving7500Wofpowerwhichisgoingintoincreasingthekineticenergyofthewater.

31. A3.0hpenginepullsa245kgblockatconstantspeedupa12.0m30.0incline.Howlongdoesthistake?Ignorefriction.

First,3.0hp746W/hp=2238W.

Next,theworkdonebytheenginemustequaltheworkdonebytheweight,fromNewton'sThirdLaw.Theworkdonebygravityis

Wgravity=mgh=mgLsin=1.442104J.

Sincepowerisdefinedasworkovertime,P=W/t,thetimeittakesis

t=W/P=(1.442104J)/(2238W)=6.44s.

Questions?mike.coombes@kwantlen.bc.ca

http://www.kwantlen.ca/physicsfile:///D:/welcome.htmfile:///D:/handouts.htmfile:///D:/P1120_Problems/P1120_Problems.htmfile:///D:/P1120_Solutions/P1120_Solutions.htmmailto:mike.coombes@kwantlen.cahttp://www.kwantlen.ca/