CH 07 HW
CH 07 HW
CH 07 HW
Ch 07 HW
Ch07HW
Due:11:59pmonTuesday,October14,2014
Youwillreceivenocreditforitemsyoucompleteaftertheassignmentisdue.GradingPolicy
WorkonaSlidingBox
Aboxofmassmisslidingalongahorizontalsurface.
PartA
Theboxleavespositionx
= 0
withspeedv 0 .Theboxisslowedbyaconstantfrictionalforceuntilitcomestorestatpositionx
= x1
FindFf ,themagnitudeoftheaveragefrictionalforcethatactsonthebox.(Sinceyoudon'tknowthecoefficientoffriction,don'tincludeitinyour
answer.)
Expressthefrictionalforceintermsofm,v 0 ,andx1 .
Hint1.Howtoapproachtheproblem
Usetheworkenergytheorem.Asappliedtothispart,thetheoremstatesthattheworkdonebyfrictionisequaltothechangeinkineticenergyof
thebox:
W f = K = Kf Ki
Hint2.Findtheinitialkineticenergy
WhatisKi ,thekineticenergyoftheboxatpositionx
= 0
ANSWER:
Ki
1
2
mv0
Hint3.Findthefinalkineticenergy
WhatisKf ,thekineticenergyoftheboxwhenitreachespositionx
= x1
ANSWER:
Kf
= 0
Hint4.Findtheworkdonebyfriction
FindW f ,theworkdonebyfrictiononthebox.Notethattheworkdonebyfrictionisalwaysnegative(i.e.,frictionalwaysdissipatesenergy).
ExpressyouranswerintermsofFf andothergivenvariables.
ANSWER:
Wf
F f x1
ANSWER:
Ff
mv0
2x 1
Correct
PartB
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Ch 07 HW
Aftertheboxcomestorestatpositionx1 ,apersonstartspushingthebox,givingitaspeedv 1 .
Whentheboxreachespositionx2 (wherex2
> x1
),howmuchworkW p hasthepersondoneonthebox?
Assumethattheboxreachesx2 afterthepersonhasaccelerateditfromresttospeedv 1 .
Expresstheworkintermsofm,v 0 ,x1 ,x2 ,andv 1 .
Hint1.Howtoapproachtheproblem
Again,usetheworkenergytheorem.Inthispartoftheproblem,boththepersonandfrictionaredoingworkonthebox:
W f + W p = K = Kf Ki
Hint2.Findtheworkdonebyfriction
WhatisW f ,thetotalworkdonebyfrictionontheboxasthepersonpushesitfrompositionx1 topositionx2 ?
Answerintermsofgivenvariables.(YouranswershouldnotincludeFf .)
Hint1.Findingtheforceoffriction
ThenormalforceontheboxisunchangedfrompartA.Therefore,theforceoffrictionisthesameinthispartasinpartA.
ANSWER:
Wf
(m v0
)(x 2 x 1 )
2x 1
Hint3.Findthechangeinkineticenergy
WhatisK ,thechangeinkineticenergyoftheboxfromthemomentitisatpositionx1 tothemomentitisatpositionx2 ?
ANSWER:
K
1
2
mv1
ANSWER:
Wp
1 mv0
2
x1
)( x 2 x 1 ) +
1
2
mv1
Correct
IntroductiontoPotentialEnergy
LearningGoal:
Understandthatconservativeforcescanberemovedfromtheworkintegralbyincorporatingthemintoanewformofenergycalledpotentialenergythat
mustbeaddedtothekineticenergytogetthetotalmechanicalenergy.
Thefirstpartofthisproblemcontainsshortanswerquestionsthatreviewtheworkenergytheorem.Inthesecondpartweintroducetheconceptofpotential
energy.Butfornow,pleaseanswerintermsoftheworkenergytheorem.
WorkEnergyTheorem
Theworkenergytheoremstates
Kf = Ki + W all ,
whereW all istheworkdonebyallforcesthatactontheobject,andKi andKf aretheinitialandfinalkineticenergies,respectively.
PartA
Theworkenergytheoremstatesthataforceactingonaparticleasitmovesovera______changesthe______energyoftheparticleiftheforcehasa
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Ch 07 HW
componentparalleltothemotion.
Choosethebestanswertofillintheblanksabove:
ANSWER:
distance/potential
distance/kinetic
verticaldisplacement/potential
noneoftheabove
Correct
Itisimportantthattheforcehaveacomponentactinginthedirectionofmotion.Forexample,ifaballisattachedtoastringandwhirledinuniform
circularmotion,thestringdoesapplyaforcetotheball,butsincethestring'sforceisalwaysperpendiculartothemotionitdoesnoworkand
cannotchangethekineticenergyoftheball.
PartB
Tocalculatethechangeinenergy,youmustknowtheforceasafunctionof_______.Theworkdonebytheforcecausestheenergychange.
Choosethebestanswertofillintheblankabove:
ANSWER:
acceleration
work
distance
potentialenergy
Correct
PartC
Toillustratetheworkenergyconcept,considerthecaseofastonefallingfromxi toxf undertheinfluenceofgravity.
Usingtheworkenergyconcept,wesaythatworkisdonebythegravitational_____,resultinginanincreaseofthe______energyofthestone.
Choosethebestanswertofillintheblanksabove:
ANSWER:
force/kinetic
potentialenergy/potential
force/potential
potentialenergy/kinetic
Correct
PotentialEnergyYoushouldreadaboutpotentialenergyinyourtextbeforeansweringthefollowingquestions.
Potentialenergyisaconceptthatbuildsontheworkenergytheorem,enlargingtheconceptofenergyinthemostphysicallyusefulway.Thekeyaspectthat
allowsforpotentialenergyistheexistenceofconservativeforces,forcesforwhichtheworkdoneonanobjectdoesnotdependonthepathoftheobject,
onlytheinitialandfinalpositionsoftheobject.Thegravitationalforceisconservativethefrictionalforceisnot.
Thechangeinpotentialenergyisthenegativeoftheworkdonebyconservativeforces.Henceconsideringtheinitialandfinalpotentialenergiesis
equivalenttocalculatingtheworkdonebytheconservativeforces.Whenpotentialenergyisused,itreplacestheworkdonebytheassociatedconservative
force.Thenonlytheworkduetononconservativeforcesneedstobecalculated.
Insummary,whenusingtheconceptofpotentialenergy,onlynonconservativeforcescontributetothework,whichnowchangesthetotalenergy:
Kf + U f = E f = W nc + E i = W nc + Ki + U i ,
whereU f andU i arethefinalandinitialpotentialenergies,andW nc istheworkdueonlytononconservativeforces.
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Ch 07 HW
Now,wewillrevisitthefallingstoneexampleusingtheconceptofpotentialenergy.
PartD
Ratherthanascribingtheincreasedkineticenergyofthestonetotheworkofgravity,wenow(whenusingpotentialenergyratherthanworkenergy)
saythattheincreasedkineticenergycomesfromthe______ofthe_______energy.
Choosethebestanswertofillintheblanksabove:
ANSWER:
work/potential
force/kinetic
change/potential
Correct
PartE
Thisprocesshappensinsuchawaythattotalmechanicalenergy,equaltothe______ofthekineticandpotentialenergies,is_______.
Choosethebestanswertofillintheblanksabove:
ANSWER:
sum/conserved
sum/zero
sum/notconserved
difference/conserved
Correct
PotentialEnergyCalculations
LearningGoal:
Tounderstandtherelationshipbetweentheforceandthepotentialenergychangesassociatedwiththatforceandtobeabletocalculatethechangesin
potentialenergyasdefiniteintegrals.
Imaginethataconservativeforcefieldisdefinedinacertainregionofspace.Doesthissoundtooabstract?Well,thinkofagravitationalfield(theonethat
makesapplesfalldownandkeepstheplanetsorbiting)oranelectrostaticfieldexistingaroundanyelectricallychargedobject.
Ifaparticleismovinginsuchafield,itschangeinpotentialenergydoesnotdependontheparticle'spathandisdeterminedonlybytheparticle'sinitialand
finalpositions.Recallthat,ingeneral,thecomponentofthenetforceactingonaparticleequalsthenegativederivativeofthepotentialenergyfunction
alongthecorrespondingaxis:
Fx =
dU (x)
dx
Therefore,thechangeinpotentialenergycanbefoundastheintegral
U =
2
1
F d s ,
andds isasmalldisplacementoftheparticlealongitspathfrom1to2.
Evaluatingsuchanintegralinageneralcasecanbeatediousandlengthytask.However,twocircumstancesmakeiteasier:
1. Becausetheresultispathindependent,itisalwayspossibletoconsiderthemoststraightforwardwaytoreachpoint2frompoint1.
2. Themostcommonrealworldfieldsarerathersimplydefined.
Inthisproblem,youwillpracticecalculatingthechangeinpotentialenergyforaparticlemovinginthreecommonforcefields.
^ istheunitvectorintheradial
Notethat,intheequationsfortheforces,^i istheunitvectorinthexdirection,^
j istheunitvectorintheydirection,and r
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Ch 07 HW
directionincaseofasphericallysymmetricalforcefield.
PartA
Considerauniformgravitationalfield(afairapproximationnearthesurfaceofaplanet).Find
U (y ) U (y ) =
f
yf
y
F g d s ,
where
^
F g = mg j
andds = dy ^
j.
Expressyouranswerintermsofm,g,y 0 ,andy f .
Hint1.RelativedirectionsofF g andds
F g ds = mg dy
ANSWER:
U (y
) U (y
mg(y
Correct
PartB
ConsidertheforceexertedbyaspringthatobeysHooke'slaw.Find
U (x f ) U (x 0 ) =
xf
x0
F s ds ,
where
^
F s = kx i ,
^
ds = dx i
andthespringconstantkispositive.
Expressyouranswerintermsofk,x0 ,andxf .
Hint1.RelativedirectionsofF s andds
F s ds = kx dx
ANSWER:
=
U (xf ) U (x0 )
.5k(xf
x0
Correct
PartC
Finally,considerthegravitationalforcegeneratedbyasphericallysymmetricalmassiveobject.Themagnitudeanddirectionofsuchaforcearegiven
byNewton'slawofgravity:
FG =
^ G,m 1 ,andm 2 areconstantsandr
whereds = dr r
> 0
Gm 1 m 2
r2
^,
r
.Find
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U(
) U( 0) =
d
f
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Ch 07 HW
U (rf ) U (r0 ) =
rf
r0
F G ds .
Hint1.RelativedirectionsofF G
andds
NotethatF G
andds areparallel,andtheirdotproductissimplytheproductoftheirmagnitudes.Thatis,
F G ds =
Gm 1 m 2
r2
dr.
Hint2.Integrating1/r2
Recallthat
dr
r
1
r
Carefullyaccountforallthenegativesignsinyourcalculations.
ANSWER:
=
U (rf ) U (r0 )
G m1 m2 (
1
rf
1
r0
Correct
Asyoucansee,thechangeinpotentialenergyoftheparticlecanbefoundbyintegratingtheforcealongtheparticle'spath.However,thismethod,
aswementionedbefore,doeshaveanimportantrestriction:Itcanonlybeappliedtoaconservativeforcefield.Forconservativeforcessuchas
gravityortensiontheworkdoneontheparticledoesnotdependontheparticle'spath,andthepotentialenergyisthefunctionoftheparticle's
position.
Incaseofanonconservativeforcesuchasafrictionalormagneticforcethepotentialenergycannolongerbedefinedasafunctionofthe
particle'sposition,andthemethodthatyouusedinthisproblemwouldnotbeapplicable.
PhETTutorial:EnergySkatePark
LearningGoal:
Tounderstandconservationofmechanicalenergyinvolvinggravitationalpotentialenergyandkineticenergy.
Forthisproblem,usethePhETsimulationEnergySkatePark.Thissimulationallowsyoutoexplorethemotionandenergeticsofaskaterridingalonga
track.
Startthesimulation.Whenyouclickthesimulationlink,youmaybeaskedwhethertorun,open,orsavethefile.Choosetorunoropenit.
Watchtheskaterskateupanddownthetrack.Youcanclickanddragtheskatertoanylocationandreleasetheskaterfromrest.
ClickontheEnergyvs.Positionbuttontoseeaplotofthekinetic,potential,thermal,andtotalenergiesasafunctionoftheskater'sposition.TryBar
Graphtoseethesameinformationasabargraph.
Forthistutorial,youshouldhavetheCoefficientofFriction(undertheTrackFrictionbuttonatlowerright)setto"None,"whichmeansnothermalenergy
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Ch 07 HW
isgenerated.YoushouldalsodeselecttheThermalenergychoiceatbottomrightfromtheEnergyvs.Positionplot.
Playaroundwiththesimulation.Whenyouaredone,clickontheResetbuttonbeforeansweringthequestions.
PartA
ClickontheEnergyvs.Positionbutton,anddeselectallformsofenergyexceptkineticenergy.
Whereonthetrackistheskater'skineticenergythegreatest?
Hint1.Howtoapproachtheproblem
ThekineticenergyofanobjectisgivenbyK
= (1/2)mv
,wheremisthemassoftheobjectandv isthespeedoftheobject.Thus,the
skaterismovingthefastestwheretheskater'skineticenergyisthegreatestandviceversa.
LookattheKineticEnergyvs.Positiongraph,andidentifythelocation(s)wheretheskater'skineticenergyreachesitsmaximumvalue.
ANSWER:
atitsmaximumvalueatthelowestpointofthetrack.
Theskater'skineticenergyis
atitsmaximumvalueatthelocationswheretheskaterturnsandgoesbackintheoppositedirection.
thesameeverywhere.
Correct
ThekineticenergyofanobjectisgivenbyK
= (1/2)mv
,wherev isthespeedoftheobjectandmisthemassoftheobject.Thus,theskater's
kineticenergyisgreatestatthelowestpointofthetrack,wheretheskaterismovingthefastest.
PartB
ChangetheEnergyvs.Positiongraphtodisplayonlypotentialenergy.
Astheskaterisskatingbackandforth,wheredoestheskaterhavethemostpotentialenergy?
Hint1.Howtoapproachtheproblem
ThegravitationalpotentialenergyofanobjectisgivenbyU
= mgy
,wherey istheobject'sheightabovethepotentialenergyreference,which
iscurrentlytheground.Thus,theskater'spotentialenergyisgreatestwheretheskateristhehighestabovethereferenceline.
LookatthePotentialEnergyvs.Positiongraph,andidentifythelocation(s)wheretheskater'spotentialenergyreachesitsmaximumvalue.
ANSWER:
atitsmaximumvalueatthelocationswheretheskaterturnsandgoesbackintheoppositedirection.
Theskater'spotentialenergyis
thesameeverywhere.
atitsmaximumvalueatthelowestpointofthetrack.
Correct
ThegravitationalpotentialenergyofanobjectisgivenbyU
= mgy
,wherey istheobject'sheightabovethepotentialenergyreference,whichis
currentlytheground.Thus,theskater'spotentialenergyisgreatestatthelocationswheretheskaterturnstogobackintheoppositedirection,
wheretheskateristhehighestabovethereferenceline.
Noticethattheskater'spotentialenergyisgreatestwherethekineticenergyisthelowest,andviceversa.
PartC
Becauseweareignoringfriction,nothermalenergyisgeneratedandthetotalenergyisthemechanicalenergy,thekineticenergyplusthepotential
energy:E = K + U .
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Ch 07 HW
DisplaythetotalenergyintheEnergyvs.Positiongraph.Astheskaterisskatingbackandforth,whichstatementbestdescribesthetotalenergy?
ANSWER:
Thetotalenergyis
greatestatthelocationswheretheskaterturnsandgoesbackintheoppositedirectionandsmallestatthelowestpointofthetrack.
thesameatalllocationsofthetrack.
smallestatthelocationswheretheskaterturnstogobackintheoppositedirectionandgreatestatthelowestpointofthetrack.
Correct
Themechanicalenergy(kineticpluspotential)isconserved.(Sincethereisnofriction,themechanicalenergyisequaltothetotalenergy.)When
thekineticenergyisrelativelysmall,thepotentialenergyisrelativelylarge,andviceversa.
PartD
Ignoringfriction,thetotalenergyoftheskaterisconserved.Thismeansthatthekineticpluspotentialenergyatonelocation,sayE 1
mustbeequaltothekineticpluspotentialenergyatadifferentlocation,sayE 2
expressedasE 1
= E2
= K2 + U 2
= K1 + U 1
.Thisistheprincipleofconservationofenergyandcanbe
.Sincetheenergyisconserved,thechangeinthekineticenergyisequaltothenegativeofthechangeinthepotentialenergy:
K2 K1 = (U 2 U 1 ), or K1 = U 2
SelecttheShowGridoption.Then,pullthebottomofthetrackdownsuchthatitis1mabovetheground(clickanddragonthebluecircleonthe
bottomofthetrack).Confirmthatthemassoftheskaterissetto75.0kg(selectChooseSkatertoviewmass)andthattheaccelerationofgravityis
setat9.81N/kg.
Clearallplots,placetheskateronthetrack7mabovetheground,andlookattheresultingmotionandthegraphshowingtheenergetics.
Matchtheapproximatenumericalvaluesontheleftwiththeenergytypecategoriesontherighttocompletetheequations.
Dragtheappropriatenumericalvaluestotheirrespectivetargets.
ANSWER:
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Ch 07 HW
Correct
PartE
Basedonthepreviousquestion,whichstatementistrue?
ANSWER:
equaltotheinitialpotentialenergy.
Thekineticenergyatthebottomoftherampis
equaltothetotalenergy.
equaltotheamountofpotentialenergylossingoingfromtheinitiallocationtothebottom.
Correct
Becausethetotalenergyisconserved,thekineticenergyatthebottomofthehillplusthepotentialenergyatthebottomofthehillmustequalthe
initialpotentialenergy(sincetheinitialkineticenergyiszero):Kbottom + U bottom = U initial .Solvingforthekineticenergy,weget
Kbottom = U initial U bottom
,orKbottom
.Moregenerally,thechangeinthekineticenergyisequaltothe
negativeofthechangeinthepotentialenergy.
PartF
Iftheskaterstartedfromrest4mabovetheground(insteadof7m),whatwouldbethekineticenergyatthebottomoftheramp(whichisstill1m
abovetheground)?
Hint1.Howtoapproachtheproblem
UsetheEnergyvs.PositiongraphlikeyoudidinPartD.
ANSWER:
4410J
735J
2940J
2205J
Correct
Noticethattheskateronlydropsdown3m,ascomparedto6mforthecasewhentheskaterstarted7mabovetheground.Inthiscase,the
skaterloseshalfasmuchpotentialenergyasinthepreviouscase,sothekineticenergyatthebottomisonlyhalfasmuchasthepreviouscase.
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Ch 07 HW
PartG
Onecommonapplicationofconservationofenergyinmechanicsistodeterminethespeedofanobject.Althoughthesimulationdoesntgivethe
skater'sspeed,youcancalculateitbecausetheskater'skineticenergyisknownatanylocationonthetrack.
Consideragainthecasewheretheskaterstarts7mabovethegroundandskatesdownthetrack.Whatistheskater'sspeedwhentheskaterisatthe
bottomofthetrack?
Expressyouranswernumericallyinmeterspersecondtotwosignificantfigures.
Hint1.Findthekineticenergyatthebottomofthetrack
Whatisthekineticenergyoftheskateratthebottomofthetrack?
ANSWER:
4410J
5145J
735J
2205J
Hint2.Findthespeedatthebottomofthetrack
BecauseK
= (1/2)mv
Expressyouranswernumericallyinmeterspersecondtotwosignificantfigures.
ANSWER:
11 m/s
ANSWER:
11 m/s
Correct
Fora75kgobjecthavingapproximately4410Jofenergy,thespeedmustberoughly11m/s.
PartH
Whentheskaterstarts7mabovetheground,howdoesthespeedoftheskateratthebottomofthetrackcomparetothespeedoftheskateratthe
bottomwhentheskaterstarts4mabovetheground?
Hint1.Findthekineticenergyatthebottomofthetrack
Howmuchmorekineticenergydoestheskaterhaveatthebottomofthetrack?
ANSWER:
fourtimesasmuch
thesameamount
twiceasmuch
Hint2.Findtheratioofthespeeds
Becausekineticenergyisproportionaltothesquareofthespeed,theratioofthekineticenergyisequaltotheratioofthespeedssquared:
2
K7m /K4m = (v 7m /v 4m ) .Whatistheratiov 7m /v 4m ofthespeeds?
Expressyouranswernumericallywithtwosignificantfigures.
ANSWER:
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Ch 07 HW
v 7m /v 4m
= 1.4
ANSWER:
twiceasfast.
Thespeedis
fourtimesasfast.
thesame.
higher,butlessthantwiceasfast.
Correct
Thepersonwillhavetwiceasmuchkineticenergy.Becausekineticenergyisproportionaltothespeedsquared,theratioofthespeedsisequalto
thesquarerootoftheratioofthekineticenergies.Inthiscase,sincetheratioofthekineticenergiesis2,theratioofthespeedsisequaltothe
squarerootof2,orroughly1.4.
PartI
Changethepotentialenergyreferencelinetobe7mabovetheground(selectthePotentialEnergyReferenceoption,andclickanddragonthe
dashedbluehorizontallinetothe7mgridline).Placetheskateronthetrack7mabovetheground,andlettheskatergo.
ANSWER:
greaterthanzero.
Thetotalenergyoftheskateris
lessthanzero.
equaltozero.
Correct
Initially,thekineticenergyiszerobecausetheskaterisatrestandthepotentialenergyiszerobecausetheskaterisstartingatthepotential
energyreferenceline.Becauseenergyisconserved,theskater'stotalenergyremainszero.
PartJ
Atthebottomofthehill,howdoesthekineticenergycomparetothecasewhenthepotentialenergyreferencewasthegroundandtheskaterwas
released7mabovetheground?
ANSWER:
lessthanthecasewhenthepotentialenergyreferencewastheground.
Thekineticenergyis
greaterthanthecasewhenthepotentialenergyreferencewastheground.
thesameasthecasewhenthepotentialenergyreferencewastheground.
Correct
Thekineticenergyatthebottomofthetrackisequaltotheamountofpotentialenergylostingoingfromtheinitialpositiontothebottom.Even
thoughthetotalenergyandinitialpotentialenergyaredifferentfromwhenthereferencewastheground,theskaterstilllosesthesameamountof
potentialenergyingoingfrom7mdowntothebottomofthetrack.Thus,whenapplyingconservationofenergy,anypotentialenergyreference
canbeused!
PartK
ClickTracksintheupperleftcornerofthewindow,andselectDoubleWell(RollerCoaster).Then,clickanddragonthebluecirclestostretchand/or
bendthetracktomakeitlooklikethatshownbelow.
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Ch 07 HW
Iftheskaterstartsfromrestatposition1,rank,inincreasingorderfromleasttogreatest,thekineticenergyoftheskateratthefivepositionsshown.
Rankfromsmallesttolargest.Torankitemsasequivalent,overlapthem.
ANSWER:
Correct
PhETInteractiveSimulations
UniversityofColorado
http://phet.colorado.edu
VideoTutor:ChinBasher?
First,launchthevideobelow.Youwillbeaskedtouseyourknowledgeofphysicstopredicttheoutcomeofanexperiment.Then,closethevideowindow
andanswerthequestionatright.Youcanwatchthevideoagainatanypoint.
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Ch 07 HW
PartA
SupposeourexperimenterrepeatshisexperimentonaplanetmoremassivethanEarth,wheretheaccelerationduetogravityisg = 30
m/s
.When
hereleasestheballfromchinheightwithoutgivingitapush,howwilltheball'sbehaviordifferfromitsbehavioronEarth?Ignorefrictionandair
resistance.(Selectallthatapply.)
Hint1.Howtoapproachtheproblem
Toanswerthisquestion,youmustevaluatehowachangeingaffectsthependulumstrajectory(choicesAandB),period(choicesCandD),
andmass(choiceE).
Trajectory:Here,thekeyisenergyconservation.Thinkabouthowkineticandpotentialenergyinterchangeasapendulumswings.Forthe
pendulumtoreturntoapositionhigherthanitsstartingpoint,thesystemofthependulumandplanetwouldhavetogainmechanicalenergy.
Wherewouldthatenergycomefrom?(ThesameargumentappliestochoiceB.)
Period:Youcananswerthisquestionbyseeingwhethertheequationfortheperiodofapendulumincludesg.Oryoucanusereasoning.A
greaterdownwardaccelerationgwillcausethependulumtoreachthebottomofitsswingmorequickly.Whatdoesthatsayaboutthe
pendulumstotalperiod?
Mass:Thependulumwillweighmoreonthemassiveplanet.Willithavemoremass?
ANSWER:
Itwillsmashhisface.
Itsmasswillbegreater.
Itwilltakelesstimetoreturntothepointfromwhichitwasreleased.
Itwillstopwellshortofhisface.
Itwilltakemoretimetoreturntothepointfromwhichitwasreleased.
Correct
Thependulumwillswingbackandforthmorequickly(withashorterperiod)becauseitisoscillatinginastrongergravitationalfieldthanthaton
Earth.
Exercise7.32
Whilearooferisworkingonaroofthatslantsat37.0 abovethehorizontal,heaccidentallynudgeshis89.0Ntoolbox,causingittostartslidingdownward,
startingfromrest.
PartA
Ifitstarts4.10mfromtheloweredgeoftheroof,howfastwillthetoolboxbemovingjustasitreachestheedgeoftheroofifthekineticfrictionforceon
itis22.0N?
ANSWER:
v
= 5.34 m/s
Correct
Exercise7.35
Aforceparalleltothexaxisactsonaparticlemovingalongthexaxis.ThisforceproducesapotentialenergyU (x) givenbyU (x)
4
=1.44J/m .
= x
where
PartA
Whatistheforcewhentheparticleisatpositionx=0.810m?
ANSWER:
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Fx
Ch 07 HW
= 3.06 N
Correct
EnergyRequiredtoLiftaHeavyBox
Asyouaretryingtomoveaheavyboxofmassm,yourealizethatitistooheavyforyoutoliftbyyourself.Thereisnoonearoundtohelp,soyouattachan
idealpulleytotheboxandamasslessropetotheceiling,whichyouwraparoundthepulley.Youpullupontheropetoliftthebox.
Usegforthemagnitudeoftheaccelerationduetogravityandneglectfrictionforces.
PartA
Onceyouhavepulledhardenoughtostarttheboxmovingupward,whatisthemagnitudeF oftheupwardforceyoumustapplytotheropetostart
raisingtheboxwithconstantvelocity?
Expressthemagnitudeoftheforceintermsofm,themassofthebox.
Hint1.Whatforcemustbeappliedtotheboxtokeepitmovingataconstantspeed?
Onceyouhavepulledhardenoughtostarttheboxmovingupward,whatisthemagnitudeoftheforcethatthepulleymustexertontheboxso
thatitmovesataconstantspeed?
Expressyouranswerintermsofthemassofthebox.
ANSWER:
Fp
mg
Hint2.Whatforcedoesthepulleyexertonthebox?
IfyoutakethetensionintheropetobeT ,whatisFp ,themagnitudeofthenetupwardforcethatthepulleyexertsonthebox?
ExpressyouranswerintermsofT .
ANSWER:
Fp
2T
Hint3.Findthetensionintherope
FindthetensionintheropeintermsofF ,theforcewithwhichyouarepullingupward.
ANSWER:
T
Hint4.Puttingitalltogether
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Ch 07 HW
Onyourownorusingtheprevioushints,youshouldhavefoundequationsforhefollowing:
1. theforceneededtolifttheboxatconstantvelocity,intermsofitsmass,
2. therelationshipbetweentheforceontheboxduetothepulleyandthetensionintherope,and
3. therelationshipbetweentheforceappliedtotheropeandthetensionintherope.
Usetwooftheseequationstoeliminatetheforceappliedbythepulleyandthetensionintherope.Youshouldthenbeabletoexpresstheforce
appliedontheropeintermsofthemassofthebox.
ANSWER:
F
4.9m
Correct
PartB
Considerliftingaboxofmassmtoaheighthusingtwodifferentmethods:liftingtheboxdirectlyorliftingtheboxusingapulley(asintheprevious
part).
WhatisW d /W p ,theratiooftheworkdoneliftingtheboxdirectlytotheworkdoneliftingtheboxwithapulley?
Expresstherationumerically.
Hint1.Definitionofwork
Ineachcase,theamountofworkW youdoisequaltotheforceF youapplytimesthedistancedoverwhichyouapplytheforce:
W = Fd
Hint2.Ratiooftheforces
Whatistheratiooftheforceneededtolifttheboxdirectlytotheforceneededtolifttheboxusingthepulley?
Expressyouranswernumerically.
ANSWER:
Fd
Fp
= 2
Hint3.Ratioofthedistances
Whatistheratioofthedistanceoverwhichforceisappliedwhenliftingtheboxdirectlytothedistanceoverwhichforceisappliedwhenlifting
theboxwiththepulley?
Expresstheratioofdistancesnumerically.
Hint1.Findthedistancewhenusingthepulley
FindDp ,thedistanceoverwhichyoumustapplyforcewhenliftingtheboxusingthepulley.
Expressyouranswerintermsofh,thetotalheightthattheboxislifted.
Hint1.Pullingtheropeashortdistance
Usingthepully,imaginethatyoupulltheendoftheropeashortdistancedxupward.Theboxwillactuallyriseadistancedx/2.
(Drawapictureifyouhavetroublevisualizingthis.)
ANSWER:
Dp
2h
Hint2.Findthedistancewhenliftingdirectly
Dd
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Ch 07 HW
Whenliftingtheboxdirectly,thedistanceoverwhichforceisapplied,Dd ,isequaltotheverticaldistancehthattheboxisraised.
ANSWER:
Dd
Dp
= 0.500
ANSWER:
Wd
Wp
= 1
Correct
Nomatterwhichmethodyouusetoliftthebox,itsgravitationalpotentialenergywillincreasebymgh.So,neglectingfriction,youwillalwaysneed
todoanamountofworkequaltomghtoliftit.
PotentialEnergyGraphsandMotion
LearningGoal:
Tobeabletointerpretpotentialenergydiagramsandpredictthecorrespondingmotionofaparticle.
Potentialenergydiagramsforaparticleareusefulinpredictingthemotionofthatparticle.Thesediagramsallowonetodeterminethedirectionoftheforce
actingontheparticleatanypoint,thepointsofstableandunstableequilibrium,theparticle'skineticenergy,etc.
Considerthepotentialenergydiagramshown.Thecurverepresentsthevalueofpotentialenergy
U asafunctionoftheparticle'scoordinatex.Thehorizontallineabovethecurverepresentsthe
constantvalueofthetotalenergyoftheparticleE .ThetotalenergyE isthesumofkinetic(K )
andpotential(U )energiesoftheparticle.
Thekeyideaininterpretingthegraphcanbeexpressedintheequation
Fx (x) =
dU (x)
dx
Inansweringthefollowingquestions,wewillassumethatthereisasinglevaryingforceF actingontheparticlealongthexaxis.Therefore,wewillusethe
termforceinsteadofthecumbersomexcomponentofthenetforce.
PartA
TheforceactingontheparticleatpointAis__________.
Hint1.Signofthederivative
Ifafunctionincreases(asxincreases)inacertainregion,thenthederivativeofthefunctioninthatregionispositive.
Hint2.Signofthecomponent
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Ch 07 HW
Ifxincreasestotheright,asinthegraphshown,thena(onedimensional)vectorwithapositivexcomponentpointstotheright,andviceversa.
ANSWER:
directedtotheright
directedtotheleft
equaltozero
Correct
ConsiderthegraphintheregionofpointA.Iftheparticleismovingtotheright,itwouldbe"climbingthehill,"andtheforcewould"pullitdown,"
thatis,pulltheparticlebacktotheleft.Another,moreabstractwayofthinkingaboutthisistosaythattheslopeofthegraphatpointAispositive
therefore,thedirectionofF isnegative.
PartB
TheforceactingontheparticleatpointCis__________.
Hint1.Signofthederivative
Ifafunctionincreases(asxincreases)inacertainregion,thenthederivativeofthefunctioninthatregionispositive,andviceversa.
Hint2.Signofthecomponent
Ifxincreasestotheright,asinthegraphshown,thena(onedimensional)vectorwithapositivexcomponentpointstotheright,andviceversa.
ANSWER:
directedtotheright
directedtotheleft
equaltozero
Correct
PartC
TheforceactingontheparticleatpointBis__________.
Hint1.Derivativeofafunctionatalocalmaximum
Atalocalmaximum,thederivativeofafunctionisequaltozero.
ANSWER:
directedtotheright
directedtotheleft
equaltozero
Correct
Theslopeofthegraphiszerotherefore,thederivativedU /dx
= 0
,and|F | = 0 .
PartD
TheaccelerationoftheparticleatpointBis__________.
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Ch 07 HW
Hint1.Relationbetweenaccelerationandforce
TherelationbetweenaccelerationandforceisgivenbyNewton's2ndlaw,
F = ma
ANSWER:
directedtotheright
directedtotheleft
equaltozero
Correct
Ifthenetforceiszero,soistheacceleration.Theparticleissaidtobeinastateofequilibrium.
PartE
IftheparticleislocatedslightlytotheleftofpointB,itsaccelerationis__________.
Hint1.Theforceonsuchaparticle
TotheleftofB,U (x) isanincreasingfunctionandsoitsderivativeispositive.Thisimpliesthatthexcomponentoftheforceonaparticleatthis
locationisnegative,orthattheforceisdirectedtotheleft,justlikeatA.Whatcanyousaynowabouttheacceleration?
ANSWER:
directedtotheright
directedtotheleft
equaltozero
Correct
PartF
IftheparticleislocatedslightlytotherightofpointB,itsaccelerationis__________.
Hint1.Theforceonsuchaparticle
TotherightofB,U (x) isadecreasingfunctionandsoitsderivativeisnegative.Thisimpliesthatthexcomponentoftheforceonaparticleat
thislocationispositive,orthattheforceisdirectedtotheright,justlikeatC.Whatcanyounowsayabouttheacceleration?
ANSWER:
directedtotheright
directedtotheleft
equaltozero
Correct
Asyoucansee,smalldeviationsfromequilibriumatpointBcauseaforcethatacceleratestheparticlefurtherawayhencetheparticleisin
unstableequilibrium.
PartG
Namealllabeledpointsonthegraphcorrespondingtounstableequilibrium.
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Ch 07 HW
Listyourchoicesalphabetically,withnocommasorspacesforinstance,ifyouchoosepointsB,D,andE,typeyouranswerasBDE.
Hint1.Definitionofunstableequilibrium
Unstableequilibriummeansthatsmalldeviationsfromtheequilibriumpointcreateanetforcethatacceleratestheparticlefurtherawayfromthe
equilibriumpoint(thinkofaballontopofahill).
ANSWER:
BF
Correct
PartH
Namealllabeledpointsonthegraphcorrespondingtostableequilibrium.
Listyourchoicesalphabetically,withnocommasorspacesforinstance,ifyouchoosepointsB,D,andE,typeyouranswerasBDE.
Hint1.Definitionofstableequilibrium
Stableequilibriummeansthatsmalldeviationsfromtheequilibriumpointcreateanetforcethatacceleratestheparticlebacktowardthe
equilibriumpoint.(Thinkofaballrollingbetweentwohills.)
ANSWER:
DH
Correct
PartI
Namealllabeledpointsonthegraphwheretheaccelerationoftheparticleiszero.
Listyourchoicesalphabetically,withnocommasorspacesforinstance,ifyouchoosepointsB,D,andE,typeyouranswerasBDE.
Hint1.Relationbetweenaccelerationandforce
TherelationbetweenaccelerationandforceisgivenbyNewton's2ndlaw,
F = ma
ANSWER:
BDFH
Correct
Youranswer,ofcourse,includesthelocationsofbothstableandunstableequilibrium.
PartJ
Namealllabeledpointssuchthatwhenaparticleisreleasedfromrestthere,itwouldacceleratetotheleft.
Listyourchoicesalphabetically,withnocommasorspacesforinstance,ifyouchoosepointsB,D,andE,typeyouranswerasBDE.
Hint1.Determinethesignofthexcomponentofforce
Iftheaccelerationistotheleft,soistheforce.Thismeansthatthexcomponentoftheforceis__________.
ANSWER:
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Ch 07 HW
positive
negative
Hint2.WhatisthebehaviorofU (x) ?
Ifthexcomponentoftheforceatapointisnegative,thenthederivativeofU (x) atthatpointispositive.Thismeansthatintheregionaround
thepointU (x) is__________.
ANSWER:
increasing
decreasing
ANSWER:
AE
Correct
PartK
ConsiderpointsA,E,andG.Ofthesethreepoints,whichonecorrespondstothegreatestmagnitudeofaccelerationoftheparticle?
Hint1.Accelerationandforce
Thegreatestaccelerationcorrespondstothegreatestmagnitudeofthenetforce,representedonthegraphbythemagnitudeoftheslope.
ANSWER:
A
E
G
Correct
Kineticenergy
IfthetotalenergyE oftheparticleisknown,onecanalsousethegraphofU (t) todrawconclusionsaboutthekineticenergyoftheparticlesince
K = E U
Asareminder,onthisgraph,thetotalenergyE isshownbythehorizontalline.
PartL
Whatpointonthegraphcorrespondstothemaximumkineticenergyofthemovingparticle?
Hint1.K ,U ,andE
Sincethetotalenergydoesnotchange,themaximumkineticenergycorrespondstotheminimumpotentialenergy.
ANSWER:
D
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Ch 07 HW
Correct
Itmakessensethatthekineticenergyoftheparticleismaximumatoneofthe(force)equilibriumpoints.Forexample,thinkofapendulum(which
hasonlyoneforceequilibriumpointattheverybottom).
PartM
Atwhatpointonthegraphdoestheparticlehavethelowestspeed?
ANSWER:
B
Correct
Asyoucansee,manydifferentconclusionscanbemadeabouttheparticle'smotionmerelybylookingatthegraph.Itishelpfultounderstandthe
characterofmotionqualitativelybeforeyouattemptquantitativeproblems.Thisproblemshouldproveusefulinimprovingsuchanunderstanding.
LooptheLoop
Arollercoastercarmaybeapproximatedbyablockofmassm.Thecar,whichstartsfromrest,
isreleasedataheighthabovethegroundandslidesalongafrictionlesstrack.Thecar
encountersaloopofradiusR,asshown.Assumethattheinitialheighthisgreatenoughsothat
thecarneverlosescontactwiththetrack.
PartA
Findanexpressionforthekineticenergyofthecaratthetopoftheloop.
Expressthekineticenergyintermsofm,g,h,andR.
Hint1.Findthepotentialenergyatthetopoftheloop
Whatisthepotentialenergyofthecarwhenitisatthetopoftheloop?Definethegravitationalpotentialenergytobezeroath
= 0
ExpressyouranswerintermsofRandothergivenquantities.
ANSWER:
U top
mg(2R)
ANSWER:
K
mg(h 2R)
Correct
PartB
h
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Ch 07 HW
Findtheminimuminitialheighthatwhichthecarcanbereleasedthatstillallowsthecartostayincontactwiththetrackatthetopoftheloop.
ExpresstheminimumheightintermsofR.
Hint1.Howtoapproachthispart
Meaningof"stayincontact"
Forthecartojuststayincontactthroughtheloop,withoutfalling,thenormalforcethatactsonthecarwhenit'satthetopoftheloopmustbe
zero(i.e.,N = 0 ).
Findthevelocityatthetopsuchthattheremainingforceonthecari.e.itsweightprovidesthenecessarycentripetalacceleration.Ifthevelocity
wereanygreater,youwouldadditionallyrequiresomeforcefromthetracktoprovidethenecessarycentripetalacceleration.Ifthevelocitywere
anyless,thecarwouldfalloffthetrack.
Usetheabovedescribedconditiontofindthevelocityandthentheresultfromtheaboveparttofindtherequiredheight.
Hint2.Accelerationatthetopoftheloop
Assumingthatthespeedofthecaratthetopoftheloopisv top ,andthatthecarstaysonthetrack,findtheaccelerationofthecar.Takethe
positiveydirectiontobeupward.
Expressyouranswerintermsofv top andanyotherquantitiesgivenintheproblemintroduction.
ANSWER:
atop
vtop
Hint3.Normalforceatthetopoftheloop
Supposethecarstaysonthetrackandhasspeedv top atthetopoftheloop.UseNewton's2ndlawtofindanexpressionforN ,themagnitude
ofthenormalforcethattheloopexertsonthecarwhenthecarisatthetopoftheloop.
Expressyouranswerintermsofm,g,R,andv top .
Hint1.Findthesumofforcesatthetopoftheloop
Findthesumoftheforcesactingonthecaratthetopoftheloop.Rememberthatthepositiveydirectionisupward.
ExpressyouranswerintermsofN ,m,andg.
ANSWER:
F top
N mg
ANSWER:
m(
vtop
R
g)
Hint4.Solvingforh
Therequirementtostayincontactresultsinanexpressionforv 2top intermsofRandg.Substitutethisintoyourexpressionforkineticenergy,
foundinPartA,todeterminearelationbetweenhandR.
ANSWER:
hmin
2.5R
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Ch 07 HW
Correct
Forh
> 2.5 R
Forh
< R
ForR
thecarwillstillcompletetheloop,thoughitwillrequiresomenormalreactionevenattheverytop.
thecarwilljustoscillate.Doyouseethis?
,thecartwilllosecontactwiththetrackatsomeearlierpoint.Thatiswhyrollercoastersmusthavealotofsafetyfeatures.If
youlike,youcancheckthattheangleatwhichthecartlosescontactwiththetrackisgivenby = arcsin(
2
3
h
R
1)).Whereistheangle
measuredcounterclockwisefromthehorizontalpositivexaxis,wheretheoriginofthexaxisisatthecenteroftheloop.
DragonaSkydiver
Askydiverofmassmjumpsfromahotairballoonandfallsadistancedbeforereachingaterminalvelocityofmagnitudev .Assumethatthemagnitudeof
theaccelerationduetogravityisg.
PartA
WhatistheworkW d doneontheskydiver,overthedistanced,bythedragforceoftheair?
Expresstheworkintermsofd,v ,m,andthemagnitudeoftheaccelerationduetogravityg.
Hint1.Howtoapproachtheproblem
Ifnononconservativeforceswereacting,thenthetotalmechanicalenergy(kineticpluspotential)oftheskydiveruponleavingtheplanewould
beequaltothetotalmechanicalenergyoftheskydiverafterfallingadistanced.
Nowconsiderthedragforce,whichisnonconservative.Thedragforceopposesthemotionoftheskydiver,whichmeansthatitdoesnegative
workontheskydiver.Thus,thefinalmechanicalenergyoftheskydiverwillbesmallerthantheinitialmechanicalenergybyanamountequalto
theworkdonebythedragforce.
Hint2.Findthechangeinpotentialenergy
Findthechangeintheskydiver'sgravitationalpotentialenergy,afterfallingadistanced.
Expressyouranswerintermsofgivenquantities.
ANSWER:
U g
mgd
Hint3.Findthechangeinkineticenergy
Findthechangeintheskydiver'skineticenergy,afterfallingadistanced.
Expressyouranswerintermsofgivenquantities.
ANSWER:
K
mv
ANSWER:
Wd
mv
2
mgd
Correct
PartB
FindthepowerPd suppliedbythedragforceaftertheskydiverhasreachedterminalvelocityv .
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Ch 07 HW
Expressthepowerintermsofquantitiesgivenintheproblemintroduction.
Hint1.Howtoapproachtheproblem
Onewaytoapproachthisproblemwouldbetoapplythedefinitionofpowerasthetimederivativeoftheworkdone.Asimplermethodthatworks
forthisproblemistousetheformulaforthepowerdeliveredbyaforceF actingonanobjectmovingwithvelocityv :
P = F v
Hint2.Magnitudeofthedragforce
FindthemagnitudeFd ofthedragforceaftertheskydiverhasreachedterminalvelocity.
Expressyouranswerintermsoftheskydiver'smassmandothergivenquantities.
Hint1.Definitionofterminalvelocity
Onceterminalvelocityisreached,theskydiver'saccelerationgoestozero.(Thedragforceexactlybalancesthedownwardacceleration
duetogravity.)
ANSWER:
Fd
mg
Hint3.Relativedirectionofthedragforceandvelocity
WhenyoufindFd
,rememberthatthedirectionofthedragforceisoppositetothedirectionoftheskydiver'svelocity.
ANSWER:
Pd
mgv
Correct
CirclingBall
AballofmassmisattachedtoastringoflengthL.Itisbeingswunginaverticalcirclewithenoughspeedsothatthestringremainstautthroughoutthe
ball'smotion.Assumethattheballtravelsfreelyinthisverticalcirclewithnegligiblelossoftotal
mechanicalenergy.Atthetopandbottomoftheverticalcircle,theball'sspeedsarev t andv b ,
andthecorrespondingtensionsinthestringareT t andT b .T t andT b havemagnitudesTt and
Tb .
PartA
FindTb
Tt
,thedifferencebetweenthemagnitudeofthetensioninthestringatthebottomrelativetothatatthetopofthecircle.
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Ch 07 HW
Hint1.Howtoapproachthisproblem
Identifytheforcesthatactontheballasitmovesalongthecircularpath.Then,writeequationsforthesumoftheforcesontheballatthetop
andthebottomofthepath.Next,useNewton'ssecondlawtorelatethesenetforcestotheaccelerationoftheball.Noticethattheballdoesnot
movewithuniformspeedsotheaccelerationoftheballatthetopofthecircleisdifferentfromtheaccelerationatthebottomofthecircle.
Tofinishtheproblem,youmaywanttouseenergyconservationtorelatethespeedoftheballatthebottomofthecircletothespeedatthetop.
Hint2.Findthesumofforcesatthebottomofthecircle
Whatisthemagnitudeofthenetforceintheydirectionactingontheballatthebottomofthecircle?
Expressyouranswerintermsofthevariablesgivenintheproblem.Youmayusegtorepresenttheaccelerationofgravity,9.8m/s2 .
ANSWER:
| F y(bottom) |
Tb mg
Hint3.Findtheaccelerationatthebottomofthecircle
Findab ,themagnitudeoftheverticalaccelerationoftheballatthebottomofitscircle.
Expressyouranswerintermsofv b andpossiblyothergivenquantities.
ANSWER:
ab
vb
Hint4.Findthetensionatthebottomofthecircle
FindthemagnitudeofthetensionTb inthestringwhentheballisatthebottomofthecircle.
Expressyouranswerintermsofm,g,L,andthespeedv b oftheballatthebottomofthecircle.
Hint1.Whatphysicalprincipletouse
ApplyNewton's2ndlawintheydirectiontoobtainTb .
ANSWER:
2
Tb
m(g +
(v )
b
L
Hint5.Findthesumofforcesatthetopofthecircle
Whatisthemagnitudeofthenetforceintheydirectionactingontheballatthetopofitscircle?
Expressyouranswerintermsofthevariablesgivenintheproblem.Youmayusegtorepresenttheaccelerationofgravity,9.8m/s2 .
ANSWER:
| F y(top) |
Tt + mg
Hint6.Findtheaccelerationatthetopofthecircle
Findat ,themagnitudeoftheverticalaccelerationoftheballatthetopofitscircle.
Expressyouranswerintermsofv t andpossiblyothergivenquantities.
ANSWER:
at
vt
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Ch 07 HW
Hint7.Findthetensionatthetopofthecircle
FindthemagnitudeofthetensionTt inthestringwhentheballisatthetopofthecircle.
Expressyouranswerintermsofm,g,L,andthespeedv t oftheballatthetopofthecircle.
Hint1.RelationshiptosolutionforT b
FollowthesamestepsyouusedtofindTb (seeHint3),notingcarefullywherevariousdirections(signs)arereversed.
ANSWER:
Tt
m(
(vt )
L
g)
Hint8.Findtherelationshipbetweenvt andvb
Thetotalmechanicalenergyofthesystemisthesamewhentheballisatthetopandbottomoftheverticalcircle.Useconservationofenergyto
writeanexpressionforv 2t intermsofv 2b .
Youranswermayalsoincludem,g,andL.
ANSWER:
v
2
t
vb
4gL
ANSWER:
Tb Tt
6mg
Correct
Themethodoutlinedinthehintsisreallytheonlypracticalwaytodothisproblem.Ifdoneproperly,findingthedifferencebetweenthetensions,
Tb Tt ,canbeaccomplishedfairlysimplyandelegantly.
Problem7.72
Ifafishisattachedtoaverticalspringandslowlyloweredtoitsequilibriumposition,itisfoundtostretchthespringbyanamountd.
PartA
Ifthesamefishisattachedtotheendoftheunstretchedspringandthenallowedtofallfromrest,throughwhatmaximumdistancedoesitstretchthe
spring?(Hint:Calculatetheforceconstantofthespringintermsofthedistancedandthemassmofthefish.)
Expressyouranswerintermsofd.
ANSWER:
y
2d
Correct
Problem7.65
Inatruckloadingstationatapostoffice,asmall0.200kgpackageisreleasedfromrestatpointAonatrackthatisonequarterofacirclewithradius1.60
m(thefigure).Thesizeofthepackageismuchlessthan1.60m,sothepackagecanbetreatedasaparticle.ItslidesdownthetrackandreachespointB
withaspeedof4.10m/s.FrompointB,itslidesonalevelsurfaceadistanceof3.00mtopointC ,whereitcomestorest.
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Ch 07 HW
PartA
Whatisthecoefficientofkineticfrictiononthehorizontalsurface?
ANSWER:
= 0.286
Correct
PartB
HowmuchworkisdoneonthepackagebyfrictionasitslidesdownthecirculararcfromAtoB?
ANSWER:
W
= 1.46 J
Correct
ScoreSummary:
Yourscoreonthisassignmentis100%.
Youreceived14outofapossibletotalof14points.
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