Collectionofproblemssolvedinclassontheblackboard.

Chapter2

1.Anastronautstandsbytherimofacrateronthemoon,wheretheaccelerationofgravityis1.62m/s2.Todeterminethedepthofthecrater,shedropsarockandmeasuresthetimeittakesforittohitthebottom.Ifthedepthofthecrateris120m,howlongdoesittakefortherocktofall?

Solution

Wesolveitbymakinguseoftheformula:

y=y0+v0t+1/2at2

Chooseacoordinatesystem:wesety=0attherimofthecraterandchoosethepositiveydirectionpointingupwards.

Thetextoftheproblemtellsus:y0=0,v0=0,y=-120m,a=-g=-1.62m/s2.

Replacingintheequationwefind:

-(120m)=0+0-1/2*(1.62m/s2)t2.

Wesolvefort:t2=240/1.62s2àt=12.2s.

2.Aballisthrownupwardswithaspeedof16m/s.Howlongdoesittakeittoreachaheightof7.0monthewayup?

Solution

Wesolveitbymakinguseoftheformula:

y=y0+v0t+1/2at2

Chooseacoordinatesystem:wesety=0atthestartingpointoftheballandchoosethepositiveydirectionpointingupwards.

Thetextoftheproblemtellsus:y0=0,v0=16m/s,y=7.0m,a=-g=-9.81m/s2.

Replacingintheequationwefind:

7.0m=0+(16m/s)*t-1/2*(9.81m/s2)*t2.

Movingeverythingtothelefthandsideweget:

(4.9m/s2)t2-(16m/s)t+7m=0.

Solvingthequadraticequationwefindtwosolutions:t=0.52sandt=2.74s.Theyarethetimesatwhichtheballreachesaheightof7.0monthewayupanddown.Theproblemasksforthetimeatwhichtheballreaches7.0monthewayup:wechoosethesmallertime.Thesolutionisthereforet=0.52s.

Chapter3

3.Duringthesuperbowl,themiddlelinebackerfortheSeattleSeaHawksmadethefollowingmovementsaftertheballwassnappedonthirddown.First,heback-pedaledinthesoutherndirectionfor2.6meters.Hethenshuffledtohisleft(west)foradistanceof2.2meters.Finally,hemadeahalf-turnandrandownfieldadistanceof4.8metersinadirectionof240°counter-clockwisefromeast(30°WofS)beforefinallyknockingthewindoutofNewEnglandPatriots’widereceiver.Determinethemagnitudeanddirectionofhisoveralldisplacement.

Solution

Displacement=finalposition–initialposition.

Ichoosetheinitialpositiontobeattheoriginoftheaxes(x=0,y=0).Wetakethepositivexdirectiontobefromwesttoeastandthepositiveydirectionfromsouthtonorth,

Thefinalpositionisgivenbythesumofthreevectors,A,BandC.

Ax=0;Ay=-2.6m.

Bx=-2.2m;By=0.

Cx=(-4.8m)*Sin(30°)=-2.4mCy==(-4.8m)*Cos(30°)=-4.16m

Wesumthethreevectorscomponentbycomponent.WecallthefinalpositionoftheplayerE:

Ex=Ax+Bx+Cx=-2.2m-2.4m=-4.6m

Ey=Ay+By+Cy=-2.6m-4.16m=-6.76m

ThedisplacementDisthefinal–theinitialposition:Dx=Ex-0=-4.6;Dy=Ey-0=-6.76m.

Iwanttofindthemagnitudeandthedirectionofthedisplacement:

|D|2=(Dx)2+(Dy)2à|D|=8.2m; tgθ=Dy/Dx=1.47.θ=56°SouthofWest.

4.VectorAis5.5cmlongandpointsalongtheEast.VectorBis7.5cmlongandpointsat+30°NorthofWest.Determinethesumofthesetwovectorsintermsofmagnitudeanddirection.

A)2.0cmat15°abovethex-axis

B)3.9cmat75°abovethex-axis

C)7.8cmat33°abovethex-axis

D)13cmat17°abovethex-axis

E)7.5cmat30°abovethex-axis

Solution:B.

5.Aplaneisheadedeastwardataspeedof156m/srelativetothewind.A20.0m/swindisblowingsouthwardatthesametimeastheplaneisflying.Findthevelocityoftheplanerelativetotheground.

Solution

Weusetheequationforrelativemotion:

vPG=vPW+vWG

where:vPGisthevelocityoftheplanerelativetotheground,vPWisthevelocityoftheplanerelativetothewindandvWGisthevelocityofthewindrelativetotheground.

WefixthecoordinatesystemsuchthatthepositivexdirectiongoesfromWesttoEastantthepositiveydirectiongoesfromSouthtoNorth.

Fromthetextwehave:vPW=156m/sx,vWG=-20.0m/sy(weindicatewithxandytheunitvectorsinthepositivexandydirections).

Weget:

vPG=156m/sx-20.0m/sy.

Themagnitudecanbeobtainedthrough:|vPG|2=(156m/s)2+(20.0m/s)2

à|vPG|=157m/s.

Thedirectionisobtainedthrough:

tgθ=-20/157àθ=7.31°SouthofEast.

Chapter4

6.Atruckmovesfromthepoint–(2.0m)x+(6.0m)ytothepoint(3.0m)x-(5.0m)yin6.0s.Whatisthedirectionoftheaveragevelocityofthetruck?

Solution

Sincetheaveragevelocityisdefinedasthedisplacementovertime,itsdirectionwillbethesameasthedisplacement.

Thedisplacementisdefinedasfinalposition–initialposition.

Inourcasewehave:ri=–(2.0m)x+(6.0m)y

rf=(3.0m)x-(5.0m)y

Thedisplacementis:rf-ri=(3.0m+2.0m)x+(-5.0m-6.0m)y=(5.0m)x+(-11.0m)y.

Thedirectionofthedisplacementis:tgθ=-11.0/5.0àθ=65.6°SouthofEast.

7.Acatjumpsoffadeskwithahorizontalvelocityvm/s.Theheightofthedeskis1.4mandthehorizontalrangeofthecatfromthebaseofthedeskis3m.Whatistheinitialspeedofthecat?

Solution

Thisisthetypicalcaseofaprojectilemotionwithzerolaunchangle.

Wefixthecoordinatesystemsuchthattheoriginisatthebaseofthedeskandthatthepositivexdirectionisfromwesttoeastandthepositiveydirectionispointingupwards.

Wetakethegeneralequationsofmotioninthexandydirections:

x=x0+v0xt+1/2axt2

y=y0+v0yt+1/2ayt2

andwerememberthaninthecaseofprojectilewithzerolaunchanglewehave:

x0=0;v0x=vm/s(itiswhatweneedtofindtosolvetheproblem);ax=0.

v0y=0;ay=-g=-9.81m/s2.

Thereforetheaboveequationsreduceto:

x=v0xt

y=y0-1/2gt2

Thefinalverticalpositionofthecatisonthefloor:y=0.Theinitialoneisonthedesk:y0=1.4m.Thefinalhorizontalpositionofthecatisx=3m.

Isolvethesecondequationtofindthetimeittakesthecattoreachthefloor:

0=1.4m–(4.9m/s2)t2àt=0.534s.

Replacingitinthefirstequationwefind:

3m=v*0.534sàv=3m/0.534s=5.6m/s.

8.Akidthrowsastonefromthetopofa20.0mtalltower.Ithasaspeedof3.0m/sandislaunchedatanangleof35.0°abovethehorizontal.Howfarfromthebaseofthetowerdoesthestonehittheground?

Solution

Thisisacaseofaprojectilelaunchwithgeneralangle.

Wefixthecoordinatesystemwiththeoriginatthebaseofthetower,positiveydirectionpointingupwardsandpositivexdirectionfromwesttoeast.

Wetakethegeneralequationsofmotioninthexandydirections:

x=x0+v0xt+1/2axt2

y=y0+v0yt+1/2ayt2

andwerememberthaninthecaseofprojectilewithgenerallaunchanglewehave:

v0x=v0Cosθm/s;ax=0.

v0y=v0Sinθ;ay=-g=-9.81m/s2.

Thereforetheaboveequationsreduceto:

x=v0Cosθt

y=y0+v0Sinθt-1/2gt2

Fromthetextoftheproblemwehave:x0=0;v0=3.0m/s;θ=35°;y0=20.0m.

Fromthesecondequationwehave:

0=20.0m+(3.0m/s)Sin35°-1/2gt2à(4.9m/s2)t2-(1.72m/s)t-20.0m=0.

Solvingfortweget:t=2.2s.

Tofindthehorizontalpositionofthestoneasithitsthegroundwereplacetinthefirstequationandget:

x=(3m/s)Cos35°*2.2s=5.4m.

9.Thehorizontalandverticalcomponentsoftheinitialvelocityofafootballare16m/sand20m/s,respectively.Howlongdoesittakeforthefootballtorisetothehighestpointofitstrajectory?

Solution

Thisisanexampleofprojectilemotionwithgenerallaunchangle.Wearenotgiventheinitialangle,butwearegiventhehorizontalandverticalcomponentsoftheinitialvelocity.

Wefixthecoordinatesystemwiththeoriginatthestartingpointoftheball,positiveydirectionpointingupwardsandpositivexdirectionfromwesttoeast.

Wetakethegeneralequationsforthevelocityinthexandydirections:

vx=v0x+axt

vy=v0y+ayt

Wehave:ax=0.

ay=-g=-9.81m/s2.

Thereforetheaboveequationsreduceto:

vx=v0x=16m/s

vy=v0y-gt

Thehigherpointofthetrajectoryisthepointatwhichvy=0.

Weget,fromthesecondequation:

0=20m/s-(9.81m/s2)tàt=2.0s.

Chapter5

10.Twocarts,incontactwitheachother,areonatablewithnegligiblefriction.Aforceisappliedtocartnumber1.Iftheaccelerationofthesystemis2m/s2andthemassesofthecartsarem1=3kgandm2=2kg:

A.FindtheintensityoftheforceFandofthecontactonebetweenthecarts.

B.Findtheintensityofthecontactforcebetweenthecarts,iftheexternalforceactsoncartnumber2.

C.Findtheintensityoftheexternalforceandofthecontactforcebetweenthecarts1and2,ifathirdcartofmassm3=1Kgisaddedtothesystem,andtheexternalforceactsoncartnumber3.Theaccelerationisstill=2m/s2.

Solution

CaseA:

TheforceFactsoncart1.Cart1pushescart2withaninternalforceT.Bothcart1andcart2movestickingtogetherwithanaccelerationof2.0m/s2.Cart2pushesbackoncart1withaninternalforceequalinmagnitudetoTbutoppositeinsign.

Wetakethepositivexdirectionfromlefttoright.RememberNewton’ssecondlaw:thesumofforcesactingonanobjectisequaltoitsmasstimestheacceleration.Therefore:oncart1wehave

F-T=m1a. WeneedtofindFandT.Wehavetoomanyunknowntosolvethisequation.WeuseNewton’sequationforcart2:

T=m2a.

Sincem2=2kganda=2m/s2weget:T=4N.

Fromthefirstequationwehave:F=T+m1a=m2a+m1a=10N.

CaseB

TheforceFactsoncart2.Cart2pushescart1withaninternalforceT.Bothcart1andcart2movestickingtogetherwithanaccelerationof2.0m/s2.Cart1pushesbackoncart2withaninternalforceequalinmagnitudetoTbutoppositeinsign.

Wetakethepositivexdirectionfromrighttoleftinthiscase.RememberNewton’ssecondlaw:thesumofforcesactingonanobjectisequaltoitsmasstimestheacceleration.Therefore:oncart2wehave

F-T=m2a. WeneedtofindFandT.Wehavetoomanyunknowntosolvethisequation.WeuseNewton’sequationforcart1:

T=m1a.

Sincem1=3kganda=2m/s2weget:T=6N.

Fromthefirstequationwehave:F=T+m1a=m2a+m1a=10N.

So,theexternalforceactingonthetwocartsremainsthesameinbothcases.However,theinternalforceTbetweenthecartschangesdependingonwhethertheforceactsoncart1orcart2.

CaseC

TheforceFactsoncart3.Cart3pushescart2withaninternalforceT23.Cart2pushesbackoncart3withaninternalforceequalinmagnitudetoT23butoppositeinsign.Cart2pushescart1withaninternalforceT12.Cart1pushesbackoncart2withaninternalforceequalinmagnitudetoT12butoppositeinsign.Allthreecartsmovewithacceleration=2m/s2.

Wetakethepositivexdirectionfromrighttoleftinthiscase.RememberNewton’ssecondlaw:thesumofforcesactingonanobjectisequaltoitsmasstimestheacceleration.Therefore:oncart3wehave

F-T23=m3a. WeneedtofindFandT.Wehavetoomanyunknowntosolvethisequation.WeuseNewton’sequationforcart2:

T23-T12=m2a.

Againwedon’tknoweitherT23orT12.

WewriteNewton’sequationforcart1:

T12=m1a.

Sincem1=3kganda=2m/s2weget:T12=6N.

Fromthesecondequationwehave:T23=T12+m2a=(m1+m2)a=10N.

Fromthefirstequationwehave:F=T23+m3a=(m1+m2+m3)a=12N.

11.Thetwoforcesshowninthefigureactona3.00-kgobject.Whatistheaccelerationoftheobject?

Solution

WeapplyNewton’slawinthexandydirections:

ΣFx=max; ΣFy=may

ΣFx=4N+1N=5Nàax=Fx/m=1.67m/s2;ΣFy=4N-3N=1Nàay=Fy/m=0.33m/s2

12.A12-kgsuitcaseispulledbya45N-force,atanangleof20°abovehorizontal.Calculatethenormalforceandtheaccelerationofthesuitcase.

Solution

Wedrawthefreebodydiagram:

andapplyNewton’slawinthexandydirections:

Intheydirectionthereisnoacceleration:thesuitcaseonlymoveshorizontally.

ΣFy=0àF*Sin(20°)+N-mg=0

Fromtheaboveequationwecanimmediatelycalculatethenormalforce:

N=mg-F*Sin(20°)=(12*9.81-45*Sin(20°))N=103N.

WeapplyNewton’slawinthexdirection:

ΣFx=maxàFCos(20°)=maxàax=FCos(20°)/m=42.3N/12kg=3.5m/s2.

13.A70-kgskierslidesdownafrictionlessslopewhichformsa30°angleabovehorizontal.Calculatethenormalforceandtheaccelerationoftheskier.

Solution

Wedrawthefreebodydiagram.

Wechoosethecoordinatesystemwiththexaxisparalleltotheslopeandtheyaxisperpendiculartotheslope.

WeapplyNewton’slaw(θ=30°):

ΣFy=0àN-mgCosθ=0àN=mgCosθ=594N.

ΣFx=maxàmgSinθ=maxàax=gSinθ=4.91m/s2.

Chapter6

14.A250kgcrateisplacedonanadjustableinclinedplane.Ifthecrateslidesdowntheinclinewithanaccelerationof0.7m/s2whentheinclineangleis25°,thenwhatshouldtheinclineanglebeforthecratetoslidedowntheplaneatconstantspeed?

Solution

Wedrawthefreebodydiagram:

Theforcesactingonthecrateare:

-Theweight,whichhasanxandaycomponents:

Wx=mgSinθ; Wy=-mgCosθ

-ThenormalforceN,intheydirection.

-Thekineticfrictionalforce,inthenegativexdirection:

fK=-μKNx

Inthefirstpartoftheproblem,wehaveafixedanglebetweenthesurfaceandtheground.WeapplyNewton’slawinthexandydirections.Initially,thecratehasanaccelerationalongx.Weuseallthedatathatwehavetoextracttheoneunknownquantity:thecoefficientofkineticfrictionbetweenthesurfaceandthecrate.Thiscoefficientisacharacteristicofthematerialsoutofwhichthesurfaceandthecratearemade,anditwillstillbethesamewhentheanglebetweenthesurfaceandthegroundchanges.

ΣFy=0àN-mgCosθ=0àN=mgCosθ.

ΣFx=maxàmgSinθ-μKN=maxàmgSinθ-μKmgCosθ=max

Fromthesecondequation,wecanextractthecoefficientofkineticfrictionbetweenthecrateandthesurface.

μK=(gsin25°-ax)/(gcos25°)=0.39.

Theproblemasksustofindtheangleatwhichthecrateslidesdowntheslopewithconstantspeed.Thismeansthattheaccelerationalongxhastobe=0inthiscase:thefrictionalforcehastoperfectlybalancethecomponentofweightalongx.

ΣFx=0àmgSinθ-μKN=0 àmgSinθ-μKmgCosθ=0 àtgθ=μK àθ=21°.

15.Amanpullsa30.0kgcrateacrosslevelgroundatconstantvelocitywithalightropethatmakesanangleof20.0°abovehorizontal.Thetensionintheropeis40.0N.Whatisthecoefficientoffrictionbetweenthesledandtheground?

Solution

Wedrawthefreebodydiagram:

WeapplyNewton’slawsinthexandydirections.Noticethat,intheydirection,besidestheweightandthenormalforcewealsohavetheycomponentofF,thetensionoftherope:thenormalforceinthiscaseisnotequaltotheweight.

ΣFy=0àN-mg+Fsinθ=0 àN=mg-Fsinθ=281N.

Thevelocityisconstant:thereisnoaccelerationinthexdirection.

ΣFx=0àFCosθ-μKN=0 àμK=Fcosθ/N=0.134.

16.A20kghangingblockanda5.0kghangingblockareconnectedbyalightstringoveramassless,frictionlesspulley.Whatistheaccelerationofthesystemwhenreleased?

Solution

Wedrawthefreebodydiagramforthetwoblocksseparately.

WeapplyNewton’slawforeachblockseparately:

T-m1g=-m1a

T-m2g=m2a

Thetwoblockshavethesameacceleration,butblock1acceleratesdownwardswhileblock2acceleratesupwards.

Wesubtractequation1fromequation2aboveandfind:

(m1-m2)g=(m1+m2)a à a=(m1-m2)g/(m1+m2)=5.9m/s2.

17.Ablockofmassm1slidesonafrictionlesstabletop.Itisconnectedtoastringthatpassesoverapulleyandsuspendsamassm2.Iftheaccelerationofthesystemisg/3,whatistheratiom1/m2?

Solution

WeapplyNewton’slawinthexdirectionforthetwoblocksseparately.Noticethat,inthecaseofblock2,wehaverotatedthecoordinatesystemsuchthatnowthepositivexdirectionisvertical,pointingdownwards.

Wehave:

T=m1ax forblock1.

m2g-T=m2ax forblock2.

Addingthetwoequationswehave:

m2g=(m1+m2)ax

Thetexttellsusthatax=g/3.

m2g=(m1+m2)g/3àm1/m2+m2/m2=3àm1/m2=2.

18.Theexamplewiththecaronabankedroadwaycanbefoundonthetextbook(Example6-9onpage173).

19.Acarofmass800.0kgismovingwithaspeedof50.0m/satthetopofahillof200mheight.Whatisthetotalenergyofthecaratthetopofthehill?

Solution

Thetotalenergyofthecaristhesumofitspotentialandkineticenergy.Thepotentialoneisgivenby:

U=mgh=800.0kg*9.81m/s2*200m=1.57*106J

Thekineticoneis:K=1/2mv2=1/2*800.0kg*(50.0m/s)2=1*106J

Thesumofthetwois:

E=U+K=2.57*106J.

20.Underthesameconditions,whatisthespeedofthecarafteritrolleddowntothebottomofthehill?

Solution

Themechanicalenergyofthecarisconserved:atthebottomofthehilltheentiremechanicalenergywillbekineticenergy:

Ei=Ui+Kiàinitialmechanicalenergy,calculatedinthepreviousexercisetobe2.57*106J.

Atthebottomofthehillwehave:

Ef=Kf=1/2mvf2 à vf=80.2m/s.

21.A5.00-kgblockismovingalongahorizontalfrictionlesssurfacetowardanidealmasslessspringwithaspringconstantof45N/mthatisattachedtoawall.Aftertheblockcollideswiththespring,thespringiscompressedamaximumdistanceof0.68m.Howfastwastheblock?

Solution

Theforceofaspringisaconservativeforce:therefore,mechanicalenergyisconservedfortheblock.

Initially,theblockonlyhaskineticenergy.Attheend,itonlyhaspotentialenergy.

Weget:

Ei=Ki=1/2mvi2; Ef=1/2kx2 wherekisthespringconstantandxtheamountbywhichthespringiscompressed.

Weget:

Ei=Ef à 1/2mvi2=1/2kx2 à vi2=kx2/m àv=2.0m/s.

22.Anobjectofmass2.00kgstartsatrestfromthetopofaroughinclinedplaneofheight20.0m.Iftheworkdonebytheforceoffrictionis-150J,whatisthespeedoftheobjectasitreachesthebottomoftheinclinedplane?

Solution

Inthiscase,themechanicalenergyoftheobjectisnotconserved:themechanicalenergyatthebottomoftheinclinedplaneisgivenbytheinitialone,minustheworkdonebyfriction(non-conservativeforce).

Ef=Ei+Wnc.

Theinitialenergyisentirelypotentialenergy:

Ei=Ui=mgh=2.00kg*9.81m/s*20.0m=392J

Thefinalenergyisentirelykinetic:

Ef=Ei+Wnc=1/2mvf2à vf2=2/m*(Ei+Wnc) àvf=15.6m/s.

23.Acarofmass1000.0kgismovingwithaninitialspeedof20.0m/satthehighestpointofa100mlongrampinclinedat20degreestothehorizontal.Whatisthespeedofthecarwhenitreachesthebottomoftheramp?

Solution

Theheightoftherampisgivenby:h=Lsinθ=34.2m.

Atthehighestpointoftheramp,themechanicalenergyofthecaristhesumofpotentialandkineticenergies.Sinceonlygravityactsonthecar(frictioncanbeneglected),themechanicalenergyofthecarisconserved.

Wehave:

Ef=Ei

Where: Ei=Ui+Ki=mgh+1/2mvi2=5.4*105J.

Atthebottomoftheramp,theentiremechanicalenergyisgivenbythekineticenergy:

Ef=Kf=1/2mvf2=5.4*105J à vf=32.7m/s.

24.If3.8Jofworkisdoneinraisinga179gapple,howfarisitlifted?

Solution

Theworkdoneinraisingtheappleisgivenbyitsweighttimestheheightbywhichitislifted.Noticethatthemassoftheappleisgiveningramsandweneedtoconvertittokgfirst.Wehave:

m=179g=0.179kg.

W=mgh=3.8Jà h=W/(mg)=2.17m.

Chapter11

25.An85-kgpersonstandsona lightweight ladder,asshown.The floor is rough;hence,itexertsbothanormalforcef1andafrictionalforcef2ontheladder.Thewallis frictionless. It exerts only a normal force f3. Using thedimensions given in thefigure,findthemagnitudesoff1,f2andf3.

Solution

Wesetboththenetforceandthenettorqueequaltozero.Inthiscase,thenetforcemustbesetequaltozeroinboththexandydirections.

KnownMassofperson,m=85kg;verticaldistance,a=3.8m;horizontaldistance,b=0.70m.Noticethataistheleverarmfortheforcef3andbistheleverarmfortheweightoftheperson.

UnknownForcesexertedonladder:ƒ1=?;ƒ2=?;ƒ3=?

1.Setthenettorqueactingontheladderequaltozero.Usethebottomoftheladderastheaxis:

τnet=ƒ3a -mgb=0Wemultiplyeachforcebythecorrespondingleverarm.Noticethatthetorqueduetotheweightofthepersonisnegativebecauseitwouldmaketheladderrotatecounterclockwise.

2.Solveforƒ3:ƒ3=mgb/a=150N

3.Sumthexcomponentsofforcesandsetequaltozero:f2-f3=0

4.Solveforf2:f2=f3=150N

5.Sumtheycomponentsofforceandsetequaltozero:ƒ1-mg=0

6.Solveforƒ1:

ƒ1=mg=830N

INSIGHT

If the floor is quite smooth, the ladder might slip—it depends on whether thecoefficientofstaticfrictionisgreatenoughtoprovidetheneededforceƒ2=150N.Inthissystem,thenormalforceexertedbythefloorisN=ƒ1=830N.Therefore,ifthecoefficientofstaticfrictionisgreaterthan0.18[notingthat0.181830N2=150N], the ladderwill stayput.Laddersoftenhave rubberizedpadson thebottom inordertoincreasethestaticfriction,andhenceincreasethesafetyoftheladder.

26.

Achildofmassmissupportedonalightplankbyhisparents,whoexerttheforcesF1andF2asindicatedinthesketch.Findtheforcesrequiredtokeeptheplankinstaticequilibrium.Usetherightendoftheplankastheaxisofrotation.

Solution

Wesetboththenetforceandthenettorqueequaltozero.Theexpressionforthenettorqueisdifferentthistime,duetothenewlocationfortheaxisofrotation,butthefinalresultfortheforceswillbethesame.

Known:Massofchild,m;distanceforforce1,L;distanceforforce2,0;distanceforweight(mg)ofchild,L/4.

UnknownF1=?;F2=?

1. Setthenetforceactingontheplankequaltozero:F1+F2-mg=0

2. Setthenettorqueactingontheplankequaltozero.[NoticethattheforceF1rotates the plank clockwise (negative torque), and the weight mg rotates itcounterclockwise(positivetorque)abouttheaxis.]:-F1L+mgL/4=0

3. SolvethetorqueconditionfortheforceF1:F1=1/4mg

4. SubstituteF1intotheforceconditiontofindF2:F2=mg-1/4mg=3/4mg.

27. A 34.0-kg child runs with a speed of 2.80 m/s tangential to the rim of astationarymerry-go-round.Themerry-go-roundhasamomentofinertiaof512kgm2 and a radius of 2.31 m. When the child jumps onto the merry-go-round, theentire system begins to rotate. What is the angular speed of the system? Ignorefrictionandanyothertypeofexternaltorque.

Solution

Angular momentum is conserved in this system, so we set the initial angularmomentum equal to the final angularmomentum to find the final angular speed.Theinitialangularmomentumistheangularmomentumoftherunningchild.Thefinalangularmomentumisduetotherotationofthemerry-go-roundandthechildabouttheaxisofrotation.

1. Writetheinitialangularmomentumofthechild:Li=rmv2. Writethefinalangularmomentumofthesystem,notingthatthemomentof

inertiaof themerry-go-round is Iand themomentof inertiaof the child ismr2:Lf=(I+mr2)ω

3. SetLf=Liandsolvefortheangularspeed:ω=rmv/(I+mr2)=0.317rad/s.

Chapter12.

28. Planet A has acceleration due to gravity 6 times that of Planet B andmass 3timesthatofplanetB.IfplanetBhasradiusr,whatistheradiusofplanetA?

Solution.

TheaccelerationduetogravityonplanetAis:

aA=GmA/(rA)2

whileforplanetBwehave:

aB=GmB/(rB)2

WehavethataA=6aBandmA=3mB,andthatrB=r

aA=GmA/(rA)2=G3mB/(rA)2=6aB=6GmB/(r)2

Thus,

G3mB/(rA)2=6GmB/(r)2fromwhich:

1/(rA)2=2/(r)2

andwecandeducerA=r/sqrt(2).

29.ShowthatthegravitationalaccelerationontheMoonisabout1/6oftheoneonEarth.

Solution.

WeusethefollowingdatafortheEarth:

mE=5.972×1024kg

rE=6370km

andwefind:gE=GmE/(rE)2=9.81m/s2.

NextwecalculatetheonefortheMoon:

mM=7.35×1022kg

rM=1737km

andwefind:gM=GmM/(rM)2=1.62m/s2.Wethusget:

gE/gM~6.

Chapter13.

30. The position of a mass that is oscillating on a spring is x=(18.3cm)cos[(2.35rad/s)t].Whatisthemaximumspeedofthemass?

Solution.

Welookatthegenericexpressionforthepositioninasimpleharmonicmotionasafunctionoftime:

x(t)=Acos(ωt)

andbycomparisonwiththeaboveexpressionweget

A=18.3cm.

ω=2.35rad/s.

Wenowlookattheexpressionofthevelocityasafunctionoftime:

v=-Aωsin(ωt).

Themaximumvelocitycorrespondstosin(ωt)=1,forwhichwehavevMAX=Aω.

Fromourdataweget:vMAX=43cm/s.