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1、6.1:a)(2.40N)(1.5m)=3.60Jb)(-0.600N)(l50m)=-0-900Jc)3・60J—0・720J=2・70J.6.2:a)“Pullingslowly"canbetakentomeanthatthebucketrisesatconstantspeed,sothetensionintheropemaybetakentobethebucket^weight.Inpullingagivenlengthofrope,fromEq.(6.1),W=Fs=mgs=(6.75kg)(9.80m/s2)(4.00m)=264.6J.b)Gravityi
2、sdirectedoppositetothedirectionofthebucketsmotion,soEq.(6.2)givesthenegativeoftheresultofpart(a),or一265J.c)Thenetworkdoneonthebucketiszero.6.3:(25.0N)(12.0m)=300J.6.4:a)Thefrictionforcetobeovercomeisf=M=gng=(0.25)(30.0kg)(9.80m/s2)=73.5N,or74Ntotwofigures.b)FromEq.(6.1),Fs=(73.5N)(4.5m)
3、=331J.Theworkispositive,sincetheworkerispushinginthesamedirectionasthecrate'smotion.c)Since/andsareoppositelydirected,Eq.(6.2)gives-fi=一(73.5N)(4.5m)=-331J.d)Boththenormalforceandgravityactperpendiculartothedkectionofmotion,soneitherforcedoeswork,e)Thenetworkdoneiszero.6.5:a)SeeExercise
4、5.37.Theneededforceiscos0—“ksig(0.25)(30kg)(9.80m/s2)=Q92Ncos30°一(0.25)sin30°keepingextrafigures-b)Fscos(f)=(99.2N)(4.50m)cos30°=386.5J,againkeepinganextrafigure・c)Thenormalforceismg+Fsin0,andsotheworkdonebyfrictionis-(4.50m)(0.25)((30kg)(9.80m/s2)+(99.2N)sin30°)=-386.5J.d)Boththenormal
5、forceandgravityactperpendiculartothedirectionofmotion,soneitherforcedoeswork,b)Thenetworkdoneiszero.6.6:FromEq.(6.2),用cos©=(180N)(300m)cos15.0°=5.22xl04J.6.7:2Fscos=2(1.80x106N)(0.75x103m)cos14°=2.62x109J,or2.6xl09Jtotwoplaces.6.8:Theworkyoudois:Fs=((30N)f—(40N)j)•((—9.0m)；—(3.0m)j)=(30
6、N)(-9.0m)+(-40N)(-3.0m)=-270N•m+120N•m=-150J6.9:a)(i)Tensionforceisalwaysperpendiculartothedisplacementanddoesnowork.(ii)Workdonebygravityis一mg(y2一y{).Whenyx-y2.Wmg=0.b)(i)Tensiondoesnowork.(ii)LetIbethelengthofthestring.Wmg=-mg(y2-y})==-25.1JThedisplacementisupwardandthegravityforceisd
7、ownward,soitdoesnegativework.6.10:a)FromEq.(6.6),3.6km/h丿=1.54x10’J.K=-(1600kg)(50.0km/h)；Lb)Equation(6.5)givestheexplicitdependenceofkineticenergyonspeed;doublingthespeedofanyobjectincreasesthekineticenergybyafactoroffour.6.11:FortheT-Rex,K=y(7000kg)((4km/hr)=4.32x103J.Thepers
