physics1课后习题答案3.pdf

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1、E13-1Iftheprojectilehadnotexperiencedairdragitwouldhaverisentoaheighty2,butbecauseofairdrag68kJofmechanicalenergywasdissipatedsoitonlyrosetoaheighty1.Ineithercasetheinitialvelocity,andhenceinitialkineticenergy,wasthesame;andthevelocityatthehighestpointwaszero.ThenW=
U,sothepotentialenergywo

2、uldhavebeen68kJgreater,and
y=
U=mg=(68103J)=(9:4kg)(9:81m=s2)=740mishowmuchhigheritwouldhavegonewithoutairfriction.E13-2(a)Theroadinclineis=arctan(0:08)=4:57.Thefrictionalforcesarethesame;thecarisnowmovingwithaverticalupwardspeedof(15m=s)sin(4:57)=1:20m=s.Theadditionalpowerrequiredtodri

3、veupthehillisthen
P=mgv=(1700kg)(9:81m=s2)(1:20m=s)=20000W.Theytotalpowerrequiredis36000W.(b)Thecarwillcoast"ifthepowergeneratedbyrollingdownhillisequalto16000W,orv=(16000W)=[(1700kg)(9:81m=s2)]=0:959m=s;ydown.Thentheinclineis=arcsin(0:959m=s=15m=s)=3:67:Thiscorrespondstoadownwardgradeof

4、tan(3:67)=6:4%.E13-3Applyenergyconservation:1212mv+mgy+ky=0;22sopv=2(9:81m=s2)(0:084m)(262N=m)(0:084m)2=(1:25kg)=0:41m=s:E13-4Thecarclimbsaverticaldistanceof(225m)sin(10)=39:1mincomingtoastop.Thechangeinenergyofthecaristhen1(16400N)25
E=(31:4m=s)+(16400N)(39:1m)=1:8310J:2(9:81m=s2)

5、E13-5(a)Applyingconservationofenergytothepointswheretheballwasdroppedandwhereitenteredtheoil,1212mvf+mgyf=mvi+mgyi;221212vf+g(0)=(0)+gyi;22pvf=2gyi;p=2(9:81m=s2)(0:76m)=3:9m=s:(b)Thechangeininternalenergyoftheball+oilcanbefoundbyconsideringthepointswheretheballwasreleasedandwheretheballreac

6、hedthebottomofthecontainer.
E=Kf+UfKiUi;1212=mvf+mgyfm(0)mgyi;2213232=(12:210kg)(1:48m=s)(12:210kg)(9:81m=s)(0:55m0:76m);2=0:143J157E13-6(a)U=(25:3kg)(9:81m=s2)(12:2m)=3030J.i(b)K=1(25:3kg)(5:56m=s)2=391J:f2(c)
Eint=3030J391J=2640J.E13-7(a)Atatmosphericentrythekineticenergyis14

7、3212K=(7:910kg)(8:010m=s)=2:510J:2ThegravitationalpotentialenergyisU=(7:9104kg)(9:8m=s2)(1:6105m)=1:21011J:Thetotalenergyis2:61012J.(b)Attouchdownthekineticenergyis14128K=(7:910kg)(9:810m=s)=3:810J:2E13-8
E=
t=(68kg)(9:8m=s2)(59m=s)=39000J=s.E1