A2Physics物理出國英語Work,EnergyandPowerQP

1.Acarengineappliesaforceof5000Ntomoveacaradistanceof800m.Howmuchworkisdonebytheengine?

Answer:

Theworkdone(W)byaforce(F)overadisplacement(s)isgivenbytheformula(W=Ftimess).

Here,(F=5000N)and(s=800m).

So,(W=5000times800=4times10^{6}J).

2.Aforceof200Nisappliedatanangleof30°tothehorizontaltomoveaboxadistanceof15malongthehorizontalsurface.Calculatetheworkdone.

Answer:

Theworkdoneformulawhentheforceisappliedatanangle(theta)tothedirectionofdisplacementis(W=Ftimesstimescostheta).

Given(F=200N),(s=15m)and(theta=30^{circ}),(cos30^{circ}=frac{sqrt{3}}{2}approx0.866).

(W=200times15times0.866=2598Japprox2600J).

3.Amanliftsa50kgmassverticallyupwardsthroughaheightof2m.Howmuchworkdoeshedoagainstgravity?(Take(g=9.8m/s^{2}))

Answer:

Theforcerequiredtoliftthemassagainstgravityisequaltotheweightofthemass,(F=mg).

Here,(m=50kg)and(g=9.8m/s^{2}),so(F=50times9.8=490N).

Theworkdone(W=Ftimess),where(s=2m).

So,(W=490times2=980J).

4.Aspringhasaspringconstant(k=100N/m).Itisstretchedfromitsequilibriumpositionby0.2m.Calculatetheworkdoneinstretchingthespring.

Answer:

Theworkdoneinstretchingaspringisgivenby(W=frac{1}{2}kx^{2}),where(k)isthespringconstantand(x)istheextension.

Given(k=100N/m)and(x=0.2m).

(W=frac{1}{2}times100times(0.2)^{2}=2J).

5.Acyclistpedalsabicyclewithaforceof120N.Thebicyclemovesadistanceof200m.Ifthefrictionalforceactingonthebicycleis30N,whatisthenetworkdoneonthebicycle?

Answer:

Thenetforce(F_{net})actingonthebicycleis(FF_f),where(F=120N)istheappliedforceand(F_f=30N)isthefrictionalforce.

(F_{net}=12030=90N).

Theworkdone(W=F_{net}timess),with(s=200m).

So,(W=90times200=18000J).

6.Ablockispushedupaninclinedplane.Theforceappliedparalleltotheinclineis80Nandtheblockmovesadistanceof6malongtheincline.Calculatetheworkdonebytheappliedforce.

Answer:

Usingtheworkdoneformula(W=Ftimess),where(F=80N)and(s=6m).

So,(W=80times6=480J).

7.Acraneliftsaloadof1500kgtoaheightof10m.Calculatetheworkdonebythecrane.(Take(g=9.8m/s^{2}))

Answer:

Theforcerequiredtolifttheloadis(F=mg),where(m=1500kg)and(g=9.8m/s^{2}).

(F=1500times9.8=14700N).

Theworkdone(W=Ftimess),with(s=10m).

So,(W=14700times10=1.47times10^{5}J).

8.Aforceof40Nisappliedtoanobject,andtheobjectmovesinthedirectionoftheforce.Ifthepoweroutputis80W,howlongdoesittakefortheobjecttomoveadistanceof10m?

Answer:

First,calculatetheworkdone(W=Ftimess),where(F=40N)and(s=10m),so(W=40times10=400J).

Power(P=frac{W}{t}),so(t=frac{W}{P}).

Given(P=80W)and(W=400J),then(t=frac{400}{80}=5s).

9.Amachinedoes5000Jofworkin20s.Whatisthepowerofthemachine?

Answer:

Thepower(P)isgivenbytheformula(P=frac{W}{t}),where(W=5000J)and(t=20s).

So,(P=frac{5000}{20}=250W).

10.Acarenginehasapoweroutputof120kW.Howmuchworkdoesitdoin5minutes?

Answer:

First,convertthetimetoseconds.(t=5minutes=5times60=300s).

Power(P=frac{W}{t}),so(W=Ptimest).

Given(P=120times10^{3}W)and(t=300s).

(W=120times10^{3}times300=3.6times10^{7}J).

11.Apersonofmass60kgclimbsaflightofstairsofheight4min8s.Calculatethepoweroutputoftheperson.(Take(g=9.8m/s^{2}))

Answer:

Theforcerequiredtoclimbthestairsisequaltotheweightoftheperson,(F=mg).

Here,(m=60kg)and(g=9.8m/s^{2}),so(F=60times9.8=588N).

Theworkdone(W=Ftimess),where(s=4m),so(W=588times4=2352J).

Power(P=frac{W}{t}),with(t=8s).

So,(P=frac{2352}{8}=294W).

12.Apumpisusedtoliftwaterfromawell.Itcanlift200kgofwatertoaheightof10min40s.Calculatethepowerofthepump.(Take(g=9.8m/s^{2}))

Answer:

Theforcerequiredtoliftthewateris(F=mg),where(m=200kg)and(g=9.8m/s^{2}).

(F=200times9.8=1960N).

Theworkdone(W=Ftimess),with(s=10m),so(W=1960times10=19600J).

Power(P=frac{W}{t}),where(t=40s).

So,(P=frac{19600}{40}=490W).

13.Amotorhasapowerof300W.Itisusedtoliftamass.Ifitcanliftthemasstoaheightof5min10s,whatisthemassoftheobject?(Take(g=9.8m/s^{2}))

Answer:

Power(P=frac{W}{t}),so(W=Ptimest).

Given(P=300W)and(t=10s),then(W=300times10=3000J).

Theworkdoneinliftingthemassis(W=mgh),where(h=5m)and(g=9.8m/s^{2}).

(m=frac{W}{gh}=frac{3000}{9.8times5}approx61.2kg).

14.Aforceof50Nisappliedtoanobjectmovingwithavelocityof2m/s.Whatisthepowerexertedbytheforce?

Answer:

Thepower(P)intermsofforce(F)andvelocity(v)isgivenby(P=Ftimesv).

Given(F=50N)and(v=2m/s).

So,(P=50times2=100W).

15.Acarismovingwithaconstantvelocityof30m/s.Theengineofthecarprovidesaforceof2000N.Whatisthepoweroutputoftheengine?

Answer:

Usingtheformula(P=Ftimesv),where(F=2000N)and(v=30m/s).

So,(P=2000times30=60000W=60kW).

16.Acyclistismovingataspeedof12m/s.Thetotalresistiveforceactingonthecyclistandthebicycleis40N.Whatisthepowerrequiredtomaintainthisspeed?

Answer:

Power(P=Ftimesv),where(F=40N)istheresistiveforceand(v=12m/s).

So,(P=40times12=480W).

17.Ablockofmass2kgisinitiallyatrest.Aforceof10Nisappliedtotheblockfor5s.Calculatetheworkdoneontheblock.

Answer:

First,findtheacceleration(a)usingNewton'ssecondlaw(F=ma).

(a=frac{F}{m}=frac{10}{2}=5m/s^{2}).

Usingtheequationofmotion(s=ut+frac{1}{2}at^{2}),with(u=0m/s),(a=5m/s^{2})and(t=5s).

(s=0timest+frac{1}{2}times5times5^{2}=frac{1}{2}times5times25=62.5m).

Workdone(W=Ftimess),with(F=10N)and(s=62.5m).

So,(W=10times62.5=625J).

18.Aballofmass0.5kgisthrownverticallyupwardswithaninitialvelocityof20m/s.Calculatethemaximumheightitreaches.(Take(g=9.8m/s^{2}))

Answer:

Atthemaximumheight,thefinalvelocity(v=0m/s).

Usingtheequation(v^{2}u^{2}=2gh)(negativebecausetheaccelerationduetogravityactsintheoppositedirectionofmotion).

(0(20)^{2}=2times9.8timesh).

(400=19.6h).

(h=frac{400}{19.6}approx20.4m).

19.Arollercoastercarofmass800kgstartsfromrestatthetopofahillofheight30m.Calculateitsspeedatthebottomofthehill,assumingnoenergylosses.(Take(g=9.8m/s^{2}))

Answer:

Usingtheconservationofmechanicalenergy.Theinitialpotentialenergy(U=mgh)isconvertedintokineticenergy(K=frac{1}{2}mv^{2})atthebottomofthehill.

(mgh=frac{1}{2}mv^{2}).

Canceloutthemass(m)frombothsidesoftheequation.

(v=sqrt{2gh}),with(h=30m)and(g=9.8m/s^{2}).

(v=sqrt{2times9.8times30}=sqrt{588}approx24.2m/s).

20.Apendulumbobofmass0.2kgispulledtoonesideandreleasedfromaheightof0.1maboveitslowestpoint.Calculateitsspeedatthelowestpoint.(Take(g=9.8m/s^{2}))

Answer:

Bytheconservationofmechanicalenergy,theinitialpotentialenergy(U=mgh)isconvertedintokineticenergy(K=frac{1}{2}mv^{2})atthelowestpoint.

(mgh=frac{1}{2}mv^{2}).

Cancelout(m)frombothsides.

(v=sqrt{2gh}),with(h=0.1m)and(g=9.8m/s^{2}).

(v=sqrt{2times9.8times0.1}=sqrt{1.96}=1.4m/s).

21.Aspringloadedgunfiresa0.05kgbullet.Thespringhasaspringconstant(k=500N/m)andiscompressedby0.1m.Calculatethespeedofthebulletasitleavesthegun,assumingnoenergylosses.

Answer:

Theelasticpotentialenergyofthespring(U=frac{1}{2}kx^{2})isconvertedintothekineticenergyofthebullet(K=frac{1}{2}mv^{2}).

(frac{1}{2}kx^{2}=frac{1}{2}mv^{2}).

(v=sqrt{frac{k}{m}x^{2}}),with(k=500N/m),(m=0.05kg)and(x=0.1m).

(v=sqrt{frac{500}{0.05}times(0.1)^{2}}=sqrt{100}=10m/s).

22.A1kgballisdroppedfromaheightof5m.Justbeforehittingtheground,whatisitskineticenergy?(Take(g=9.8m/s^{2}))

Answer:

Bytheconservationofmechanicalenergy,theinitialpotentialenergy(U=mgh)isconvertedintokineticenergy(K)justbeforehittingtheground.

(U=mgh),with(m=1kg),(h=5m)and(g=9.8m/s^{2}).

(K=U=1times9.8times5=49J).

23.Acarofmass1200kgismovingataspeedof20m/s.Calculateitskineticenergy.

Answer:

Thekineticenergy(K=frac{1}{2}mv^{2}),where(m=1200kg)and(v=20m/s).

(K=frac{1}{2}times1200times20^{2}=240000J=240kJ).

24.Ablockofmass3kgismovingwithakineticenergyof24J.Whatisitsspeed?

Answer:

Thekineticenergyformulais(K=frac{1}{2}mv^{2}).

(v=sqrt{frac{2K}{m}}),with(K=24J)and(m=3kg).

(v=sqrt{frac{2times24}{3}}=sqrt{16}=4m/s).

25.Aballofmass0.1kgisthrownverticallyupwardswithaninitialkineticenergyof5J.Howhighwillitrise?(Take(g=9.8m/s^{2}))

Answer:

Atthemaximumheight,thekineticenergyisconvertedintopotentialenergy.

(K=U=mgh).

(h=frac{K}{mg}),with(K=5J),(m=0.1kg)and(g=9.8m/s^{2}).

(h=frac{5}{0.1times9.8}approx5.1m).

26.A5kgblockisslidingonafrictionlesssurfacewithaspeedof6m/s.Itcollideswithaspringandcompressesthespring.Ifthespringconstant(k=1000N/m),whatisthemaximumcompressionofthespring?

Answer:

Theinitialkineticenergyoftheblock(frac{1}{2}mv^{2})isconvertedintotheelasticpotentialenergyofthespring(frac{1}{2}kx^{2})atmaximumcompression.

(frac{1}{2}mv^{2}=frac{1}{2}kx^{2}).

(x=sqrt{frac{m}{k}v^{2}}),with(m=5kg),(k=1000N/m)and(v=6m/s).

(x=sqrt{frac{5}{1000}times6^{2}}=sqrt{frac{5times36}{1000}}=sqrt{0.18}approx0.42m).

27.Acarofmass1500kgisclimbingahillataconstantspeedof15m/s.Theslopeofthehillissuchthatforevery10malongtheroad,thecargainsaheightof1m.Calculatethepowerrequiredtoclimbthehill,assumingnoenergylossesduetofriction.(Take(g=9.8m/s^{2}))

Answer:

Theforcerequiredtomovethecarupthehillataconstantspeedisequaltothecomponentoftheweightofthecaralongtheslope.

Iftheslopehasaratioofheighttodistancealongtheslope(h:s=1:10),then(sintheta=frac{1}{10}).

Theforce(F=mgsintheta),with(m=1500kg)and(g=9.8m/s^{2}).

(F=1500times9.8timesfrac{1}{10}=1470N).

Power(P=Ftimesv),with(v=15m/s).

So,(P=1470times15=22050W=22.05kW).

28.Amachinehasanefficiencyof80%.Itdoes4000Jofusefulwork.Howmuchenergyisinputintothemachine?

Answer:

Efficiency(eta=frac{W_{useful}}{W_{input}}).

(W_{input}=frac{W_{useful}}{eta}),with(W_{useful}=4000J)and(eta=0.8).

So,(W_{input}=frac{4000}{0.8}=5000J).

29.Anelectricmotorhasaninputpowerof1500Wandanoutputpowerof1200W.Whatisitsefficiency?

Answer:

Efficiency(eta=frac{P_{output}}{P_{input}}).

(eta=frac{1200}{1500}=0.8=80%).

30.Apumpisusedtoliftwater.Ithasanefficiencyof70%.Iftheusefulworkdoneinliftingthewateris3500J,howmuchenergyiswastedintheprocess?

Answer:

First,findtheinputenergy(W_{input})using(eta=frac{W_{useful}}{W_{input}}).

(W_{input}=frac{W_{useful}}{eta}),with(W_{useful}=3500J)and(eta=0.7).

(W_{input}=frac{3500}{0.7}=5000J).

Theenergywasted(W_{wasted}=W_{input}W_{useful}).

(W_{wasted}=50003500=1500J).

31.Aforceof60Nisappliedtoanobjectatanangleof45°tothedirectionofmotion.Theobjectmovesadistanceof12m.Calculatetheworkdonebytheforce.

Answer:

Usingtheformula(W=Ftimesstimescostheta),where(F=60N),(s=12m)and(theta=45^{circ}),(cos45^{circ}=frac{sqrt{2}}{2}approx0.707).

(W=60times12times0.707=509.04Japprox509J).

32.Ablockofmass4kgispushedupaninclinedplanewithaforceof50Nparalleltotheincline.Theinclineis5mlongandtheblockgainsaheightof3m.Calculatetheworkdonebytheforceandtheworkdoneagainstgravity.

Answer:

Workdonebytheforce(W_{F}=Ftimess),with(F=50N)and(s=5m),so(W_{F}=50times5=250J).

Workdoneagainstgravity(W_{g}=mgh),with(m=4kg),(g=9.8m/s^{2})and(h=3m).

(W_{g}=4times9.8times3=117.6J).

33.Aspringhasanaturallengthof0.2m.Whenaforceof10Nisapplied,itstretchestoalengthof0.25m.Calculatethespringconstant.

Answer:

Theextension(x=0.250.2=0.05m).

UsingHooke'slaw(F=kx),then(k=frac{F}{x}).

Given(F=10N)and(x=0.05m),so(k=frac{10}{0.05}=200N/m).

34.Acarengineprovidesapowerof80kW.Ifthecarismovingataspeedof25m/s,whatistheforceexertedbytheengine?

Answer:

Usingtheformula(P=Ftimesv),then(F=frac{P}{v}).

Given(P=80times10^{3}W)and(v=25m/s).

(F=frac{80times10^{3}}{25}=3200N).

35.Aballofmass0.2kgisdroppedfromaheightof8m.Justbeforehittingtheground,whatisitsmomentum?(Take(g=9.8m/s^{2}))

Answer:

First,findthevelocity(v)justbeforehittingthegroundusing(v^{2}=u^{2}+2gh),with(u=0m/s),(g=9.8m/s^{2})and(h=8m).

(v=sqrt{2gh}=sqrt{2times9.8times8}=sqrt{156.8}approx12.5m/s).

Momentum(p=mv),with(m=0.2kg)and(vapprox12.5m/s).

(p=0.2times12.5=2.5kgcdotm/s).

36.Ablockofmass3kgismovingwithavelocityof4m/s.Itcollideswithastationaryblockofmass2kg.Afterthecollision,thetwoblockssticktogether.Whatistheircommonvelocity?

Answer:

Usingtheprincipleofconservationofmomentum(m_1u_1+m_2u_2=(m_1+m_2)v).

Here,(m_1=3kg),(u_1=4m/s),(m_2=2kg)and(u_2=0m/s).

((3times4)+(2times0)=(3+2)v).

(12=5v),so(v=frac{12}{5}=2.4m/s).

37.Abulletofmass0.01kgisfiredwithavelocityof300m/sintoablockofmass0.99kgatrest.Thebulletembedsintheblock.Whatisthevelocityoftheblockbulletsystemafterthecollision?

Answer:

Usingconservationofmomentum(m_1u_1+m_2u_2=(m_1+m_2)v).

Here,(m_1=0.01kg),(u_1=300m/s),(m_2=0.99kg)and(u_2=0m/s).

((0.01times300)+(0.99times0)=(0.01+0.99)v).

(3=v),sothevelocityofthesystemis(3m/s).

38.A5kgobjectmovingwithavelocityof6m/scollideselasticallywitha3kgobjectatrest.Whatarethevelocitiesofthetwoobjectsafterthecollision?

Answer:

Foranelasticcollision,weusetheequations:

(v_1=frac{(m_1m_2)}{(m_1+m_2)}u_1+frac{2m_2}{(m_1+m_2)}u_2)and(v_2=frac{2m_1}{(m_1+m_2)}u_1+frac{(m_2m_1)}{(m_1+m_2)}u_2).

Here,(m_1=5kg),(u_1=6m/s),(m_2=3kg)and(u_2=0m/s).

(v_1=frac{(53)}{(5+3)}times6+frac{2times3}{(5+3)}times0=frac{2}{8}times6=1.5m/s).

(v_2=frac{2times5}{(5+3)}times6+frac{(35)}{(5+3)}times0=frac{10}{8}times6=7.5m/s).

39.Aballofmass0.5kgismovingwithavelocityof8m/s.Ithitsawallandreboundswithavelocityof6m/s.Whatisthechangeinmomentumoftheball?

Answer:

Initialmomentum(p_1=mv_1=0.5times8=4kgcdotm/s).

Finalmomentum(p_2=mv_2=0.5times(6)=3kgcdotm/s)(negativebecausethedirectionisreversed).

Changeinmomentum(Deltap=p_2p_1=34=7kgcdotm/s).Themagnitudeofthechangeinmomentumis(7kgcdotm/s).

40.Aforceof20Nactsonanobjectfor5s.Whatistheimpulseexertedontheobject?

Answer:

Impulse(J=Ftimest).

Given(F=20N)and(t=5s).

So,(J=20times5=100Ncdots).

41.A2kgobjectisacteduponbyanimpulseof30N·s.Whatisthechangeinitsvelocity?

Answer:

Impulse(J=Deltap=mDeltav).

(Deltav=frac{J}{m}),with(J=30Ncdots)and(m=2kg).

So,(Deltav=frac{30}{2}=15m/s).

42.Acarofmass1000kgismovingataspeedof20m/s.Itstopsin5s.Whatistheaverageforceactingonthecar?

Answer:

Initialmomentum(p_1=mv_1=1000times20=20000kgcdotm/s).

Finalmomentum(p_2=0kgcdotm/s).

Changeinmomentum(Deltap=p_2p_1=20000kgcdotm/s).

Impulse(J=Deltap=F_{avg}timest).

(F_{avg}=frac{Deltap}{t}),with(Deltap=20000kgcdotm/s)and(t=5s).

(F_{avg}=frac{20000}{5}=4000N).Themagnitudeoftheaverageforceis(4000N).

43.Aballofmass0.3kgisthrownverticallyupwardswithaninitialvelocityof15m/s.Whatisthemaximumpotentialenergyitcangain?

Answer:

Atthemaximumheight,thekineticenergyiscompletelyconvertedintopotentialenergy.

Initialkineticenergy(K=frac{1}{2}mv^{2}),with(m=0.3kg)and(v=15m/s).

(K=frac{1}{2}times0.3times15^{2}=frac{1}{2}times0.3times225=33.75J).

So,themaximumpotentialenergy(U=33.75J).

44.Ablockofmass4kgisplacedonaroughhorizontalsurface.Aforceof30Nisappliedtotheblock,anditmovesadistanceof10m.Thefrictionalforceactingontheblockis10N.Calculatethenetworkdoneontheblock.

Answer:

Thenetforce(F_{net}=FF_f),where(F=30N)and(F_f=10N).

(F_{net}=3010=20N).

Networkdone(W=F_{net}timess),with(s=10m).

So,(W=20times10=200J).

45.Amachinehasapowerratingof500W.Itoperatesfor30minutes.Howmuchenergydoesitconsume?

Answer:

First,convertthetimetoseconds.(t=30minutes=30times60=1800s).

Power(P=frac{W}{t}),so(W=Ptimest).

Given(P=500W)and(t=1800s).

(W=500times1800=9times10^{5}J).

46.Aforceof40Nisappliedtoanobjectatanangleof60°tothehorizontal.Theobjectmovesahorizontaldistanceof8m.Calculatetheworkdonebytheforce.

Answer:

Usingthef