PARTIIProblems
Problem1
Inthediagrambelowa30kgcrateisbeingmovedatconstantvelocitybyaforceF
appliedat30°belowhorizontal.Thecoefficientofkineticfrictionis µ k = 0.25 .A)
FindthemagnitudeoftheappliedforceF.B)Howmuchworkisdonebythisforce
afterthecratehasmoved4.5m?C)Howmuchworkisdonebyfrictionforthesame
displacement?D)Howmuchworkisdonebythenormalforcen,andbygravitymg?
E)Findthenetworkdoneonthecrate.
F
y
n
v
30°
x
30.0kg
ƒk
mg
ANSWER: A) F = 99.2 N; B) 387 J; C) W f = − fk ( 4.5m ) = − ( 85.9N ) ( 4.5m ) = −387J ;
D) Zero; E) Zero HINT: Normal force is FN = mg + F sin30!
Problem 2** This problem considers the same system as problem 5 of part I. Inthe
diagram,blockAhasamassof4.00kg,blockBhasmass12.00kg,andblockCis
12.9kg.ThecoefficientofkineticfrictionbetweenblockBandthetableisµk=0.25.
a
v
Theropeis
B
masslessand
thereisno
frictioninthe
pulley
C
A
A)CalculateworkdoneonblockAbygravity.B)CalculateworkdoneonblockBby
gravityandbytheforceoffriction.C)CalculatetheworkdonebygravityonblockC.
D)WhatistheworkdoneonthesystembythetensionintheropeconnectingBand
C, TBC andtheworkdonebythetensionintheropeconnectingAandB, TAB .HINT:
Youdonotneedtodoanycalculations,butyoumustjustifyyouranswer.E)Addthe
workdonefrompartAtoDtofindthetotalwork.Nowassumethattheinitialspeed
ofBis v0 = 1mi s −1 ,andusethework-energytheoremtocalculatethespeedofB
afterChasfallen0.2m.NOTE:youmustusetheworkenergy-theorem.
ANSWER: v F = 1.34mi s −1 Problem3**
Infigurebelowa14kgstoneslidesdownasnowcoveredhill,leavingpointAwitha
speedof12m/s.There’snofrictionbetweenpointAandB,butthere’sfrictionafter
pointB,whereitreachesthespringandcompressesittillitcomestoastop.A)Find
thespeedatthebottomofthehill(pointB).B)Findthemaximumcompressionof
thespring.
ANSWER:A) v B = 25.2mi s −1 ;B)x=29.7m.
atthebottomofthehillUB=0,thenatthetopofthehillUA=mgh(h=25m).
1
1
K A + U A = K B + U B → mvA2 + mgh = mvB2 + 0 2
2
vB = vA2 + 2gh =
(12m / s )2 + 2 ( 9.8m / s 2 )( 25m ) = 25.2m / s Afteritreachesthebottomofthehill,therocktravelsadistanceof100m+x,
wherexisthemaximumcompressionofthespring.Thedissipativeworkdoneby
frictionis
W f = −mgµ k (100m + x ) = − (14kg ) 9.8m / s 2 ( 0.2 ) (100m + x ) = −2744W − ( 27.44N ) x (
)
Hereuseconservationofenergywithexternalforce
W = ΔEmech = Uel ,max −Uel ,B + K Max − K B , K B + U el, B + W f = K max + U el,max f
1
Atthemaximumcompression K max = 0,U el,max = kx 2 = (1.00N / m ) x 2 ,and U el, B = 0 2
1
2
Thisgives (14kg ) ( 25.2m / s ) − 2744W − ( 27.44N ) x = (1.00N / m ) x 2 2
Thisgivesthequadraticequation x 2 + 27.44x − 1701 = 0 Thesolutionis x =
−27.44 ±
( 27.44 )2 − 4 ( −1701)
= 29.7m,−57.2m 2
Thephysicalsolutionisamaximumcompressionof29.7m.
Problem4
A640-Nhuntergetsaropearounda3200-Npolarbear.Theyarestationary,20m
apart,onfrictionlesslevelice.Whenthehunterpullsthepolarbeartohim,thepolar
bearwillmove:
A)1.0mB)3.3mC)10mD)12mE)17mANSWER:B,COMproblem
Problem5
Athickuniformwireisbentintotheshapeoftheletter"U"asshown.Whichpoint
indicatesthelocationofthecenterofmassofthiswire?ANSWER:B
Problem6
Thecenterofmassofaasystemofparticlesremainsatthesameplaceif:
A)itisinitiallyatrestandtheexternalforcessumtozeroB)itisinitiallyatrestand
theinternalforcessumtozeroC)thesumoftheexternalforcesislessthanthe
maximumforceofstaticfrictionD)nofrictionactsinternallyE)Noneoftheabove
ANSWERA
Problem7
A0.3kgrubberballisdroppedfromthewindowofabuilding.Itstrikesthe
sidewalkbelowat30m/sandreboundsupat20m/s.Themagnitudeoftheimpulse
duetothecollisionwiththesidewalkis:
A)3.0N-SB)6.0N-sC)9.0N-sD)15N-sE)29N-sANSWERD
J = ΔP = mv F − mv I = 0.3kg × 30mi s −1 − 0.3kg −20mi s −1 = 15kgi mi s −1 Problem8
Ahockey puckofmassm=2kgtravelingat4.5m/salongthex axishitsanother
identicalhockeypuckatrest.A)Ifafterthecollisionthesecondpucktravelsata
speedof3.5m/satanangleof30°abovethexaxis,calculatethefinalvelocityof
the first puck? B) Calculate the change in kinetic energy, ΔK . Is the collision
elastic? Briefly explain. ANSWER: A) Velocity: v = 2.28 m/s at 50° below the
m
m
!
horizontalor v = 1.47 iˆ − 1.75 ĵ ;B)-2.8J
s
s
PROBLEM9NEXTPAGE
(
)
Problem9
Abulletwithamassof8.00×10-3kgstrikesandembedsitselfinablockwithmass
1.25kgthatrestsonafrictionlesssurfaceandisattachedtoacoilspringwitha
N
forceconstantof 315 .Theimpactcompressesthespring15.0cm.
m
A)Findthespeedoftheblock+bulletjustaftertheimpact.B)Whatwasthe
initialspeed(v)ofthebulletjustbeforeithitstheblock?C)Calculatetheimpulse
(magnitudeanddirection)ontheboxduetoitscollisionwiththebullet.
ANSWER:A)2.38m/s;B)373m/s;
m⎞
kg i m
⎛
C) J x = P2 − P1 = MvF − 0 = (1.25kg ) ⎜ 2.38 ⎟ = 2.975
totheright.
⎝
s⎠
s
(a)Forthispart,useconservationofenergy.Initiallytheelasticpotentialenergyis
zero,whileatthemaximumcompressionthekineticenergyiszero.Hence,
(KEofblock+bulletbeforeimpact)=(elasticPEatmax.compression)
1
1
2
= kx 2 mbullet + Mblock vtot
2
2
(
)
(
)
k
315N /m
vtot =
x=
0.15m = 2.38m/ s mbullet + Mblock
0.008kg +1.25kg
(b)Thispartusestheconservationofmomentum.Beforeimpacttheblockdoes
nothaveamomentum,onlythebullethasanonzeromomentum.Afterimpactthe
block+bulletmovestogetherwithspeedvtot=2.60m/s.
(momentumofbullet)=(momentumofbullet+block)
mbullet v = (mbullet + M block )vtot v =
mbullet + Mblock
mbullet
vtot =
(
)
1.258kg
2.38m/ s = 373m/ s .
0.008kg