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Unformatted text preview: Q) I'hysicsZ_21, Exam 2. Answers Oct 2000 cﬁmaﬁamaaanﬁcﬁaacasx
maﬁaswmmammeBM$mmm$w
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nunyamywywnaaMﬁmnmwm wheeze/Wzm .M% ii; A 1530 kg elevator is lowered it] in by a vertical cable attached to the top of the elevator. The elevator
is at rest at theirliiial andﬁrt‘al positions. The work time by the cable 0:1 the elevator during this
process is k]. *iFk. —300 8.450 C. 0. D. iii} E. 30.0 Note that the force applied by the cable on the elevator is upwards, but the displacementol‘ lhs: elevator
is 'dmmwards, in the oppositedirection. Therefore me work “’5 done by H1: force applied by the cable
on the elevator must be negative. We don't know what the applied tome versus the position of the I
elevator is. However. from thematickinetic energy theorem we haveAK = Kf K1 : Wm = W5 +.
W3 and we can caiculaje the worng done by the gravitational force. Since AK =.i}— 0': 0. we have w11 = —ws = [ tarotr0] = (1530 kg)[9.3 moron m] = —3.0x 105.1 = . 300 kJ.. 32. A person pushes a 102 kg box in a straight line along a horizontal ﬁonr at a constant speed of Uéilrnt's
by applying a horizontal force. If themefﬁcient of kinetic [ticlion of the box with the ﬂeet is 0.20, die
rate at which Lite person does work on the box is W. im—Z 11—100 C. U IfD. 100 E. 200 The bolt movesiwith constant. velocity. The horizontal forces on the box are the horizontal applied
force F, and the horizontal It'u'ieLit: friction force 73. Since the hex moves with constant velobity
horizontally, me net horizontal force on the box must. be zero. Thus is; must hueequal in magnitude but
opposite in tihectiou to 'fk (of magnitude ft = pkN. where N = mg. is the normal force on the box). which itself is in the opposite direction to the velocity {if the box. Therefom the applied force is in the
direction of the velocity of the box and the rateof work P, done by the'fome applied by thepcnmn is Pa = ﬁa‘tm = 11km“: pkmgvm = (0.201(10‘2 kg)(9.2im’52}(0.50 W?) = we w. 33. A ll} kg block slides down it frictionless ramp from a heightof 15.2 m above me ground to a height. of
5.0 m above the ground. Lime initial speed of the blockis 5.0 rule, the ﬁnal Speed of the block is _ A. 5 BJO I'C. 15 D. 20 E. 25 The normal force'on. die block does no work on the block and the gﬁwlmional force is a conservative:
force. 'l'itetelbre we can use Ute principle ofconserratipu of mechanical energy hams = his: 1— AIJ =
0. Thus; Emug. = Ema“ or equivalenll)‘ K. +Ui = K; +. L}, so that .E :31 vi? + mg'ﬁ = .1, rn'vr1 + logsE. Solving [or the ﬁnal Speed Vl' of the block gives “ w = WJIV12+ 2m, — ,1 = ﬂit] nus}! + ates misimil m  5.0 m] = 15:0 mrs . Note that ur does notdepend on the mass of thc'bloitk. @ 34. A springpowered my gun ispointed vertically upward. The spring constantof'the spring is 100 Mm
and the spring is compressed by 0.10 rn when the guri is loaded with a 10. g projectile. After thegun is
ﬁred and held at rest, the pmjectile reaches a maximum height of _ in above the tap ot' the relaxed spring. (Neglect the influence of ﬁle aimn the motion of the projectile)
IPA. 5 B. it} C. 15 D. 20 E. 25 The gravitational and elastic (spring) forces on the projectile are conservative. so we can use the
principleof conservation of mechanical energy as in problem 33,except that here the irtitinl'potentia‘tt
energy contains the elastic potentiai'energy of thc projectile—completed spring system in addition to
the gravitational potentiui energy of theprojectilc—Eerth system. The initial and ﬁrml speeds of th:
projectile are zero. so wejnst have Ui = Uf, The coordinate of theprojeolile in the vertical direction is
called v. Setting y = {l 'atthe top of the relaxed spring, the initiﬁl position of'tljeprojwtilc is S'i =  0.10
m, the initial compressional” the Spring is d = [1.10 in. and we have i]: = Us; + up = ﬁlth2 + Logo and U; = Us‘H UBJ = (1+ mgyf.
Setting U; = Ur and solving for yr. the ﬁnal height of the projectile a'hove therelaited spring, gives _ . toil. _ . (toolNimggmomtz _
3" ' “Wane ’ 0‘10“” 2(0.0l0kg](9.31m53) ‘ 5'0 ' Po) 0” 35. A panicle undergoes one~djmeneionai motion
along the): rods under the inﬂuence of a net
force_F(x) applied along the same axis. The
force Fm] is plotted versus the particleposition
x in the graph to the right. As the particle
moves from x = 1.0 m to x = 3.0 m, the net
work done on the particle by the net forceir' J. A. 2 B,—i C. 0 IrD. 1 E. 2 If
The work detach}! the force F0.) on” the particle is. by definition, W = “.J F(x} dz , which is just the
. l ' area between the plot of Fix) and [he it axis. between the initiai and final positions. From the. geometry
oi'ihepiuL this area baween xi = 1.0 m and 1f: 3.0 rrt is 1 Not = 1 .T. fww :2 52222651245220 @ _.,_'F—' 36. 223.01:ng is located a'L{x,y.}=(ﬂ.4rﬂ) and :20 kg 3km.
musIs located at {3.0). What: “mild a. 4.0 kg nim
have it: bqplmdmpmthzmofmassfarmm
3 masses at the origin? 1&5
A midwbmmezksmaskgm B (— 1.5 —3.0
02,.302 MM'MJX3J1+M7‘*M‘EK4 E C—2.5.—3.u) O :[3. ‘O]+ 2.:im24h2ﬂq noncoftheabwe _ _ M Wm. nmudm “34
g.) : qusﬁn+1hfom+4hl 11* : —_r',h—a£  0m.
0%”: 4k», 3
37. AIUUDkgcardﬁvescastatdvﬂkmfhmdalsmumdﬁmmaﬂthmMmiryof
themtemfmaxsofthisZarsystmhasmagnuude hwhandisdimcted
A. 13,0251
B 10. was
. looé) 5'01” 
@2ﬂ 53 $32
can'ltellfrumthisi‘nfammﬁom ‘fOan/J: 3‘} WM
ﬁW “= 7\./T NET}: ‘2‘ Mgr1F.
”W W (”WWWWS“°W2
~r _J__,(zge m 22% 4552025223 fm: 1500%
5 sW/i g—Mm/g W" 15019qu ($135.22! 5,3352 mpggﬂoo 38. A 1.000 Iran: mﬁformiymlmﬁom 501mb towdthceastto SD Inn's toward Muslin a time period of 9.0 sec. The average force on the ca'r has magu‘tmde N, and is dixemd
10.000, m _. ' — .
Wm a — ELL; mum
c. £00,000, west ‘1‘; r_ I L'
:1 8050.090, cam  ' _ \
13.. can‘ttellﬁumthisinfurmnﬁan ' a : 30?:50 /$:_/dy7/SL
S
j . . ~ :3
F mar/MHUOM/s
F : g 0) mo M M
1600 Eu,
39. A 5,000 kg trucktrawling east at semis strikes ‘EO “/4
a 2,000 kg Iruok traveling south :1: 40 m"s andThe
twu vehiclcs stick together aﬁer collision. The
17300 La directidn 9f the wreck is A. 7’2“ Scrum of east [join/F.)
B 53° south of has: 5 x , R m
(9 23a south ofeast flit ENE" .
D. due east ' w h 13—
5. can't [611 fmm this Mnrmalion M m1
’L)“ "Ti—1‘ (m. vimx + M2011) film—11H EWMIBOME+O a LEMMM/S:zr.% “/5 40. A buckey puck, moving £8.51: at 3.0 mfs strikes
an identical hockey puck initially at reﬁt After
collision the. puck (initially attest) moves at
2.0 mfg at an angle 30" north of east. 'l‘hejspaed
Elf the inCidanI: puck aﬂér coliision is_ mfs. A. 1.0 B. 1.3 ©m n. 2.2 E; can‘t tell withoul knowing the
mass ufthe hockey puck H4): ”
WW8. F3 mimﬁ'ﬁfmlwaﬁ— mrwna 1’ 21):.43
C) + = 15A3+2mm9 41. Abuﬂet of 0.0132 kg mam naming a1 30:: m."s strikes and L” M94 I
betelﬁes embadded in a 4.0 kg wooden block initially at rest. '3 El 1 a Timvelocityofﬂm blockmdbuILeIuﬂercollisiunjs mm. 6 B [
— roam ,
It rem 8. 0.045 @045 D. 90 W P
.  . K E. can‘ttellfmmthis inforrnatien .
ml W3+M1U25 : (Mlf M1) 1);!
0307330055. +0 : {1‘00th 1374, .4; ﬂ .
%:%2‘ 0‘15m/5 .. LU ” E
42. A Wmﬁactdiﬂi starts from restand accelerates M __ _§'.T :3
at constant angular acoeiemtiou to an anguiar TF — L 5: S 5 velocity of 52 rad'5 in 5.5 seconds. The
angularwiclcmﬁon is [60% D< 2 9‘ S ”/gL A. 1.5 9.5 C. 12.0 D. 93
E.. can't tell from this information E  ~ — — o
43. Two 3 kg poinl mama are” cormc'cte'd by a masless 1.1 m 4
rod 4 111 lung inﬂichoriznnta! piano. The moment of “2..
inertiaahom a vertical axis through one afﬁne halts is 1
kgmz‘ I: m; + Mar}. A. zero B. 12 c. 24 I: O +3bﬁﬂmjl
®4S E. can't tell from the Mommien T ﬂ 4gb]? m1 . F'{N) 5?
44. A 020 kg hall initially at rest is struck bya 4.0 kg but such
thanhc force on the ball vs time curve is given '31 the rightlfdoa
The velocity of the balljust after impact 1'5 mfs. A10.“ (3. so D. 4.0m 00 com 000?.
a can'tteilﬁumﬁijsinfommion ' ' f [(5) I=jJ=nat :yzéoozsﬂoﬁoh’ :30 131%; farpi wingv 7)“:ng : limo/9 45. A 20 N force pulls on nagm ofa whee] 0:50 at inradjus _ 101V
and I = 4.0 kg :112 momentr of'merlia assttown, The taiagondal
acceleration ofrhe rim ofthe wheel is 111533. The
rotaﬁongl axis'of the wheel is 11:11:] at. reﬁt & @125 B. 2.50 a 5.0 o. 10 E. can't tell from this information ._ mw'nwﬁr ..~,. 1
mo“ TI”— ”Sm/5 46. A 1. 0 kg block is initially at rest on a hoﬁz'ouml surface. The coefﬁoient. of mo {ﬁction
betweenthe smfoce andtl'l: block isﬂ.60audthecoeﬁiciemof1unencﬁ1cuon 'IsOAO. [fa
horizontal force of mimde 10 N' ts applied to 'the block, the magiumde of the acceleration of mebioclsis__mfsz. Fe 3.1— c.2_ u.3 _ at _ ,_
y/usm : (Home are; Itwwmmmorf 4?. Humvectmiudﬁnegimm§= (— 20m)i+(3.l]m)jnndh= (5om)T+(— 90111)?
When thetwu vectorsare planedtali 101811 theanglebetweenthetwo vem'n'S‘Is ° A. n B. 45 c. 9.0 D. BS (3 180 4S. Apartielc executesunifomlcircnhrmoﬁnnwiﬂiaperiodo‘fﬁ. 3 s. Iftheradiusofﬂie ciioleis
I D m. the magnitude of the acoelmtion ofthe particle 15 this: A. 0' .1 c 2
4r... turr' :t_n____tom :M/ _ "UMAEZ’
___T_::__5____’_DL
49. Arnigwheetomdim= 0.50misinitiallyntrestandlhen 3am
rolls Mdiout slipping siowo an incline set 5130“ to the horizontat
foradistanceofSDm IfthEEpeedpfthecenoerofmnssinfound inks”:
to be 4. ﬁll this at the bottom, th'omomentofinerh'a of the wheel 35"
about theaxis hf rotation of the wheel Is _kg 1112.
.04: 3.13.54 3. I171 D. 1.10 ,2: 3 GMIWSOZ'L
E. oan‘thellfrumthisinfommion EM 2 W© famu‘ tern) +m3j~=§jmv +/Itu +0z~iMLr+/QI_1J£ O t o+£hattizit5m=L1mq®QI+ZI .o
9333: IEI+3QIT I: 1W49haM23 Oilﬁrth} .SH‘QL PW 221 W2 SAUL; M20019 50. A uniﬁmn 110de disk with moment ofinertial=100kgm3 _
rotateswith angular velocity Mi] radii abouta wﬁcal axis. If i 3
a 10 kg mass drops vertically anti the'wheclmamdfns «2.0m
and sticksﬁen tbs ﬁnal angular velocity will be mdfs, "L
A. 0.15 a. 0.24 @035 D. 0.42 I , 193M314 m: 0 ”11
E. cm‘tteilﬁ'émthjsinfmmim Anisodbh’i'" 1’ (11‘ ha; A uniform circuiﬁlr'djsk ofradius r = 0.50 m 113 mass, m=3LGkg. mmmnrormerﬁaaboummsperpendicwa: _ T + mi?"
to the disk ante edge ofthediskis kgm3. I ”—m ‘7,
'2, ' ‘1
1.
A. om B. 93:1 (@113 13.51:“ + m rzgﬁqﬁg
13.2.26 E. can‘tteﬂﬁomthismfomﬁou I q/ ’13 ERIN?" 52. A 3.0 kg batlmcwing east with speed 4.0 my“: strikes 3 mm
ball maﬁngeast'with sp'ecd Ll) mfs. The speed ofthe Gems:
ofmass aﬁel coliisiunis mfg. mﬂﬁhtLFWBi‘l
W} “0‘7 +1“; 1
2. ml+ ML
A. 12 B. m c. 4.0 .11 _ 3514% $55.14... 5
2. cm tell ﬁum‘this infom‘ation “ 9k 6:
rm = 2.1 W5 53. PM linear motion, the physical qu'anﬁty "impulse" 1m ths'same
dimion's or uniﬁes A. force B. power C. energy linear mom'entm'n E. displacement J“Pc‘Fa‘WMmW 2’24 54. Which oftﬁe following smtemetmi is (are) always true? 1. Kinetic energy is conserved in a collision. —~ W 2. Linear magnum is conserved in a collision.— W :5. g I
3. If one end. of a Hooke's Law spring is held ﬁxed, the '0 energy storedinﬂlesmtngispmporlionalmmeseuare E _ oflhe elor'lgatibn of the spring ﬁ'om‘it‘s relaxed length. 4. Fn‘clian fences are conservative forces. I .JV'W Alonly ®Zand3o4ﬁy c.3011]? D.2,3and4onjy E. alleremle 55. Two blocks arejoined by aLhread andtravel together in the me straight line directionowr a
frictionless surfers: wilt speed v. The thread is then out. and the ﬁnal speeds V1 and V1 are A. v1 = am; we we Eli—El” V I
_ V1=v.n’3;V1*2vi’3 * ' _ £131.“? M
v1=v.m;v,=3m 341% . J. 1) = lf =V‘ B
C.
o. v1=3w4gvl=w4 outﬁt 'H'CLW'M J 2
® noneofthese 56. A projectile in ﬂighmxplodee into several fragments. The total linear mommhnnimrﬁedialely
after the explosion is @ the some as the total linear momenmjost beﬁn‘e explosion (394”5410 WAG
C
D
E less than the total lmee: mommomjusl before explosion.
more than ﬂeeiotai linear Immenm just before'explosion.
msfomedinto kinetic energy. transformed into thermal ”energy. S7. A radian is about A. 2'5“ B. 3?” C. 45.” a 5?” E. 90“ .' P_“' Z1! ZWWGO' 55. A whwlmﬁngm angular speed 3.u rims takes a tinie afabom s11: make one
revolution. 6 : Lift
A.o.33 13.0.6? 01.0 D. 13 @21 211339.»? I
t : Z .1 ALL
59. A wheel of radius, 1' = 050 mmierates from test with angula: acceleraﬁon U = 6.0 M53. The
angumr velocity arm 2.!) sec is des.
® LOP "‘ LU' L. + OCT
17. B 6 C. 3 D. 1.5 FL E75 ' I 3 Q
 I  6. i 2 ML—
a)?  O 4 . s w; 17—5131" 9 60. A thin walled boiluw cyﬁnddcal tube nullsx Wiﬁmm: slipping along me ﬂoor. The ratio of 1H:
transiaﬂonal ldncﬁc snug}? nfthe male: of 11135910 thekihetic energy af rotation about the tube ax'sthm Ihsceum ofmassis 
. I “Eh 4/1)": wfm‘dﬁ‘ﬁ‘
.1 B. 2 C. 3 D. 132 F. 11'} 1
[ 1. ...
H: . .LMaUJ‘ “I —: m'U' =. in“: :I ..___.._——... ZJZJ W2 MELM . 5.:le @ ﬁg Which gag of the following ﬁve statements is m? A. The wo'rItakinetic energy theorem predicts that if a net ptjsitrivo work is done on a particle, theltimlic
energy of the particle decreases. B. The workdohe on a moving particle by a kinetic friction force is always positive. (3. The kinetic friction force on a moving particle 15 a conservative force. fD. The net work done by the gravitational force on a particle which goes completely around any
closed path is zero. E. The principle ofconservation oFmechanicalenergy states that die mechanicalenergy of a particle
never changes with time. Answer A is false because the work—kinetic energy theorem predicts that if a net positive work is done
on at particle, thekinetic energy of the particle increases. not decreases. _ Answer B is false because the work done on a moving particle by a kinetic friction. force is always
negative‘ not positive, because the kinetic Friction force is always in the opposite direction to the
particiefs velocity. Answer C is false because the kinetic friction force is a neoconservative. not conservative, force. That
is. the work done by the kinetic friction force on a particle moving from its initial position witsﬁnal
position depends on the path taken by the particle between those positions. For a conservative ﬁerce.
the work done by rhetorce does not depend on ivhich path is taken. Answer D is true'hecsmse the gravitational force is a conservative force. for which the work done on a
particle around any closed path is zero. _ _ Answer E is [also because Lhe ptiitciple of conservation of mechanical energy holds if only conservative forces, andror forces which do no work on the particle. act on the particle. If'n neoconservative force is also presént and does work on the particle, the mechanical energy changes with time. For
example. ifn kinetic friction'force does (negative) workon the particle in addition to work done no
the article h conservatit'e forces, the mechanical energy decreases with time. A. if a particle's velocity changes from +1.0 oils to 2.0 nits along the x axis. the ﬁnal kinetic energy of
the particle is smaller than the initial kinetioenergy. l3; If a box is litted from the ﬂoor to a shelf above the floor, the work done on the box by the
gruvitaLiOnal force is positive. (3. The work done on a particle by a static friction force has a negative nonzero value. D. The mechanical energy of ahlock sliding down 2t ﬁ'ictionle'rs ramp increases wiLh time. 1‘13. As a particleslides along any stationary surface, the rate of work done on the particle by the
normal force is zero. Answer A is false because the kinetic energy increases as the particle's speed (the magnitude of the
velocity) inorgases. Answer 13 istalse because the. gravitational force is downward 'and thedisplacement is upwardso the
Work doneon the boat by [he gravitational force is negativemot positive. Answer C is false because a force only does nonzero work on a particle 'if'mc particle's displacement
component in the dirtiction of the time is nonzero during the time. that the.'force is acting on the
particle. Frorn' the definition of a static friction force. this displacement component or a particle is
zero during the time that the static friction force is applied. _ _ Answer 1') is false because when Oniycooservnrive'ﬁircea such antire gravitational force. and forces such
as the normal force which do no work. are precept, the principle of eonsen'ation of mechanical
energy holds true, i.e., the mechanical energy does not change with time. Answer E is true because a neutral force on a particle is always perpendicular to the stationary surface.
whereas the particle moves parallel to the surface. Since the normal Force is perpendicular to the
velocity of the particle at each instant of time; any increment of workdw =ﬁ' d? and also the instantaneous rate of work done. or power P =thdt =N d Flor = N at. are zero. However. if the
surface is not stationary. the normal force can do workon the particle. For example. the normal
force exerted by the floor afar: elevator on a person inside the elevator does a positive rate of work
'on the person if the elevator isnioving upwards. 63. The humic energy of 212.0 kg particle moving at a speed of 5.0 mfs is J. @ 4L 5 B. 15 @25 D. 5) E. 100
K= €va = 4m: bow/55‘: 2s 3" U( '64. A plot ofthe potential energy Ufa parﬁde
versus the position ofﬂw particle along the
x axis. due no a emain conservative force
acting on Ihe'parﬁcle, is 51mm at fhs right
No 0111:: fumes ”act on ﬂmpmﬁqfc. H'rhc
value 9f the mechanical energy ufﬁlc panicle 0
am: Lthejccoordjxi'ateoflbeparticle o 2_ q 6x 014)
must be between _ A.1ma:1d2m ®lmaud3m C.‘2ma.nd3m D.2m:md4m E. Smami‘im [(00 =' Emu — um and the kinetic energy Km can never be negative. From the ﬁgure, the regian in
which Km} is Positive, [or Em: = 2.01.15 for x values beLween 1 m and 3m. 55. Ifdm elastic pummel energy of g particlemag system'is I .l] I when the spring i=3 stretched by.
0.113 m from its ralaxcd léngﬂl, the spring can'stant ofthe spring is H11. 3. 300 c. .400 D. 500 E. 1500 200
__L.z .vﬂu‘sﬁzmﬁ U5 _ 111:: M K: —E fLlezrzaoAﬂw
66. The vectors a and b are shown in each n'fthcﬁve diagrams gﬂuwgh E below. In which ﬂofﬂm ﬁve diagramsistbewctorg=g — 3me shown? g 7 ‘9. Which one of the following. five Statements is tag? A. If particle #i exerts a certain force on particle #2. then particle #2 exerts an identical force on
particle #1. B. If the net force on aparlicle' is zero then according to Newton: 5. ﬁrst law d1: PJIILlCiC must be :at test. it“ it a  .0 kg part1c1e initially.I at rest is raised fmm at height. oil 0 TR above the ground to a position at
rest that Is 5 O m shove the ground. the network done in this process by all forces on the particle is
zero. D. [fa car accelerates in. a straight line. a'force occurs which pushes a passenger backwards into the
5&t. E_. An inertial reference frame is one that. accelerates downward a1 a rate of 93 min2 with respect to a
reference frame attached to the surface of the Earth Answer A is false because according an Newton' 3 3rd law, the two forces ittetttioncd must be equal 1n
magulultle but point in opposite directions. ”Thus the two forces. cannot be identical Answer B is false because Newton’ 5 first law predicts that the pin'tiele must he moving at constant
Velocity, which includes nonzero velocity as well as zero velociw. Answer C ts true, as is predicted from the work —ki1u:tii: energy theorem when the initial and ﬁnal kinetic
energies of the particle are [lie same. Here. the initial and fatal kinetic energies are both zero Answer D is false because more is no force pushing the passenger backward into the seal“ The
horizontal force that' [3 actually exerted on the passenger is the normal force in [he fnrward direction
of the seat on the passenger. This normal force accelerates the passenger forwards at the same rate asthe car. but the passenger perceives this normal force as a (fictitious) force in the opposite (tackwartl) ditcCtiaii because the passenger is in a noninertial lactelerattng] reference frame. Answer E is false because an inérliul reference frame is one in which Newtan' e law's lIdld These Saws
do not hold for an observer' in an accelerating reference frame like the one mentioned @145 A particle goes around a circle of radius r at a constant speed 1r. Whichn an; of the following ﬁve
statements 15m]: A. The acceleration of the partiele' 19 perm. *B. The acceleration of the particle' is in the opposite. direction to the direction of the vector from the
center of the Circle to the particle C The acceleration of the particle 13 in the startle direction :15. the velocity of the pardcle._
D. [f v is doubled the. ace: lcnition of the particle 15 doubled Ll. If r is doubled, the acceleration of the pardcle' is doubled A partittetrtoving in a circle at constant speed executes uniform circular modem. The acceleraLion of a
particle in uniformchcuiar nmdon has magnitude Vllr and points i'rorn'the particle to the. center of the
circle at each inittztnt in time. Thus. answers A. C, D. and E are false and answer B is true. Wat €42»an maﬁzm 69. A circular disk Dom: abgiut éuaxis perpendiuﬂa: with: disk andthmugh ﬁn: center of mm If
11:: radius uf_ thedjsk £312 In, the mass is '12 kg, the momenl afinértia ahom this axisthmugh
the cum: ofmass is 12 kg m3. and trueanswer velocity is E law's, than the angular momenmm
about the center of mm is kg m2.'s. A. m‘ 63 C. 69 D. 138 E. 143 '3'0. A uﬁfm‘m square shcm'ofmallias achmkcm out
from a cnmarin the 4th quadrmIL The camier of
mass is loomed in quadrant Al 2 C.3 D. 4' ecan'mllﬁnminfamation given.
\ E ’24“ ‘1‘)COM “01' ”IRENE
n. If}:=T(7.)_+](3}md§=T(4)+§(5}'ﬂ1€n§‘§iS E6 2A B+Aﬁ 511 Am) +3as3 Biomﬂua @23 F.
D. 2210 E. mneofthwe A l' B 2 q _ position of a palhell: undergoing onedimensicnzﬂ mouou 15): = 2+3t+1¢t2 where 11 Ls
inmelersand t isinseconds.ﬂuently:velcncityofthepaniclcvxatI= Isaak mars. .A..24 .19 11.15 113 E20" 3% 3+3:
4);: 34% =lW/g 'r'3.‘l'hemagmtwieoftbeveqtor1k=i2+j3is A. 5.0 B. 6.0 C. 3.5 D. 1 E. mneofﬁm‘ie
m; 422+? ' 21M 3 = 31(9
?4. The angle belmmtheunitvacmr i mudthc vacmr A = i (2) + j (3)is ___d_egrees .A. m B. 42° c. 43“ D. W O as" 614343 '3 $6. ...
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 Spring '08
 HerreraSiklody

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