hwsol9 - Homework Solutions #9 Spinning the Wheels: An...

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Unformatted text preview: Homework Solutions #9 Spinning the Wheels: An Introduction to Angular Momentum Part A: Which of the following is the SI unit of angular momentum?:r.::.i"~ m 'J.W.:::.JY.s1" hSince..L-=TiJJe S.:r:.V n it--.5C\...re)S~m:;Part B: An object has rotational inertia 1. The object, initially at rest, begins to rotate with a constant angular acceleration of magnitude (}.What is the magnitude of the angular momentum L of the object after time t?Jjn0{C. e....w(t).::: cA-tL ;: I.W(t)[1:t1Part C: A rigid, uniform bar with mass mand length b rotates about the axis passing through the midpoint of the bar perpendicular to the bar. The linear speed of the end points of the bar isu What is the magnitude of the angular momentum Lof the bar?L :;.J:.u->\.V e.Ji r. (3Vv"::c::I;), m,-J...L.L.2..c~~e.e. iDa (j /"')Lu:::- ~r::..JL..P/~~+-Au:5:L-=..Lmb~.~)),.J91+ 1~vbbVPart D: The uniform bar shown in the diagram has a length of 0.80 m. The bar begins to rotate from rest in the horizontal plane about the axis passing through its left end. What will be the magnitude of the angular momentum IJ of the bar 6.0 s after the motion has begun? The forces acting on the bar are shown. , ') Torl1 lJ e.S --<. L> [J, is bo..r IS ~xpe,..ie.VlGlnj[, =-J;)'N(O.brr'l)T;)..=- -8 tv ( 0 . go0(yY\)Ll=-O~gNVV\=01.= (l N''!' Ig-rQ.-ho u t-,-'45( +-s f'vo-r-13M.L~L=-Io<.t- =-O_~tJ'm(65);;:;'-/. I", J 8(Y\ 'J.JPart E: Each of the four bars shown can rotate freely in the horizontal which diagrams is the net torque equal to zero?plane about its left end. ForFor-B<t'(.:2-r :::.0Part F: Consider the figures for Part E. For which diagrams is the angular momentum constant?BC}c_rf-ZL+-he-nL :;.aY\S 1-o-n cYPart G: Each of the disks in the figure has radius r. Each disk can rotate freely about the axis passing through the center of the disk perpendicular to the plane of the figure, as shown. For which diagrams is the angular momentum constant? In your calculations, use the information provided in the diagrams.ForAZr::::::/P-r -4r ;lg r=- 0J LjFor 8For L2. f:::: rX ~ -r tr: +- 4 r;;2.. L==. L(rr-g',-l')...r -:;;::..;:l..IA vrFo-r0~r:;;; ?$r+-b r+<tr-~r-6';)..-Part H: Three disks are spinning independentlyon the same axle without friction. Their respective4I . . . ') rotatlOna I' lOertms an did angu ar spee s are l,w( c Ioc k wise ) ; 2ft&"';( counterc Ioc k wise ; an d ,W/2 (clockwise). The disks then slide together and stick together, forming one piece with a single angular velocity. What will be the direction and the rate of rotation wDdofthe single piece?The--rofO..lyv\ 0 tn en-ruyY'\J'YluS-r be co f} Se r veol::.. 3 I{)..JL.::: - I ILfw+-b Xw -;l r wt- li :r:.:= (I+ ;).I)UJJ-L i =- L.(!.37w_ ~7/-UJ+lw 7Problem 10.86 A unifornl rod of mass 111 land length rotates in a horizontal plane about afixed (L'(isthrough its center and perpendicular to the rod. Two small rings, each with massm'2, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance ron each side from the center of the rod, and the system is rotating at an angular velocity w. Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. Part A: What is the angular speed of the system at the instant when the rings reach the ends of the rod?Lj:::L~(1:1 In LI'l..j.-?.(fYl;).rl)fI\I)W:::'(Iim, L ?, +- 2( yY',,-LW,;)j) w{Jjf(L. '). +.IJ...' rY\ILa. ()').. +-:t:> TV' r~) 0---. .L L'J...)yY\:lPart B: What is the angular speed of the rod after the rings leave it?-r), e.. Ctn3 u lOf5 DfYl~t-()Sj'Je..ed-I::>+AeIh.s6.111e ..Ie 0(e.o -f+het-CA..-}yY\ 0 J?1 e-11r lJ/..:sca.-I" f'o If vJ; 1-' h +--A e. r; /7f)S ()... .S .Wf---(~):2" .J-Jm~r1~ _W(. yn,~ ~J~+- Lm d-L <)~The Center of it All Part A: Two particles of masses fTlland m2('111 < m2) are located 10 meters apart. Where is the center of mass of the system located?0fYl,CvY'-rhecwre(ofyY);). }Yno.SsISG)Gloserne.o..vier+0fh ~par-hele-"I h P art B : F or t h e sys t em 0 f th ree partlc es sown, where is the center of mass located?"d" w h" h h ave masses j\il , 2M , an d 3M as III Ica t e d , IC0Ih e- c.e...n+e rwj/}fIv{~.s.sI,M.o).1"\.eM.oo31'\be.Lt ndbe-rwe.eY);2.JVl3 M .Part C: For the system of particles described in the Part B, find the x coordinate:r.o'lofthe center ofmass" Assume that the particle of mass At is at the origin and the positive x axis is directed to the right.Xcm:::(tM)[O)!.(;)./.J.)( L) +- (3 M) (.:<.,-)(IMt:2M-r3!'-'t) --- --:lML-t-bN.Lbt1.Xc.. W\:::-8~Lbj1-- case. The system includes three particles of equalPart D: Let us now consider a two-dimensionalmass M located at the vertices of an isosceles triangle as shown in the figure. Which arrow best shows the location of the center of mass of the system? Do not calculate.lhe.ceJllfe.r\\CJ{- VVL~5i5ClrJoca.+~OI1[j}e+-h~ . Jocot--ed.(j) br:- 1-CC-t Y't' tb e...Q-t(J)b e..,e-a. u .s e..vY1Gc,<,S\'~d--is(pUtlvJ\1f\e-Io.>e+o(0cf~'()e>or't)'hy -rAe-.yA (l.r~.Part E: What is the x coordinatexCfriofhe center of mass of the system described in Part D? t/rem':::'(:1M)(O)+- Lf-t3M-~I - L ~'-Part F: A system of four buckets forms a square as shown in the figure. Initially, the buckets have different masses (it is not known how these masses are related). A student begins to add water gradually to the bucket located at the origin. As a result, what happens to the coordinates of the center of mass of the system of buckets?GI.9co~ ~emj' "I' 4JI~i+-i Gt. Jl Y'~II!Cem0.{-;(lO-I/i!() fleA5e p a-ruf./-( / L crnCLS~')b e- <:'0rYl e5mOreMO-SS i ve...+-6-thecenterMOVe-3C/6<5e..-pur+Ao.-+tie-IePart G: Find the x coordinate figure.xc:mofthe center of mass of the system of particles shown in theXc,," -(30 hc:J)CO) +-'/00(70KJ)(6 - 3)N\)/~Ae.-YYl ~().56YY)Part H: Find the y coordinate previous part.of the center of mass of the system of particles described in the"YeM:;::"(10 h~+-'10 ;"'y()I DO+-- 50 V,:]K)(o-Z;'>'\)YC Y)'\':::'-O,~YYlPart I: Let us now consider an extended object. A wire is bent at its midpoint through a right angle as shown in the figure. Which arrow best shows the location of the center of mass of the system? Do not calculate.1 h~-rhe-c-- C)" -I- e yo+-(Y\U-<S~<)i j'e.sU?Otl\" ~ 3~"ofH ne co (\" ect; ~ +A e- c1 e-e-/l-re (/'..5 yYlCLS-~ lor -r-h e. sep erv.-fe.. hCA z<)'!IIif..0Part J: A straight rod has one end at the origin and the other end at the point (L, 0)and a linear A = aJ? density given by , where (lis a known constant and Xis the x coordinate. Since this wire is not JM __ dm M uniform, you will have to use integration Find xCtllforthis rod. to solve this part. Use to find the total mass .XcI{=- MJI~(;q::::;d:rf\ XJ /\ cAoV\...--'"11LX::._---30....L3.'V 1\('"m .-I r:t ~xjL:l;X 0\ X :::;; oM.~Jof tjL -X'1 o'xc Nt ---_L Lja...l/"CtL3 ----.3 =-Tipping Crane Part A: While watching the crane in operation, an observer mentions to you that for a given load there is a maximum angle Olll""thatthe crane arm can make with the horizontal without tipping the crane over. Is this correct?NOI,Part B: Later that week, while watching the same crane in operation, a different observer mentions to you that there is a maximum load the crane can lift without tipping, and you can find that maximum load by observing the minimum angle Omlnthat he crane arm makes with the horizontal. t Is this correct?Yes)Part C: Select the correct explanation for why you can determine the maximum load given that 8ml:n is the minimum angle the crane arm can make with the horizontal.~ +~<9 Yl'\ iY\+h.e lev'er ()..rfYI ,0- r'\ ct(IGose)+0 r~is ~t)'& mo.-X';~)V\Ol,xfmUyY'I-r he.y-(l. -ore /u e.n, j z. eo{ .tc...osBPart D: You know that the torques must sum to zero about equilibrium. if an object is in statica Y f 0 Y\tJ'\j0 11 0 r0+F rheho cA'IPart E: This implies that you can pick the point about which to sum the torques to simplify the calculation. Often it is best to pick the point where an unknown force acts, so that the torque due to that force is zero. In this problem the simplest equations result if you take torques aboutfh. e roi '"+- vJ A ere. + A e.f'h e. .(!ro n .J.--/-t re..5 1-0 v ~ A5'0IJ h(;)... .Part F: Given the angle Olllill, what is the maximum weight (or load) WL,ml'Uthathis crane can lift t without tipping forward? (Recall that weight has units of force.)Sa I Ve.\tV&.. )1-0 r W L}M~)c'moxw(%)W~ (0. cos G~i1\)t\NL-w~aX.COj(e-mil"l)Part G: What is W~llr", the largest weight of the load that is safe to lift regardless of the angle of the crane's arm?fA e..I'YI/).)(;m Ii M+0r ~u e rse.-xe. (' ted.-.w he11B ~O\tvl%: )::0\,Alb Rw:::. Sw~...., :lQ.Part H: Notice that we have the weight of the crane exerting a torque about the front wheels of the same crane. To create a torque, a force must be present, so it would seem that somehow the weight of the crane is exerting a force upon its front wheels. However, the crane is one object, and it follows from Newton's laws that an object cannot exert a net force upon itself. This crane seems to be defying Newton's laws. What's going on here?Th.e- e..o.. r- T h. . ( j e. 3ro..Il ; +/) e-x e. ,rf-s }-n e- c r (A "e a ol -r It e. 10 o..cJ.. .. OilY\{orc-.5Part I: Assume you get a summer job as a crane operator. On the first day you are lifting a heavy piece of machinery. Even though you have the arm at 7(}" above the horizontal, the crane begins to tip slowly forward. Consider the following possible actions:A}}0f+It e.(Y1l.1. Release the brake on the lifting cable so that the load accelerates downward.Asj-he100-0... Clc.c.eleraJ-esforc..a.ownw~ro{)I~-j-exe,fsJe.s~2. DecreaseOr) f-he cra.n so that the load accelerates downward.e...A~o.it'\}rAe. }oo.d.. OLe c elera..f-i"<j (o..nd... }or~ue)011ti~w n wa..r-ol lesselt...S+Ae lever.vhe, +orc.efJ 3. Increase while simultaneously letting out the lifting cable so that the load accelerates downward. , +J1e.. n 0. e.lerQ..h'o n 1\ ~ CL-~As 10~>r e}eJ-ccw;l}be.+-0 wa.,r tk. tA L/00.t()r olA- nd. ..4. Put the crane wheels in gear and accelerate the crane forward.Ye.s I f he-etSYtIi Y\~S bo.. c K (I ess e tl i n3 .J-Ae- I-or~(J(?)None of these solutions is ideal, but which will have the short-term effect of restoring contact of the crane's rear wheels with the ground?r~is ;05 ex. ba.ct.VV }id-elAJhoweve.rhe-co..use .fAe.. crane..CAe.nc. f ashW henyou .s 1-0pe. e 1 e ,a. t ;1'\ 3.Problem 11.60 Part A: In the figure a uniform beam of length Lis hanging from a point a distance x to the right of its center. The beam weighs and makes an angle of 1 with the vertical. At the right-hand end of the beam a concrete block weighing w1ishung; an unknown weight l/Jhangs at the other end. If the system is in equilibrium, what is w? You can ignore the thickness of the beam.!::..-?t;;).0ll'l~vForr his.s ys l-e""-tohe.jn .s 1-0.. +i Ge-~ u Ii bri U MOY)z...,-;::. 0le..f+-Sicl~rne~heW'~(L-a-I-x)+-~ W.( ?<' )=="w~J}ne ( 1;- + x)+- WeIw e:xsit1OnriCj"tsiote.GJ,the- rwo1Sine(~-x)W, <S j1\sid. ..sty\ IJS t e..~uo../ :C Q.,n C e. /.5 0 U'jJ VV~ -e {~ - ~ ):::..W(~.s}n B( ~:r~)The 5 j n , d. ep e n.d en c e.+:.}- W<si"lt (?()'AI 1 ...;t').:::: w{-.i+-:t)+ W;x\,4/, ( ; 02Part B: If the beam makes, instead, an angle ofwith the vertical, what ix) -- W'X __ ( '-;:1. + ;%) -}-h ().r n f 0..r+- A)depenol~f'c.e0.(we, f 0 V flrAYltJffA It-r+- A e(;}-LCLYlCe-ls ou+)wi Jlt-nefelore..o..oti~efn+Vo./ve..f)ec+--rJ.. eCt. n.s 141 e 1'.\AI;::::..W, (~-X) - \VA(~+x)Problem 11.72 You are trying to raise a bicycle wheel of mass Tnand radius Rup over a curb of height h. To do this, you apply a horizontal forceFPart A: What is the least magnitude of the force that will succeed in raising the wheel onto the curb when the force is applied at the center of the wheel?FIi\AI e-TYlust{,'nd- fh eComponents0fFq..tnjfntAt ().re fer pen oI..icu Io..r J;U~. h '),fo~A e. Ie va/' a f' /Y\n .:eIt'\F.L = FSineFo.,rm'elle"p.-hm3.L.=m3cos8s,n8FJ. z. m3J-F ~fYljc~se_>F > rn ')jJ.ll.h-h'Part B: What is the least magnitude of the force that will succeed in raising the wheel onto the curb when the force is applied at the top of the wheel?F~FJ~-hFJ..;; F s i rY\3 +- ::::- tY':JF .2gCOS~ J-8~1., ::::. 1YlM 0\ C 0 s~ ~") J s,'n9r h'lj--j ~ R.h - h 'J.::1.R.-),JAt{ 'h ~h),... eJ.f\-~Part C: In which case is less force required?fJ fJ./+B - where- -role.e i's aft/,'e-d1-0+he ~fop,A Matter of Some Gravity Part A: Two particles are separated by a certain distance. The force of gravitational interaction between them is Fh. Now the separation between the particles is tripled. Find the new force of gravitational interaction Fl.Ie..rf;:Ibe. +he iYlJrio.' 3'0 ts ~Ae. fiYlO;..Ir6sefo..l'o.-fjoY\,se.pa./'CA..fiOT\r. -=0"""l1yr().:l.mt4.c;. ~(3 ro)1JmMC-r_-Po-;;;-roJ.Part B: A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is Fo Then the satellite moves to a different orbit, so that its altitude is tripled. Find the new force of gravitational interaction F2F :::.omj.{ G;_(d. rJ'J..yVJ(-~ Ljrn f{r~~Yl1!1GJ ~F =;l.t1 r e);l.f{G_=--l/6 - rt ,. -G -ifPart C: A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is Fo Then the satellite is brought back to the surface of the planet. Find the new force of gravitational interaction F4F ::..o~J{G(~1'e.)~~If::: !'" f{ C?. ::::-(rJ~1tjF.lPart D: Two satellites revolve around the Earth. Satellite A has mass 'mand has an orbit of radius r. Satellite B has mass 6mand an orbit of unknown radius ft,. The forces of gravitational attraction between each satellite and the Earth is the same. Findt'I,.r ~ mMG r:tPI(8;~mrM Cit:l.. bo,'n Ce.Fit ==-'8,.b .;).br~rb'" ='/~JbrJPart E: An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far rfrom the center of the Earth should an elephant be placed so that its weight equals that of the elephant shrew on the surface of the Earth? The radius of the Earth is 6400kill.(1 tOll = 103 kg.)m.shre,w ~~ ---------f'~e-m ele.pna.nf-~j(:XI0 X'?;3r~(0.05(,.)1~) <1,x ItJ6m)?._..s-r~--------{!-:::.:1 ~ J 0 6 /A,w}Part F: Find the net gravitational force F;letacting on the Earth in the Sun-Earth-Moon system during the new moon (when the moon is located directly between the Earth and the Sun).-(he- f.orc.t. he-tween +Ilee.CA.J"fJ..ancJ ,"oon:::-;>PM::"M~.MMra!3:l.:20(YlFM.=-1.9'6;</OrJthe fot'cebetlN'eetJF =-F",TFs:::rrite. etA.rfil).l1d S.UI17r-_r 5 ---:.----::::-.M.,s Me ~5 'f )rIO:;J~N)3.5 D )( /D'1:>,rs ~Fs ~ 3.521 g'NO :J'LNPart G: Find the net gravitational force Ft;'alactingon the Earth in the Sun-Earth-Moon system during the full moon (when the Earth is located directly between the moon and the sun).F.:::.Ps -F M ~(tJ1Problem 12.24 When in orbit, a communication satellite attracts the earth with a force of F and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is _ U. Part A: Find the satellite's altitude above the earth's surface.r he.p (' 0 b Ie fY\U;::.Sf e.c;~ie5UQ..-PmM.G,F.:::-mM~rr~e.wnere.r:::. o./+i-rucJe -r r ~u r= :::r.:=. rfhvs,-' ---+ ~ If,' tVcJ.~exff-itUc{e. .=:Y,::-re.,Part B: Find the mass of the satellite.tJ 0te.+ h oJ'forU A.:::.yY) '1.W\ ~M 'J. G :l.r'l.U ~r~-SotVit\Jm?..M :J. q:l.-u ~ lU)M:I."G.~'J. _U &j1,lp-:::; F f1"i:LfYl;:;MGrFu'J. -Problem 12.58 Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs IVon the earth weighs Wlat the north pole of Planet X and only W"latits equator. The distance from the north pole to the equator is L, measured along the surface of Planet X. Part A: How long is the day on Planet X?A+ +ht- pDIe.s)yourAto...ccel~ro.flonf.scJ.ve.sl1Ylply-to~rt>-vd .. . y.+}o,e e~ vlAtor) fJ,c.ro.cAja/w-m~RIaccelero.-tiol1 Y0l) e)(pel"ieYlce (fne..-l'la.ne+-w~v/~ co. u.s~ you +-0 w~}~n S/ljh+ly less._ w~ WIyo!-afjJ1;jJ_ mv Rm:::.":l"-/-~w, _ w~ - ,., v_.N\~RNo+' c e..A.. .:::.. ;).,Lrr ::::;.) ~\ILj ~= ;?1fRwe~nowVJhus=-I;)'~L -:) (wla/~ ~(on ~().("fh) rrlAnl"A~LTT ~(W, -W.1)JoJ.bur w.e..~J~no}Vr-(Ic1o.y);::;ttL-~x.::-vtD..X-=- ~7frforId-o..vPart B: If a msatellite is placed in a circular orb.t orbital period?/:~L(W/-V~)hC1~1Y\x r?"Ci1Y\.x _=-'tLove the surface of Planet X, what will be its/We, l~now:---r~V2.vJ he,re..-v=~rfl1 US:S6Jve.-Forr I:(o/)G yY\x. _ r{~n)J.r~~-rJ..T'l.-::::(lTT)'},r3-GmyT.:: /'trT'>-r'&m.x-Problem 12.43 At the Galaxy's Core. Astronomers have observed a small, massive object at the center of our Milky Way galaxy (Section 12.8). A ring of material orbits this massive object; the ring has a diameter of about 12.0 kmfE light years and an orbital speed of about 160 . Part A: Determine the mass of the massive object at the center of the Milky Way galaxy. Give your answer in kilograms.forCirc.vlo.f"orb'/tMp(G. _rX-rv~r11\:;0.; r v~r:;:. }'l li~bt')...yeo-csQ1-'16/ ,X/I)?../5r(\V;:::. :ZooJ{~;:::..~x/o5~M-:::- ('1.7305)(/0;[J!J)(Wi'1X/OI0?)~b.6 73 Nm;,h o..ctv---:, -:::-2..836)<1036~/1.=1/'K" ~ /' /0Part B: Give your answer in solar masses (one solar mass is the mass of the sun).1 m o.s s ; V ~ 0 b j e. c:+ he.,M~ J.} Sol().T0...m~.sS0r836)(/656X /0s6/-<5Nt. (). s s:::' /. 90/J~)1e y t A+0 rf2.36'J . 8 3 (, X10 'j =:=- / /. '1 J. 5 / . '! Cf )(/0 36 1'$9- I,) it'Srs,x 1()6S61a.r;J.. a.5 S e..5Part C: Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star?No .../\"jOb.f o..re.-x c e. d..Q.56() IPart D: Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzschild radius of this black hole be?1he-5CAWCArz,sch;/etTCAditJSis. cAe..~"Y1eo{o-.s .As::"-- 16tH.'),v~(\c.~-----C).Part E: Would a black hole of this size fit inside the earth's orbit around the sun?V-e. Sl , IProblem 12.66 Part A: Calculate how much work is required to launch a spacecraft of mass 71lfromthe surface of the earth (mass mE, radiusRE)and place it in a circular low earth orbit, that is, an orbit whosealtitude above the earth's surface is much less than Re. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6380 km.) You can ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation.EbE;:jc::: -~mM.r~EfGml'\ . re.,~ (' e.-Gfn/i ;l r~Ef- i-(- ; 1-1)Part B: Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. You can ignore the gravitational effects of the sun, the moon, and the other planets.Ihe.e..1\e.r~y,Inorb,. f-J'j-0~Gm~~r~F -00~E.=- 0ac-=E&::J -Eof"'Problem 12.70 If a satellite is in a sufficiently low orbit, it will encounter air drag from the earth's atmosphere. Since air drag does negative work (the force of air drag is directed opposite the motion), the mechanical energy will decrease. According to Eq. (12.15), if E decreases (becomes more negative), the radius Tofthe orbit will decrease. If air drag is relatively small, the satellite can be considered to be in a circular orbit of continually decreasing radius. Part A: A satellite with mass mis initially in a circular orbit a distance hlabove the earth's surface. Due to air drag, the satellite's altitude decreases to It.!. Calculate the initial orbital speed.Vlt Hno\AIJliv').. _~f"'YAt11--1'1 ,;y(C,-)(Re.+h }~v =- L.ME. Gi'liEt' hIPart B: Calculate the increase in orbital speed.h<!.i'Jl-tth"J~;;:jt1GI?of-hJ...50 wLLo../CU/Ulref -Vi .:::.fheAt thJChO-fle{'11lieI oc/ fy ..LW ~ V/j1)e. g:' _":3.(jte.GJJf.f,.:. "/Part C: Calculate the initial mechanical ~,; ::; J". E r P j;.=Tf\.VhIG. tt, 1Yt h , J.lie)butwLjrJ,s+Ii. =- -\}IrAn+whl-cAI~fhe(Y1e chan" QJ Ie. l1e/'(j Y :-k m tI ~ ;::::m V.f. ~) /~Im M G ~ {AT:;+-h2Part D: Calculate the change in kinetic energy.6 KE;;.,/;AIrn. v., ~J ---- ----;;: ME 1 ;;lrnbi(fi.E.-fh}Part E: Calculate the change in gravitational potential energy.rl1;~io.lly:u- :::: Ijm f{r; G-h, tfEFiJl\(XlIy ~-VMMEGJhJ.-rf'--U +6. u~ Uf- U''=-JN1ffE Gh,tA[yY1/VIE GhJ.+-(\EPart F: Calculate the change in mechanical energy.fr 0 I'"rhUSf 0.'+ C ~E-, ;;.. - 0 M ~2eAr-I'"+h,)I; 1-\ e vV"ISC[f -".'-.1 - - G I,tjV\------:J. (f{ErJ,,:z.)6.. E =. -G /1E-:t.M (IAef- /"1._,I_ )At ..,.. },,\Part G: Calculate the work done by the force of air drag.theVJor-l'-,iJ1d..Dne-bl f,ic+ioncaIe)1"I)t..Ltu(Li-1-0 +he-e, hCA "'c:J e-rnQ..Ch Cl () ;e,rSy-,vJ::- - G ME. ~ (':l.IAe+hL_1)I\e,+hl...
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