Assignment 3 Solutions

Assignment 3 Solutions - Assignment 3 Start Date: Due Date:...

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Unformatted text preview: Assignment 3 Start Date: Due Date: 13 Cct 2008 at 11:00 AM 27' Oct 2008 at 11:00 PM Student Access al'ter Due Date: ‘r’es Graded: Yes Chapter 8, Problem 16 . '-.-“iew with Assistance Figure 8-3-1 shews a thin red; eF length l. = 1.? m and negligible mass; that can pin-ct aheut ehe end te retate in a vertical circle. l-IEa'u'Iy' hall elE mass m = 9.2 kg is attached te the ether end. The red is pulled aside te angle 0.3 = 16° and released with initial uelecit'y- I?“ .3 = 0. '-.-'-.'hat is the speed eFthe leall at the lewest peint? L — Lass ---- ’— “R .l.'|' / ...... 'u'.. NLQI-lnu- PQ‘Wi' J» PC» JMWMlS I’ll Ole-Fine. m7, rel-c hence. Pou‘n‘l' ‘l’lac 1M1 fi- ohwl- avl- ‘lk F5 wk”; U to. Infi-(mllm ‘lt-L MMS 1‘5 «4' a l‘ 2 L— Loose «lawn. ‘l‘LL PbCQr‘ane. Pa.‘n~\~. 'TLk P; “Wu TL; Ma 35 has LN t‘l-‘L Q n that) In 4L1 mall = Q4 lot. keiait‘l' P‘ltnl-I'nl a'l- PK‘IA /' ‘M’UK We; Mal, ~— Mast). I‘Mi‘n‘al U‘elou'H k?- lamviz *- slmk, MM [\tO mad 1 i171!) I {o )I‘i‘f 0 Mm; 1‘! Mi- #1. (A; = 0. IN N ox.) M82. Cans (Pi/Aron 56i- Mu, kin/fiat) t.) t: -(- (A; t + Mi U]: “ iagL- 321L059 an - jazL (I- 6059) 2 Jam: M/g1)(l.'+m) [1 - as 14’] 2 Chapter 8, Problem 22 block of mass m = 1.8 kg is dropped from height h = T-‘E cm ente a spring of spring constant k: = 1?BEI Hs'm (Fig. 8-38). Find the maximum distance the spring is compressed. I l m : ........ .. i t i. '3‘ - -~---_..__-_.-- FL1 ¢ B's—.3 .I - I 745 Pet - fad- E’lr u--——~- -...—.. 6 I k2: RT} ( >1fia§¢flfigfigflfigfi \ \ \ \. \v " " ‘ L6+’$ 0184;“ M3; CMSer Indian 04' Met, [ta m'ca/ enerm. 1’“ airway. M‘ r/h1 m «an; [)oiA-i' (wimp; (’i T— D ) ‘HA. poh‘i‘ (34' Maximum Cdme Pcsh'M 9'L '“M— {Pg-Mn. Tail-H CuMoiH-fms 1 hi = O ( limppe/ ‘llam Fed") U; 7. N3 (ham) 4‘ o ( gawk} (eiasisd Quit‘ Coanu'oAS : k1: : 0 (MM'IM'AM com/Omn‘°'\ Mum [01.51; AM 55”!) (It? 7. O + JZ k'X-L (gm/51113 ( clad-V.) (o «sax/At“ 0!— Me at“ m‘cq] energy 3hr“ Ms kf" W ‘— ‘sz + £44: r. .3 l 7. P M3 CA+ 1c) ‘1 be 1) '71,ka ~ m8 )4 .. m L :0, =7 £0190 Wm) 3oz - (1-? leg) (TIM/1‘) 2c — (1'8 hymn/1% “+2 m);o =) 8’40 ,1 ~ {1.6% 7c. » 51.100? =0- §0\Vt ‘HAL zwatmu‘w‘c Z 1 _ [1.53 i /?n- 1m 4 LINN.qu 11-80 = Wham“ 1 (9.114811— m 9 )c: 0.124% m or )c‘c —o.|o‘I‘Ié W- \I/ va— ‘HM'X "8 MN.— Cdt‘t‘CU" {YAMH- ‘ (3‘40; IL A1,; (RN. )L WWM Man He. S‘Prh‘J [In 3 ALLA anm prcgsd- Chapter 8, Problem 64 In the Figure a hleek slides aleng a path that is witheut FriI:_tien until the black reaches the settien eF length l. = ISIS-’80 n1..whil:l'| begins at height P1 = 1.80m en a ramp eF angle 6' = 26.0 '. In that sectien the ceeFFicient eF kinetic Frictien is IZI.-1-1|ZI.The hleck passes threugh peintfiu with a speed ef 9.50 n'I.-'s. If the bleck can reach peint E (where the Frictien ends); what is its speed there. and if it tannetl. what is its greatest height alge'u-e .4? dr- Emu.’ L A E-l'k gnu. , i \h) ‘.M\ W gram} at g 0.4 r FCC—(Pena; P D“"+, (A; —. mm = “2 NH is”) Tint. 1:34:04 «ll ‘Cru. 4; 13 “FL 7- /’1It Eu 2 fl; M3 Cor? So wL lAM ’ ‘2. imp; + MaUchLQ‘AM +/4tm(7c‘,59l.: Emu; + O g0 lb; I‘m U‘t '. 01 2 v.2 -— L3 {L+L§.‘«s§ ~§Mk3u59 L z (iguale - ~10“ m/g‘)(los’om+(¢9-'+X m) fi'nlé') — .1 (#WHiMfi)lemmas”) Lil-1M.)— n17!" ll U‘L= (-l-i #11:. Chapter 9, Problem 46 In Fig. 91-59,. a statipnar'y- plpclc explpdes intp twp pieces .1 and F! that slide acrpss a Frictipnless than and then intp regipns with Frictipn. where they step. Piece .1. with a mass pl: 1.1 lcg. encpunters a cpeFficient pf lcinetic Frictipn pL = ISL-16 and slides tp a stpp in distance 5'; = 0.1.5 In. Piece F! encpunters a cpeFFicient pF kinetic Frictipn up = 0.3? and slides t-: a step in distance d; = 0.30 m. 'i-'-.'hat was the mass pl: the black? I. -- H If g MM. 5 3141143" w.’ l Com lolnl. Ca '\ 3e I‘VOL‘I'IID" 6’ ‘L {/1031 M1 Canwf Va‘l-‘oa 09'- Wlo [Mini-44M- firsl M I] a) flu. ,‘n \‘ulul V6 (OLHJ’QS 94' Cal-1‘. .‘e. be Pfeoe ’— .- + = 0 ‘9 0:1 2 -Fh AL So vL : JA/AL plL *- A(0‘HL\(1'YMIP)(O‘WM) '1 HUM 21- m/(. Vl‘c‘vt L '. Soww. §I|M+fw~ '- U‘RZ = J .1 (an) («Immunom 2 I. LFWM Wm WM Cansgrvdt'n 0-!- Momcniwlw‘ 1 '=) 0 7- ‘- MLUZ "’ MKV‘A Yo Mk : (—1; ML : (HLMD- ’1“) (H #10) v; (NH-WM pm) = 0.841;? kg. Ami ‘HJ. Mass 0(- ‘HL J‘oH-‘ul Hoot [‘5 Chapter 9, Problem 54 In Fig. 9-62,. a 12 g bullet maying directly upward at lIZIIZIIZI m.-'s strikes and passes threugh the center at mass eF a 6.3 kg hlcck initially at rest. The bullet emerges Frem the black meying directly upward at .'-'2III m_-'s. TI: what maximum height does the black then rise aIJCye its initial pcsitien? l '| llullm [mu l5 WI I'nela (\lu‘o Collision } Mame/\JMM Cd" SCH/all“ aim Ms 9 '5 4) 1;; "' l’zi : h? 1' l’TvF Like»; “l " Mi calm M4. lolott AMA "Z" 1' 5 Ha. AWoll- I l... = 0 (but; «4 mm 1‘: - MI erI (Nah ‘lqul' we're 0l$$uminJ Ma'lLer J44. bullet-’1 May m, in law; my, elm? alum-.0 m calm-M). mu ~C-r 4L. (glows [M Valet-Fla] o + Mil/T“ '3 Mlv‘m 1- M2 (I‘m =7 Vic = 1 (m m — wig/Le) Ml = 592 m.- — mu) = 11m" k; (law/920 Mr M. (“flab = 0.96113 M/s. \A/C- Cam MM. C-MSeNn 4h~ 0+ 94 a” v)» ‘H‘L L135“- 44». Monk 92:4 ’ or am“, Lo lc-‘v-wwh'c at, Malt-'0‘ : Lem 0le w A. {.w‘v- 94- L: v.5 = (Mlcqzz M): a? 2 010 ” luq IV [3633 CM 1- (q-K Ira/g") M Howl) ...
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Assignment 3 Solutions - Assignment 3 Start Date: Due Date:...

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