hwsol5 - Problem 5.15 Atwood's Machine A load of bricks...

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Unformatted text preview: Problem 5.15 Atwood's Machine. A load of bricks with mass m1hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass tn2is suspended from the other end of the rope, as shown in the figure. The system is released from rest. Part A: What is the magnitude of the upward acceleration of the load of bricks? () In_2 7 GJ T-m'3'=: rn r -r Ill_I 1m'5 I {).., We, 1- yY1~5== yY1 ;;l.. 0..." ~ ~ '''' n 0 ~ -r:::-r a. ;1, 1 4 0.., ..:::;1"\ T= T-=- m} (3 + 0-) so Ive {:-o r ~ '.3 -r m 0..'::::' (A ::;::. rn a. ~ 0... - fYI;J. CA.. yY1~ (~- 0..) f>... ~ (yY1~-fY\I)::::' ~ (J111-trn~) (f71.;z -fYl,) fYl} Part B: What is the tension in the rope while the load is moving. rtn;;l.. We.. us"- our fo r (YI, J 0 I ve, - '" J vo..lve. for ~ {)... (.pro nt {Jo,r fA) 1-0 1.:= oj,t-()..il'jtl0- (Ij +- ~ Cot"" fY1 0 (m:>. - tn, )) yYlJ rm,;). t' cJ.. e nom" +m(;1. n ~ 1-0 /' T= (Yl1:J( ~-r m:t yYl1 - ~) -+-m-;;. I::=. ;(~ f'Yl,tn (YI) r rn do.. - Problem 5.26 An 85-N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 mfs each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and floor. o,().. ~t t W infj +n e .fr--ee- boJy diO-DTelm F ~n + Fx Fx rn~ 'r~e, S urn e~ t) 0f r he., .po f'C.5> ~n-Fy in + hey -m~ -d~('eC +,~ 0 JI1 (hv,s-r a-I zero~ ZF=D::::- +},~ )( eli reL Tn ZF Y'I' (- o. 9 ~1js~) ::=~ F:;; C};o n) We-OI~ +0) Is ds)o ( n e.lj a. + iv ~ b e ca-llJ e.. +h e h ox w j y\ ~ eto W f1 ) m (-D. C[Y>1/s~):::: Fx - ~FN A. ~or~ CDn CiS ely rJ i V i' eX i n Cj ~~ e. ~ F n::::- F y rY'l 'J ( >I .fo rce5) y.Fn = I), n- 'fYIi' F x (X .forc.f..5-) u oJ; 11s .f-o 5 0 J v e., +- 0 r F n (j _ {!JiI tt "ls~ m Yh ~ J7;: - - + F-x' -rF~ fJv~ )n . number'S . )A -=- 0 ,'~S 2> Problem 5.28 Stopping Distance. Part A: If the coefficient of kinetic friction between tires and dry pavement is PI<, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling atu? The.. .+ e r C.e. Ju e... "r 0 .~ riG t i 0 1\ I :> ; f -; - Fn y :::. y. rn tj -/ jA ':J;::: 0( (cA e..- o..c c e }e.r o.:hOJ1 otue. +0 .friel-ion) u.si 113 hOi ~ emet+i cs V-:J.=-;Ao<.6.X V ~ ;). ( )J-~) 4x=- ~ 'K v~ - - JY"J Part B: On wet pavement the coefficient of kinetic friction may be only ]l... How fast should you drive et. on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.) 'l. L\X;;:. Y ':2jAK ~ 1-6 :s f-o f IS rAe 5 Clme d.. /~ +- 0vY\ ce. (}n we +- pOvlleftl ell.+- DX'= SDI (Vwe.:t)"1. ;).Y~W(J;t ve .fOY Vwe,t :h 3 V :l. -~ -/'~ v'":l.. ~ V we. t =- /ywet 9f /jA~i -- V J .;.- YWi-; )J'-h Problem 5.62 Block A in the figure weighs wland block B weighs W2. The coefficient of kinetic friction between all surfaces is fJk. (a) (bl Part A: Find the magnitude of the horizontalforce speed if A rests on B and moves with it (figure (a)). F necessary to drag block B to the left at constant t:f AVV\oves OLI wi+-A i3 ~ ~ -1-Ae.. friC tion .Fo('ce; K fiJ ,.--, + f~ (W, -t'W~)y 6\k L 0 " sf (j. 1\ ~ rf{) :J t Sf e e cA .: F:::;f I r: :::;;. 1nef.a:. c.+- i 113 bo tl 0 tY\ (lfV) +- w;}.) .Fc F I'\J Part B: Find the magnitude of the horizontalforce speed if A is held at rest (figure (b)). necessary to drag block B to the left at constant lj 0.. FrictiOYlctl (;)n .force afwl b 6 t h the- -/-0p o.F- (3: f{Aon!3) :::; JA", WJ :0 Jed so 0" Gtf"lw,J) Y 1'\ (\1</ I + W:l) IDl"ce J5 -the necessoJ'j F:::: PM ('JW, +W~) Problem 5.80 t~r-,. Losing Cargo. A box of mass rests on the flat floor of a truck. The coefficients of friction between the box and floor are 1l5and fJk. The truck stops at a stop sign and then starts to move with an acceleration of a. Part A: If the box is a distance xfrom the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? -f. tn~:r'F Ftl f1\0- L e+ 0< be... +Ae.- O-cce./erat?oYl of }h-e, h 0 X j n f'A t!- f-r u 01'< '3 f r o..-rfl rn~ e.. mCL-m3)AX Z'F::::-m~:::.- 0< ::. ~ - 5 jAf'<. r f' +A e.. boX)S 0- d. i s +-a nee. )< I- ,,0 m +- A e. b c>--G 1'\ :l- X-:;. !; D\ ~ d.?( t?::=. Y;z.. (CA - <j j.A /-l,) t fo / a -J)A,~ 6 lime.. (A ?-, f~ /1 0 v i- Part B: How far does the truck travel in this time? VLL r in :5 +h r s f h e- -rr lJ G J~ -fro- \I 0 (J J' ~ /'d-. -t ot~~ /; at 4- I~ ( 0- o....?C r:Jp.. I~ ) - 0. - o.,)AK Problem 5.94 Banked Curve 1. A curve with a 120-m radius on a level road is banked at the correct angle for a speed of 20 mls. If an automobile rounds this curve at 30 mis, what is the minimum coefficient of static friction between tires and road needed to prevent skidding? " V~ We. t\no w .tho..+ xr fhe ;;to nv's ) Ce yYl (AX;;' - R ~ ro..cJ.i~1 o-.c.c,elercd--ion is .tl\e horiz.Oi1 M,cd rrovidecJ.... eYl hy +t!l pon thYlOW 6t the. nor .ft> r C e. v"'l. A - fY\ VY\ 0...'::: Fn .s i n f- .... R v' ~-;;; we..- Fn ;. ~ C.05,!-S p(tj +-Qn ,8;; \I~ I-()... . ::;;-0\". n;3 J Sol -" -/ ~:> MG ran ve. for 0 13 1f we... w;Mvse I Ctdc1'+' (fR)J +h;.s ra-sul'" 5Don! i;~+ ~ mls . -rher~ .f ore,e., +0 }4.-e~p t'h e., / musf COf /If he OvV\ IOlla ..,...,iC1-i0I10-1 r 0/'1 no t e. HCl-t -rro..C h: X W l.- }It()..v <I- e.- t-o s v II< V ec.: r " ,..5 oLio,,:) F"Sln(3 Y o... ps re-sjJR--cf-/ ;/eJv~ , fJ Ix} .~ Fy ~A -~ y: .:3 uN' Fn y 0 ('(L A F- +A.e. .Ff -/- m J f( t:.JYYI rOJ1e/ih !::.!. -- 5 " 0 ) . efc/c.d.:l.' -the.. -(he,. (' CA-cJ I o-c...c elef"o-r,\'c>)I1 : , oJ~ =: ~::J 0- r o.cJ.. ". ~.. ::I . .;J. >:) +-0. n/.3 C I ea-n ;1\ 3 v P fA es e VO (Jod-,,, n"-: R A I: n ;x-: F'n Sl' I' j3 -i-y F" c l>sj3 ~ ~. :l 5 m~ +o--nj3 [, Y\ ((J S; "(2 +-" "- res" II- o.. ove) b 9"~ F;, COS? -r FnS;' 11)3 ~ m:) .qL! fj)'\ e-CYn 1-; n v ecAU5 ~h e- ?re v i 0 ,p CA-(3 e I 'l,N'8 f-6U I1 d. fY\ Fn .s i 11,13 ~ 'r y- Ff) C-6Sj3 ~ :- 'J,::< 5 ~ j +~l'j3 ~t -f~ )( fr16t- ro--A r ~ n C,OS/> - jA Fn .s j ;'l P :::: 3 yY) - .fA "j F .A " Y here C).fl F('0 IYI y +0 c> l.-l/ CO u-I ul .s '" I v' e.. h y subs h' f 'fu D l-; yt (j ,'-f- you "'" + f-"e... Jp:.. e'?J h0 1A/ Ii.-" e. r :;:;/). '" 0 0 i '" 3 oJ.. , II' ,d. e. Ii (s" _ vc.<..riol1 Y) by f'he d.~ j3 ..-~ 4Ft;C 05;3):; ~ (;J. . .2. 5 .1-0-11;3) .-'- % (Lc>Sj3 e.. h CA-V e. -ys" >'tfJ ) ~ ~ 1'\b w' W' .s j r1;e, oj- r- C-05j3 ~ ~~;2. .2.;;2. 5 +~ /3 V\ ~o>j3 -ys;Y'l;8 Solve -Po T' r: .r C03j3 ~ y .y~.:J. S T 11f3 cl,V)cl~ 5 f-o.-nj3 Lco.sj3 -ys;;;I3) :{.;;'5 TD-nj3 ~ (c 0:5;3 s; fY\{' ~ fD-Vlj3 s;,;13);:::. co~;3 - ~;~ by ~ co 13 (,.f.y : C-'o 5j3 c o~;9 (":J. . .). 5 - J-A- ( I +- ~ _ ~ 5 +o.n)3).::: . In F- I) -fa Yl /3 o-.Ity u... ~ .;r 1./01. -;----~ 5 +O.Jl/3 :J.. 5 fa Yl "/-3 ) wi -"/ +h o..f'c. -I-a.n ~ (I ..,.J.. ; i.s 13 ==- (~"I ~Rj '(:;r-rAe- s,;fn~r,fjCo.tr'on no+ J sfr;c..fly / l1ecSSo..,ry Problem 5.98 You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with a mass of msuspended from the ceiling of the bus by a string of length Lis found to hang at rest relative to the bus when the string makes an angle of 0 with the vertical. In this position the lunch box is a distance R from the center of curvature of the curve. What is the speed v of the bus? DAAW IN';:' Ff\!3- BoOY D..z::AGPA-t \1/ e, c 0J'1 w r ; e. cJ.ow n r eledi 1\ + 0 u (' ()..ro-cl e. C6 uo--+)' 0 JIl .s f (5 (' c,e..s m 0. t'Yl fit 3 .f- he- a-J...:::" :=;, Y1 T S i () e S -r cos. e (Y1~ .rCl e:= 'We.. r a.cJ.. m:J wo...rd-t-a so Ive. P1'o..rD-cL .port!J tan 8= P'Y3 we..- Ii110 '(0-1"1 W v:l.. arO-{)... - A v~ e ::- f{j n e.e..ot W e.., ! .+- 0 V ~ /fj 1{f 0Jl e 7 S 6 Iv e.. .fo r v: IM~ ::J (" 'C..; r 'J -em 'C -eJ W -:::: -e::J (l? 'J W:: ::i (:=J ~ ~ J '7 0 S"'W ..A df -;:: lJ ! -e j~! +"Y.ll( /11{ :::J :=' "<:. ~ "C YJ -e 'c J 'ell W )'1'"'Ovf -0 wt : 7r)J of y-+ !NI uo S1o) 0\j ?~ 'y/ ?"'Q.c Problem 5.112 A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 13.0 m. The physics major has a mass of 70.0 kg, and his motorcycle has a mass of 40.0 kg. Part A: What minimum speed must he have at the top of the circle to lose contact with the sphere? if the n tires of the motorcycle are not -[he, +/001' frtJv;cJ..e-s F~ .M 0- f D I' ce.o +h e.. eye II's+- B-f A v:1. 0.+- fhe +-0 p 0 .f}-he loop ~ ;rif ?:.73 rAU5 IImin > )8; (J.f /V/Ln1?S Part B: At the bottom of the circle his speed is twice the value calculated in part (A). What is the magnitude of the normalforce exerted on the motorcycle by the sphere at this point? 0-+ -/-). e-- b c tt-<~p1 +h,<-- circ./-e.. '" ra.V'+Y 5 he, ~e..e.15 +h.e.- +01' ce dueforce +0 h'5 of- r 0 .}-o.. + f 0 11 a. f yv?oi-i"on PL uS -rAe. Fro .1-0.1 -- __ (;).. Ym; 11) ::l.rn ~ ~ -+- m~ ~ Fr()....rA. F3 l-l .:J, 'F.J- 0 {--o.-I :;;.. rn( V i"1i n A +8) rnt !~ +~) A.,.. /\... 5m~ Problem 5.120 Moving Wedge. A wedge with mass Mrests on a frictionless horizontal table top. A block with mass 'mis placed on the wedge (the figure a). There is no friction between the block and the wedge. The system is released from rest. Setup --..... F (u) (Il) We- S+w+- o..~ USUCLl Joy cAf'l\\,v,;(!c-) L"'- fr~~ hocAY dio.[J,(.lJ'Y7 ~,,~ mE) fYl 0\ C050( ex- :;:; S jf'"! r: ~ ax-;. OlJ ::= 0- CO s 0(;:::- Ffl s r rl D( o-.s)" 1'1 c:I\ R e..c.cd' -Pc (' t e, fa rce.h()ve. M (A)< PCA; 6 i"! J'Yl 6l r.~ .. ~.' +h~. \IV' b l (') c'M ~ (J +h e.Ej j/)" eJ 3 e ., r ex c r .f-s 0c e. F l/ S 'fYI'1 .Q.coua-\ -r u cA <? - YV1 CAy< f{A~ /JjTicP; ~;;, 1\-1 I'J1 _ -r:" 5 ;'1 0( '7 --} ()..)t' -::;... - A / 0". A 1 rAy 1'1 A:: - F" .J; '1 <>( n1 i-y ;:- F (1 5; I) 0( 0<.-;11,~ m ()... ;;. y F/1 Cos Part A: Calculate the acceleration of the wedge. o..-x-A ;1 ~ )IY\ ,~ ~ -l:o.f\ 0\ :::::? Ot 1Iv ::: A fa.n o{- a x +'1 II 0( I c.. 0 S v( - m 3 -::::: m It + c..n 0< - I~ 0\ -la no(:::. fA (M +- M) ;. +v-. Y) eX. 0.-/ -- J'YI 0...)1 ~ A A::: -3m -reV10\ (Iv{ tJl1) +- J.4/-1 t{f\ rJ.... Part B: Calculate the horizontal component of the acceleration of the block. Sif)ce..- fY\ (). ~ ::::- - /vt A () ,X-v... ( It{ :1 /'-\ + tyl) fa. J) 0( -t- M/" /" +- (;L11 0\ Part C: Calculate the vertical component of the acceleration of the block. ()..y :::: ((A y - A) +0 Yl ()\ + he.- r f- -For e G\::; 'I __--':.Vl--------( tM. t Y\") f- ().)1 q +- tJ... CA)II 0( - 0\ (M'" J/}1 ) +Cln q /+ Problem 6.58 Chin-Ups. While doing a chin-up, a man lifts his body 0.40 m. Part A: How much work must the man do per kilogram of body mass? VI 0'-" JA, :::: -rhef?. Fo rc e. I v cA\ stc,\.Y'C e VII:::: yY1 J.-o +CA-' J--<'; I 0 \AI 0 r f'. S j (" Lf yY\ ) r j ra 111 " ~ 3~ 1).. J IAI'J -I Part B: The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a OAO-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical70-kg man with 14% body fat is about 43%.) !IV~ +()./~ e h '1.5 J- (.) +0... l We>r IV I"'; I0 ~ ' cA 1/)'1 o-n d. d; r/ d *- by fA e WOr K / J-<.; /1fj ':/-0.111 (YT1 o.c )ML).sc}c: 3..9,.J..J" " K!j v':iC I.e) 70-:) X / DO 1;; 6 % Part C: Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 J of work per kilogram of muscle mass. -rAe.. ch; IJ 01.0 <2.-5 Wli'") ~ 0+: 3), =- 1~10:J ~ 1ft U_'~; /, c;b..;} /~ Cj () ~,x/oo ?6J ",/28~- Problem 6.56 Rotating Bar. A thin, uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (Hint: Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass dm and integrate to add up the kinetic energy of all these segments.) L~+ f A.C! J1 t1. :::. +6 t- I A';; ~L 0-- m G\..S5 ~ alA:::' ~ cJ?( L ( n;,\ y m{ASS ChuJ'l/"j FjAr! CA.n~uIClr w::;. v~/oc;+-y: ro-ol i CH\5 ::::::- S (:e V () -311 35 /0 -1/ 3 re-vo,i~ti6/'1 ./ S roJ- We.. f1 Y\ 0 111/ I 1:2. YY\ V ~ --? ) /~ m (r w )'J.. f.iL!/ need TIl ~ 7 L o Wi!- OW) +0 ifr\,Le~ro.+e.. over +Ae- """"1' ~(YY\0e)5ChJYlGI"'S\\ K E; J;;,. Jm ~ J v ~ --7 W L J.. JA o.re c-l;\ (rw)'- --"I (O!l Jf~ It: w'" ,-" rJr I /~ __ '-') ~ UJ ::l,j s -ra.n 1--s.: 0 I;). 1M. -C I r '1-1r -::::.- (j.) ")..{13-)L u~rJ\ -C :3 o S-j m pI, fj j "'j M . - L, J., w '"' ..... b - II I~j {;A 0 y~(lfrr _)-').z -= - J '7 '7 J Problem 6.76 The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. Part A: Calculate the speed with which the ball leaves the barrel if you can ignore friction. Ecouo.+e.. -th.e.... ~l. IA. lAJorJ\ xoJ.;::. 0+ yY'\ -lhe V ~ Srrlll~ hJ .file X,'/I(Jt,C bl1r?,"'3Y 1;;2. 2. 14 v~~ r;; /.6 v::- V d) r:K 'X 0 :0: J :03 ,..\~ '(DO Y;6- ['0 b "')::- __ b, q 3 --- /5 Part B: Calculate the speed of the ball as it leaves the barrel on the ball as it moves along the barrel. if a constant resisting force of 6.00 N acts \tv'+-0 +r.t I h 0 ltV J n G Iud e 5 f.,,- / C + I 0 1\ 1-\ E ::::.1;).. M X 6 ). i\x; ).. - f )<:0 f' ,x 0 Jf -;;;"\0 I; .....),. I J.- I'Y\ V ---= )/J.. V :2..::: l~ :mJo -7 \1 ::::. V n;-,xc> - ~ J.;)f m Xu ~ V;:: LI-. ( '~ Part C: For the situation in part (b), at what position along the barrel does the ball have the greatest speed? (In this case, the maximum speed does not occur at the end of the barrel.) '1ha 1\ C?)r-t2-o.-te-.st 1I') 5pefliJ... OCGU$'3" 'w h e-)'I +he- n c- +- +-urc- t ),e.. b O.Ile I h)<' IS zerO r~fr;nJ::: ,:,. . t'Jon:=;" -rflC- (' +"' '7 ~ -F ==- J"';X X;::; - -;:;0, JA 0 /5 ;1') P J I) Cj -fh j 5 i 1'1 +0 .(; rcA. ( I h e VQ.,io ~) Vm~)<' =0 JI'ml c" f Y 'y' ) Xc> ."1 '---X - M :1. f (' --- 0. f f-n," roi/1 t..s Xo-/Y/ V VV\ lA.X ==- ~ -' J c;?'- 0 )/1/\ / J 5 Part D: What is that greatest speed? V mey< .:;. 6" ')..0 r'Vt/-.5 Problem 6.82 Consider the system shown in the figure. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the tabletop is JJ.k 8.00 kg = 0.250 . The blocks are released from rest. Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m. J~ E::; \tJ Uj ,-"-,) b lac f')) ~ r LA ~ j .- Wy (8 I~J bloc/',) 6.00 kg +- ! rn W 0 r )~ 5 0 n +h ~ b j~ J b 10 C /~ IY1 \IV.::;. ~ j h -;:::. I-1 (, J .' 1. Y/5 r .~" /, ') f ric h C h cl. 0 ~5> 6. Vl 0 11,. i r1 of A i!. 0 jJ rf' 6 s;j e ol; r <2C + /0 >1 \AI .:::::. yFn X -:::: Y-VVlJh j5; (. d- 5 ) ( g M j ) Cj ~ .rA e.. 1-0 i- 0.1 Warl"" [f:, fJ,j + --;.25 C5 J-\~ ~ ~ h ""-);II 1-0 t ~ 5g~8u !J g- - g J~ ~J- (lM I "yn d- ) \/ 'l v.~ ~..q tv'/S Problem 6.93 Power of the Human Heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American female (1.63 m). The density (mass per unit volume) of blood is 1.05 x 10~kg/m~ . Part A: How much work does the heart do in a day? 75DOL 0 W\ .3 t> I.. 0 5X 103 )~ yV)3 ~~ . I006L . 7J'7 5 Mj +he.... wc> r I~ ela ne- ; s :. (r, 8 ~. ~) ( 'w =- m J h ~ ( 7 0 '7 5 i~J) I b 3 rn) # lW;cL~bx/{)5J Part Bo' What is its power output in watts? ) do- Y ~ J..j 6 W many 6 s e co 1\<0; iLr e 66'-100S ill ()... 36005 hr :J..!-jAr ~y Q('~ Wo..tr-) de-}" neDl 0--5 )()uJes /S(LC fA LJ S 5 I- 167'10 oblf005 J= II, C- 'f b Wo.-tt.s] ...
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