PHYS
solutions to assignment4

solutions to assignment4 - 4.2 The acceleration given to...

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Unformatted text preview: ' 4.2 ' ' The acceleration given to the footmd from ”—1; 7-304- arias-7 V'V 10m/ - a'_"E—Q"—ffiois_q"5°m/32- Then. from Newton's 2“‘1 Law, we findF— ma _ (o .50 kg)(50 W32), 25 N. 67.- ,, --.._..J _... -_-. 4.8 (a) We resolve the forces shown mto ‘ - F =400N FR: 122'N 789 N The magnitude of the resultant force is found from the orean theoremas FR- (m)2 + (zpyfl- (122 m2} (789 mil-7933:, and tans-Q 1-3-1392; - 6.47, fromwhinh 9- 812“. Thus, the resul- umt force is at an angle of 8.8“ to the right of the forward direction. (13) The acceleration is in the same direction as PR and is given_by _ arVfi/m= 798‘NQ99932,:9;6_6_§113?: _. , _ , , __.__H.__h__,_,,__,,___h ,. _;__-;, -____.. 4.15 (a) A study of the forces on the block that are parallel to the incline reveals that m-Fcosa- mgsine-0,so n F - (2.0 kg)(9.80):‘o‘;2g. - 33.9 N. (34 N) ZFy-N- Bins - mgcoss-O, sothat N- Fsinfi + mgcose - (33.9 N}sin60° + (2 ¥_w__..,,,,, _. _ W," _ 4.20 mum m -1 . _ haveg 1 00kgandm2 5.00kg,we 1 mla-mlgV—V'I: £1123 - T- ngsine. Adding these gives ("11 + Ema-11113 - ngsine. Thus, ”(W g,“ m1 '+ mg 10.0 - 5.00 sin40‘ 3" ( 15 ) 9.80 - 4.43 m/sZ. [Taina- {g \m . 1'11 Inn nf a owes: T- m1 (5! - a) - 10.0(930 - 4.43) =- 53.7 N. 4.28 First, consider the 3.00 kg rising mass. The ' forces on it are the tension, T, and its weight. 29.4 N. With the upward direcfion as positive, the second law becomes T— 29.4. N - (3.00 kg)a. TheforcesonthefaflingSDOkgmassare its weight and T, and its acceleration is the sameasthatoftherisingmass. Caiiingthe positive direction down for this mass. gives 49 N - T:- {5.00 kg) a. (a) the tension as T- 36.81% (b) and the acceleration as (c) Consider the 3.00 kg mass- We have 4.32 (a) a- v; - vg_ 6.00 m/s t We also know that N-mg. Now app -f= m(-1.20 m/sz) 01' But, also, 1" - pkN =- pkmg. (1) (2) a - 2.45 m/sz. y= Vot+%at2 -0». i— (2.45 m/s2)(1.oo s)2 —123 m. + Equations (1) and (2) can besohred simultaneously to give 5-3110 m/ ..-120 111/32. lythesecondlaW! f-m(l.20 111/52). (1) From (1) above, we have . 2 'pknig- m(1.20 m/sz) , so P-k‘ 930 “1/3: (c) x- {r r- (-9——~) t- (Wysm s) - 45.0 m. V+Vf 2 ...
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