sol to assignment5 - Auswges ‘67 Crt 5...

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Unformatted text preview: Auswges ‘67? Crt. 5 fl-SSIeM,ME7-JT‘ 5.2 w= mg = (20.0 kg)(9.80 nth/32) and 196 N = the applied force to lift the bucket at constant w= (PcosB)s= 600x103 J Fs= (196 N) s yielding, s= 30 6 In. , 5.1 2 At the top of the arc, V; = 0, and vx = vex = 40 (203300“ = 34.64 m/s. Therefore v2 = V2}; + v y = (34.64 iii/3F, and 1 KE= 5 va =§ (0.150)(34.64)2 = 90.0 J. 5.2 1 (a) We take the zero level of potential energy at the lowest point of the arc. When the suing is held horizontal initially, the initial position is 2 m above the zero level. Thus, PE= mgy= wy= (40 N)(2 m) =- 80 J. (b)Promthesketch, weseethatat ----"-----“-0 an angle of 30° the ball is : (2.0 m)(1 - c0330“) above the lowest point of the arc. Thus, P134 (40 N)(2.0 m)(1 - 00330“) = 11 J. (c) The zero level has been selected at the lowest point the arc. Therefore,. PE = O at this point. 1 + ”1353* 2 mvfz + mgyr to obtain 0+(0.400kg)g(5.00m)= 2(0.400kg)v132+0. Fromthis, VB=990m/S. (b) At int f0 C ,with theistarting point at A, we again use mez+mgfi= 2me2+:flgy1= andobtain 5.28 (a) Liming m the the mass of the projectile, k be the spring constant, C! be 0.120 in, and H== 20.0 m, we have, using conservation of energy 2_mg_ H_ 2.!0 0211980190. 0') — 2,. _ a chf mgH, or k— _2g_ (0120): 544N/n1. )Here we have ink? =lrngd+-'21,mv2 so that 2 vat/I‘d ~ 2 gd— SMSOlezol—JLWJ 2(9.80)(O.120) =19.7 m/s. At 5.43 Pag’fl'gALtsQQQ-JA—mmg = (1.2): 106)(9.80)(50) ' =5.88x108W=590 MW. ...
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