148_Physics ProblemsTechnical Physics

148_Physics ProblemsTechnical Physics - 150 P5.64 The Laws...

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150 The Laws of Motion P5.64 (a), (b) Motion impending 5.00 kg n = 49.0 N F = 49.0 N g f s 1 15.0 kg n = 49.0 N f s 1 147 N 196 N f s 2 P fn s 1 14 7 == µ . N f s 2 0500196 980 .. N N af FIG. P5.64 Pf f ss =+= + = 12 14 7 98 0 113 N N (c) Once motion starts, kinetic friction acts. 112 7 0 100 49 0 0 400 196 15 0 196 0 100 49 0 5 00 0980 2 2 1 1 . . . . . . N k g ms k g 2 2 −−= = = = a f bg a a a a *P5.65 (a) Let x represent the position of the glider along the air track. Then zxh 22 0 2 =+ , xzh =− 2 0 2 ej , v dx dt zh z dz dt x 1 2 2 2 0 2 . Now dz dt is the rate at which string passes over the pulley, so it is equal to v y of the counterweight. vz vu v xy y = = 2 0 2 ch (b) a dv dt d dt uv u dv dt v du dt x x y y y = + at release from rest, v
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