166_Physics ProblemsTechnical Physics

166_Physics ProblemsTechnical Physics - 168 P6.25 Circular...

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168 Circular Motion and Other Applications of Newton’s Laws P6.25 FF m a g max =+ = 591 N m a g min =− = 391 N (a) Adding, 2 982 F g = N, F g = 491 N (b) Since Fm g g = , m == 491 50 1 N 9.80 m s kg 2 . (c) Subtracting the above equations, 22 0 0 ma = N ∴= a 200 . m s 2 P6.26 (a) a rr = mg mv R m R R T g R T T R g F H G I K J = × = 2 2 2 2 26 3 2 4 4 2 637 10 507 10 141 π . .. m 9.80 m s s h 2 (b) speed increase factor F H G I K J = v v R T T R T T new current new current current new h 1.41 h 2 2 24 0 17 1 . . *P6.27 The car moves to the right with acceleration a . We find the acceleration of a b of the block relative to the Earth. The block moves to the right also. F ma n mg n mg f mg a m gm aa g yy k xx k b b k
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