problem07_49

University Physics with Modern Physics with Mastering Physics (11th Edition)

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7.49: a) 2 2 other 1 1 U K W U K + = + + Let point 1 be point A and point 2 be point B. Take 0 = y at point B. , mv mv mgy 2 2 2 1 2 1 2 1 1 = + with m 0 . 20 = h and s m 0 . 10 1 = v s m 2 . 22 2 2 1 2 = + = gh v v b) Use , U K W U K 2 2 other 1 1 + = + + with point 1 at B and point 2 where the spring has its maximum compression x . s m 2 . 22 with ; 0 1 2 1 2 1 1 2 2 1 = = = = = v mv K K U U , kx mgs W W W f 2 2 1 k el other - - = + = μ with x s + = m 100 The work-energy relation gives . 0 other 1 = + W K 0 2 2 1 k 2 1 2 1 = - - kx mgs mv μ Putting in the numerical values gives . 0 750 4 . 29 2 = - + x x The positive root to this equation is . m 4 . 16 = x b) When the spring is compressed
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Unformatted text preview: m 4 . 16 = x the force it exerts on the stone is N. 8 . 32 el = = kx F The maximum possible static friction force is N. 118 ) s m 80 . 9 )( kg . 15 )( 80 . ( max 2 s s = = = mg f The spring force is less than the maximum possible static friction force so the stone remains at rest....
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  • Force, Friction, static friction force, 9.80 m, Wother, possible static friction

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