problem06_76

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

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6.76: a) Equating the work done by the spring to the gain in kinetic energy, 2 2 1 2 0 2 1 mv kx = , so . s / m 93 . 6 ) m 060 . 0 ( kg 0300 . 0 m / N 400 0 = = = x m k v b) tot W must now include friction, so 0 2 0 2 1 tot 2 2 1 fx kx W mv - = = , where f is the magnitude of the friction force. Then, . s / m 90 . 4 ) m 06 . 0 ( kg) 0300 . 0 ( N) 00 . 6 ( 2 ) m 06 . 0 ( kg 0300 . 0 m / N 400 2 2 0 2 0 = - = - = x m x m k v f c) The greatest speed occurs when the acceleration (and the net force) are zero, or m 0150 . 0 , m / N 400 N 00 . 6 = = = = k f x f kx . To find the speed, the net work is ) ( ) ( 0 2 2 0 2 1 tot x x f
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Unformatted text preview: x x k W---= , so the maximum speed is m/s, 5.20 m) 0.0150 m 060 . ( kg) 0300 . ( ) N 00 . 6 ( 2 ) ) m 0150 . ( ) m 060 . (( ) kg 0300 . ( m / N 400 ) ( 2 ) ( 2 2 2 2 max =---=---= x x m f x x m k v which is larger than the result of part (b) but smaller than the result of part (a)....
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