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problem07_56

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

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7.56: a) The energy stored may be found directly from J. 10 33 . 6 ) J 000 , 58 ( J 000 , 51 J 000 , 625 2 1 5 2 other 1 2 2 × = - - - = - + = mgy W K ky b) Denote the upward distance from point 2 by h . The kinetic energy at point 2 and at the height h are both zero, so the energy found in part (a) is equal to the negative of the work done by gravity and friction, h, h h f mg ) N 600 , 36 ( ) N 000 , 17 ) s m 80 . 9 )( kg 2000 (( ) ( 2 = + - = + - so m. 3 . 17 J 10 66 . 3 J 10 33 . 6 4 5 = = × × h c) The net work done on the elevator between the highest point of the rebound and the point where it next reaches the spring is J. 10 72 . 3 ) m 00 . 3 )( ( 4 × = - - h f mg Note that on the way down, friction does negative work. The speed of the elevator is then . s m 10 . 6 kg 2000 ) J 10 72 . 3 ( 2 4 = × d) When the elevator next comes to rest, the total work done by the spring, friction, and gravity must be the
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Unformatted text preview: negative of the kinetic energy 3 K found in part (c), or . ) m N 10 03 . 7 ( N) 600 , 2 ( 2 1 ) ( J 10 72 . 3 2 3 4 3 2 3 3 4 3 x x kx x f mg K × +-= +--= × = (In this calculation, the value of k was recalculated to obtain better precision.) This is a quadratic in 3 x , the positive solution to which is [ ] m, 746 . ) J 10 72 . 3 )( m N 10 03 . 7 ( 4 N) 10 (2.60 N 10 60 . 2 ) m N 10 03 . 7 ( 2 1 4 4 2 3 3 4 3 = × × + × + × × × = x corresponding to a force of N 10 05 . 1 5 × and a stored energy of J 10 91 . 3 4 × . It should be noted that different ways of rounding the numbers in the intermediate calculations may give different answers....
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