problem04_49

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

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4.49: a) b) The net force on the system is N 0 . 53 ) s / m 80 . 9 )( kg 00 . 15 ( N 200 2 = - (keeping three figures), and so the acceleration is , s / m 53 . 3 ) kg 0 . 15 /( ) N 0 . 53 ( 2 = up. c) The net force on the 6-kg block is N 2 . 21 ) s / m 53 . 3 )( kg 00 . 6 ( 2 = , so the tension is found from N 2 . 21 = - - mg T F , or N 120 N 2 . 21 ) s / m 80 . 9 )( kg 00 . 6 ( ) N 200 ( 2 = - - = T . Equivalently, the tension at the top of the rope causes the upward acceleration of the rope and the bottom block, so
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Unformatted text preview: a g T ) kg 00 . 9 ( ) kg 00 . 9 ( =-, which also gives N 120 = T . d) The same analysis of part (c) is applicable, but using kg 00 . 2 kg 00 . 6 + instead of the mass of the top block, or 7.00 kg instead of the mass of the bottom block. Either way gives N 3 . 93 = T ....
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