# ch05-p019 - cos 30 ◦ = 72 N(c When the string is cut it...

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19 (a) The free-body diagram is shown in Fig. 5 16 of the text. Since the acceleration of the block is zero, the components of the Newton’s second law equation yield T mg sin θ =0and F N mg cos θ = 0. Solve the first equation for the tension force of the string: T = mg sin θ = (8 . 5 kg)(9 . 8m / s 2 )sin30 =42N . (b) Solve the second equation for F N : F N = mg cos θ =(8 . 5 kg)(9 . 8m / s
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Unformatted text preview: ) cos 30 ◦ = 72 N. (c) When the string is cut it no longer exerts a force on the block and the block accelerates. The x component of the second law becomes − mg sin θ = ma , so a = − g sin θ = − (9 . 8 m / s 2 ) sin 30 ◦ = − 4 . 9 m / s 2 . The negative sign indicates the acceleration is down the plane....
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## This note was uploaded on 05/13/2010 for the course PHYSICS 12121 taught by Professor Bill during the Winter '01 term at CSU Channel Islands.

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