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problem07_61

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

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7.61: a) For a friction force f , the total work done sliding down the pole is fd mgd - . This is given as being equal to mgh , and solving for f gives . 1 ) ( - = - = d h mg d h d mg f When 0 , = = f d h , as expected, and when mg f h = = , 0 ; there is no net force on the fireman. b) N 441 ) 1 )( s m 80 . 9 )( kg 75 ( m 5 . 2 m 0 . 1 2 = - . c) The net work done is ) )( ( y d f mg - - , and this must be equal to 2 2
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Unformatted text preview: f , , 1 ) ( ) )( ( 2 1 2 -=- =--= d y mgh y d d h mg y d f mg mv from which ) 1 ( 2 d y gh v-= . When = y gh v 2 = , which is the original condition. When d y = , = v ; the fireman is at the top of the pole....
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