problem07_23

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

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7.23: a) In this case, J 000 , 625 1 = K as before, J 000 , 17 other - = W and J. 900 , 50 ) 00 . 1 ( ) s m (9.80 kg) 2000 ( m) 00 . 1 ( m) N 10 41 . 1 )( 2 1 ( ) 2 1 ( 2 2 5 2 2 2 2 = - + - × = + = mgy ky U The kinetic energy is then J 557,100 J 000 , 17 J 900 , 50 J 000 , 625 2 = - - = K , corresponding to a speed s. m 6 . 23 2 = v b) The elevator is moving down, so the friction force is up (tending to stop the elevator, which is the idea). The net upward force is then
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Unformatted text preview: , kx f mg N 400 , 138 ) m 00 . 1 ( ) m N 10 41 . 1 ( N 000 , 17 ) s m 80 . 9 )( kg 2000 ( 5 2 =--+-=-+-for an upward acceleration of . s m 2 . 69 2...
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This document was uploaded on 02/04/2008.

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