problem06_05

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

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6.5: a) See Exercise 5.37. The needed force is N, 2 . 99 30 sin ) 25 . 0 ( 30 cos ) s / m 80 . 9 )( kg 30 )( 25 . 0 ( sin cos 2 k k = ° - ° = - = φ μ mg F keeping extra figures. b) J 5 . 386 30 cos ) m 50 . 4 )( N 2 . 99 ( cos = ° = Fs , again keeping an extra figure. c) The normal force is sin F mg + , and so the work done by friction is J 5 . 386 ) 30 sin ) N 2 . 99 ( ) s / m 80
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Unformatted text preview: . 9 )( kg 30 )(( 25 . )( m 50 . 4 ( 2-= +-. d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero....
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This document was uploaded on 02/04/2008.

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