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

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6.4: a) The friction force to be overcome is , N 5 . 73 ) s / m 80 . 9 )( kg 0 . 30 )( 25 . 0 ( 2 k k = = = = mg n f μ μ or 74 N to two figures. b) From Eq. (6.1), J 331 ) m 5 . 4 )( N 5 . 73 ( = = Fs . The work is positive, since the worker is pushing in the same direction as the crate’s motion. c) Since f and s are oppositely directed, Eq. (6.2) gives
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Unformatted text preview: J. 331 ) m 5 . 4 )( N 5 . 73 (-=-=-fs 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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  • Force, Normal Force, Friction Force, net work, gravity act

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