problem07_06

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

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7.6: a) (Denote the top of the ramp as point 2.) In Eq. (7.7), J, 5 . 87 m) (2.5 N) 35 ( , 0 other 2 - = × - = = W K and taking 0 1 = U and s, m 25 . 6 J, 147 ) 30 sin m (2.5 ) s m (9.80 kg) 12 ( kg 12 J) 5 . 87 J 147 ( 2 1 2 2 2 = = = ° = = + v mgy U or s m 3 . 6 to two figures. Or, the work done by friction and the change in potential energy are both proportional to the distance the crate moves up the ramp, and so the initial speed
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Unformatted text preview: is proportional to the square root of the distance up the ramp; . s m 25 . 6 s) m . 5 ( m 1.6 m 2.5 = b) In part a), we calculated other W and 2 U . Using Eq. (7.7), J 491.5 J 147 J 5 . 87 s) m (11.0 kg) 12 ( 2 2 1 2 =--= K . s m 05 . 9 kg) 12 ( J) 5 . 491 ( 2 2 2 2 = = = m K v...
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