problem05_p101

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

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5.101: a) The rock is released from rest, and so there is initially no resistive force and . s m 6.00 kg) (3.00 N) 0 . 18 ( 2 0 = = a b) . s m 3.80 kg) 00 . 3 ( s)) m (3.00 ) m s N 20 . 2 ( N 0 . 18 ( 2 = - c) The net force must be 1.80 N, so N 2 . 16 = kv and . s m 6 3 . 7 ) m s N (2.20 N) 2 . 16 ( = = v d) When the net force is equal to zero, and hence the acceleration is zero, N 0 . 18 t = kv and s. m 8.18 m) s N (2.20 N) 0 . 18 ( t = = v e) From Eq. (5.12), m. 78 . 7 ) 1 ( m s N 2.20 kg 00 . 3 s) (2.00 ) s m 18 . 8 ( s) kg))(2.00 00 . 3 ( m) s N 20 . 2 ( ( + = - -
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Unformatted text preview: = -e y From Eq. (5.10), s. m 29 . 6 ] s)[1 m 18 . 8 ( s) kg))(2.00 00 . 3 ( ) m s N 2 . 2 ( ( =-= -e v From Eq. (5.11), but with a instead of g , . s m 38 . 1 ) s m 00 . 6 ( 2 s) kg))(2.00 00 . 3 ( m) s N 20 . 2 ( ( 2 = = -e a . f) s. 14 . 3 ) 10 ( ln so , 1 . 1 ) ( t = = = =--k m t e v v t m k...
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This document was uploaded on 02/04/2008.

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