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Unformatted text preview: 0 T = mg cos = (0.50 kg)(9.8 m/s 2 ) cos75 = 19 N (c) F x = T sin = ma x = m v 2 r = m v 2 L sin v = T L sin 2 m = (19 N)(2.0 m) sin 2 75 0.50 kg = 8.4 m/s 12. (a) C (c) F y = N  mg cos + F sin = ma y = (b) N = mg cos  F sin (d) f k = k N = k (mg cos  F sin ) Fx = mg sin + F cos  f k = ma x = = mg sin + F cos  k (mg cos  F sin ) = F = mg k cos  sin cos + k sin = mg k tan 1 + k tan (e) E t t t w +x +y T w (= mg) +y w grav Earth F tension string N contact ramp f k ramp friction +x (1 point for each force correctly identified by interaction and body providing it AND in the correct direction )...
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 LECLAIR,A
 mechanics, Force

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