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Unformatted text preview: d replaced by L and the final v 2 term being the unknown (instead of assumed zero): 1 2 m v 2 = K C ( mgL sin + k mgL cos ) . This determines the speed with which it arrives at point B : 2 2 2 2 (sin cos ) (4.98 m/s) 2(9.80 m/s )(0.75 m)(sin30 0.4cos30 ) 3.5 m/s. B C k v v g L = + = + = 64. We will refer to the point where it first encounters the rough region as point C (this is the point at a height h above the reference level). From Eq. 8-17, we find the speed it has at point C to be v C = v A 2 2 gh = (8.0) 2 2(9.8)(2.0) = 4.980 5.0 m/s....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
- Spring '08