coop59 - 1 P207 Fall 2004 Coop Solutions Coop Problem 5 1(a...

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1 P207 - Fall 2004 Coop Solutions Coop Problem 5. 1. (a) The net forces parallel and perpendicular to the incline are F par = F app cos(30) - k - mg sin(30) F perp = - F app sin(30) + N - mg cos(30) and since there is no acceleration the net forces are zero. We multiply the second equation by μ k and add them together to get 0 = F app (cos(30) - μ k sin(30)) - mg (sin(30) + μ k cos(30)) F app = mg (sin(30) + μ k cos(30)) cos(30) - μ k sin(30) = (50)(10)( 1 2 + 0 . 2 3 / 2) 3 / 2 - 0 . 2 / 2 = 439N Also N = F app sin(30) + mg cos(30) = 653N We might also have chosen to write the horizontal and vertical forces instead of forces parallel and perpendicular to the incline. Then we would have F x = F app - k cos(30) - k cos(30) F y = - N cos(30) - mg - k sin(30) Setting the net horizontal and vertical forces equal to zero gives the same result. (b) (i) W app = F app cos(30) d = (439)(0 . 866)6m = 2281J (ii) W w = - mgd sin(30) = - 50(10)(6) 1 2 = - 1500J (iii) W f = k d = 783 . 6J (c) Δ E mech = mgd sin(30) = 1500J 2. (a) Δ W = 1 2 kx 2 = 1 2 (320)(0 . 075) 2 = 0 . 9J
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2 (b) E diss = F f x = mgμ k x = (2 . 5)(10)(0 . 25)0 . 075 = 0 . 46875J (c) 1 2 mv 2 = Δ W + E diss = 1 . 37J. Then v = 1 . 05m/s 3. (a) The total energy is
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This note was uploaded on 12/18/2011 for the course PHYS 2207 at Cornell.

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coop59 - 1 P207 Fall 2004 Coop Solutions Coop Problem 5 1(a...

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