108&109 - − ∆ + − = ⇒ 2 2 2 2 2 2 5 . 1 5 . 1 1 5 ....

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2.108, 109 Spring, 2004 2.108 109 Given: In the configuration shown, the spring is initially uncompressed and the speed is 7 m/s to the left. For 108, the coefficient of friction is 0, and for 109, the coefficient of friction is 0.2 Find: What is the velocity when after it has moved 4m to the left? Solution: Using the work energy theorem KE W = block on the (1) () () ( ) () ( ) ( ) 2 2 2 0 2 / 7 81 . 9 / 200 2 / 1 ) ( 2 / 1 s m v kg v v m KE f f = = (2) A generic f.b.d is shown to the right. For both problems, since the block is not moving in the vertical direction, the normal force and weight do no work. For 108, friction is = zero, so ( ) () ( ) ( ) () m N m N m m N k W W o f = + = δ δ = = 9 . 306 77 . 2 40 0 5 . 1 4 5 . 1 / 80 2 / 1 2 / 1 2 2 2 2 2 2 2 spring block on the Putting this together with (1) and (2): s m v f / 34 . 4 200 / 81 . 9 * 2 * 9 . 306 49 = For 109 friction is not zero. N f µ = , so friction spring block on the W W W + = Summing the forces perpendicular to the plane gives () θ + = θ δ = θ = = θ + cos 5 . 1 5 . 1 cos cos 0 cos 2 2 s k mg k mg F mg N mg F N s s () 2 2 5 . 1 5 . 1 cos s + = θ () () () + = +
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Unformatted text preview: − ∆ + − = ⇒ 2 2 2 2 2 2 5 . 1 5 . 1 1 5 . 1 5 . 1 5 . 1 5 . 1 5 . 1 s k mg s s k mg N ( ) ∫ ∫ ∫ = = = = = = ∆ + − − µ = µ = = 4 2 2 4 4 friction friction 5 . 1 5 . 1 1 5 . 1 F x x x x x x dx s k mg Ndx dx W ( ) ( ) ( ) ∫ ∫ = = = = + + − = + − − − = 4 2 2 4 2 2 friction 5 . 1 180 80 2 . 5 . 1 1 / 80 5 . 1 200 2 . x x x x dx x N dx x x m N N W ( ) ( ) ( ) ( ) ( ) J J x x dx x m N W x x x 5 . 125 5 . 1 ln 180 * 2 . 64 5 . 1 180 2 . 4 16 4 2 2 4 2 2 friction − = + + − − = + − − = = = = ∫ Now, putting this in (1) and (2) gives ( ) s m v f / 56 . 2 200 / 81 . 9 * 2 * 5 . 125 9 . 306 49 ≈ + − = W N f F s θ...
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This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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