Then the work energy theorem requires that wr 11u mgyb

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Unformatted text preview: acement along each slope is == y j sin () (measured Assume that the surface is horizontal, so th~t there are no changes in the block's potential energy ~11U = 0) and the force of friction on it is !k ::::: -ILkN;::::: ~I-'kmg (opposite to the direction of motion along the x-axis). If the block crosses the entire region from al ::::: 1 m to X2::::: 2 m, the work-energy theorem deman1s that Wnc = - . or . Xl ~ 1 l a X2 I-'kmg dx ::::: 11K = ~m(v~. - vi) '. 1 1 ' j 2 e vi - v~ = 2g (ax2 + bx +c)dx b 2m 1m Xl = 2g l 3X3 + '2X2 +CX = 6.53,'! m /s 2 2 (The given values of a, b;and c were used. ) liThe minimum speed at the start of the region isi:the value FIGURE 8-39 Problem 59 Solution. .~ CHAPTER8 137 ~bottom of both slopes), so that work done by, em each slope is 11. = (_p,'!:axmg cos fJ) x 'If J= _p,raXmgy cot fJ. Then the work-energy :: ).becomes -mg(YA - YB) = _p,;:axmgx Q~+ YBcotOB), from which p,'!:a.x be can ax = (950 - 840)/(950 cot 27 + )iecl: p,r )35)= 0.036. (0.19 kg)(9.8 m/s2)(0.85 m) = -0.427 J. Sin...
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This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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