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Unformatted text preview: h ic e8 e k# 7 i b iC a aha 7 Problem 28 ( a ): The force is in the xdirection, but the proton moves in the ydirection. As these are perpendicular, the work done is 0. ( b ): This path is in the xdirection, so the work is W = integraldisplay . 30 . 10 F dx = integraldisplay . 30 . 10 x 2 dx = x 3 3 vextendsingle vextendsingle vextendsingle vextendsingle . 30 . 10 = . 10 J . ( c ): The path is same as in part ( b ), only backwards. Thus the work done is 0 . 10 J. ( d ): The force is conservative. If we take U = x 3 3 , then we find that the force is F = U x  U y  U z k = x 2 . 7 Problem 32: The force is F = dU dx = d dx ( C 6 x 6 ) = 6 C 6 x 7 . This is an attractive force. 7 Problem 37 ( a ): The force is F ( r ) = dU dr = d dr parenleftbigg a r 12 b r 6 parenrightbigg = 12 a r 13 6 b r 7 ....
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 Summer '08
 Chamberlin
 Force, Work

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