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A boxed 12.0 computer monitor is dragged by friction 7.50 up along the moving surface of a conveyor belt inclined at an angle of 35.9 above the horizontal. If the monitor's speed is a constant 2.40 , how much work is done on the monitor by friction, gravity, and the normal force of the conveyor belt?

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Below is the solution:

W done by Normal = 0. (make the incline flat, Normal force goes directly up: no work done) 
W done by gravity = w*displacement = (11kg*9.8) * 7.5sin(35) = -463J 
W done by friction is the opposite of the work done by weight because the object is not moving. Therefore W done by friction = 463J

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