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A boxed 10.0-kg computer monitor is dragged by friction 5.50 m up along the moving surface...

A boxed 10.0-kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle 34.9o above the horizontal. If the monitor’s speed is 2.10 m/s, how much work is done on the monitor by friction?

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Answer #1

Work done by friction = umg cos theta L

= mg sin theta L...(because speed is constant)

= 10*9.8* 5.5* sin 34.9 degree

= 308 J

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