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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25-11-2016
Physics

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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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Can some1 help.? I have no Idea what to write and what to say!

Аноним Аноним · Answer 1 · 2016-11-26T13:21:29+01:00

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

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