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

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 of 35.9 above the horizontal. If the monitor's speed is a constant 2.10 cm/s, how much work is done on the monitor by friction, gravity, and the normal force of the conveyor belt?
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Answer #1
the work done by normal force is zero cause the force and displacement are perpendicular to each other

the height raised by it is h = 5.50sin35.9
so work done by gravitational force is -mgh = - 316.0547 J
now for calculating work done by friction applying work energy principle
that is work done by external forces is equal to change in kinetic energy
since change in kinetic energy is zero
so work done by friction = - work done by gravity = 316.0547 J
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