A boxed 10 kg computer monitor is dragged by friction 5.5 m up along the moving surface of a conveyor belt inclined at an angle of 36.9° above the horizontal. If the monitor’s speed is a constant, how much work is done on the monitor by friction?

324 J

Explanation:

58.84 * 5.5 J = 324 J

I got 58.84 by 10 * 9.8 (gravity) * sin(36.9)

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

< 3

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be