A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

how much work is done on the monitor by (a) friction, (b) gravity

work (friction) = 453.5J

work (gravity) = - 453.5J

Explanation:

Given that,

mass = 14kg

displacement length = 5.50m

displacement angle = 36.9°

velocity = 2.30cm/s

F = ma

work (friction) = mgsinθ. displacement

= (14) (9.81) (5.5sin36.9°)

= 453.5J

work (gravity)

= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

= 126.9°

work (gravity) = (14) (9.81) (5.5cos126.9°)

= - 453.5J