Question

- A boxed 10.0-kg computer monitor is Cragged by friction
5.50 m up along the moving surface of a conveyor belt inclined at
an angle of 36.90 above the horizontal. If the monitor's speed is a
constant 2.10 cm/s, how much work is done on the monitor by
(a) friction, (b) gravity, and (c) the normal force of the conveyor belt?​

Answer 1

Work(W) = Force(F) x Distance(D)

Work done by friction.

Given: distance = 5.5 m , mass 10 kg, and incline angle(θθ) of 36.9 ∘∘

Breaking down force on computer monitor:

Fax=mgsinθFax=mgsinθ, ==>Fax=(10)(9.8)sin(36.9)Fax=(10)(9.8)sin(36.9) ==>FaxFax = 58.84118208 N

Fay=mgcosθFay=mgcosθ>>>> not needed

Fg=mgFg=mg>>>> not needed

FaxFax is the force that opposes any force that might move this box.

Therefore the work done to prevent this box from moving during 5.5 m is:

W = 58.84118208 N x 5.5 m = 323.6265015 NM or ≈≈323.63 J