Question

A 2.91 kg block is pushed 1.33 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 59.8 ◦ with the horizontal.

Answer 1

given,

mass of the block = 2.91 Kg

distance of push, L= 1.33 m

angle of inclination = 59.8◦

Assuming coefficient of friction = μ = 0.3

calculating work done by F.

force due to kinetic friction

f = μ R

from free body diagram

R = F cos θ

f = μ F cos θ

from the diagram attached below

F sin θ = w + f

m g = F sin θ - μ F cos θ

`F = (mg)/(sin\theta - \mu cos\theta)`

`F = (2.91* 9.8)/(sin 59.8^0- \mu cos 59.8^0)`

F = 39.97 N

work done

W = F sin θ x L

W = 39.97 x sin 59.8° x 1.33

W = 45.95 J

Work done by the Force is equal to W = 45.95 J