Answer
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
[tex]F = \dfrac{mg}{sin\theta - \mu cos\theta}[/tex]
[tex]F = \dfrac{2.91\times 9.8}{sin 59.8^0- \mu cos 59.8^0}[/tex]
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
