Question

A block of mass 5kg slides down a rough inclined surface. The angle of inclination is 45 degrees coefficient of sliding friction is 0.20 when the block slides them

Answer 1

F = mue × mg

So,

F= 0.2×5×10

F=10

Now,

W= mg cos (theta)

W= 10× 10 × √2\2

So ,

W = 50 √2 ( ans )

Answer 2

Hey dear,
You posted incomplete question.

◆ Complete question should be -
A block of mass 5 kg slides down a rough inclined surface. The angle of inclination is 45°. Coefficient of sliding friction is 0.20. When the block slides 5 m.
Calculate the work done by frictional force.

◆ Answer-
W = 35.35 J

◆ Explanation-
# Given-
m = 5 kg
θ = 45°
μ = 0.2
d = 5 m

# Solution-
Frictional force is given by -
F = μmg
F = 0.2 × 5 × 10
F = 10 N

Work done is -
W = F.d
W = Fdcosθ
W = 10 × 5 × 1/√2
W = 35.35 J

Work done by force of friction is 35.35 J.

Hope this helps you...