Question

a man drags a block through 10m on rough surface (mew=0.5). A force of root3 kN acting at 30 to the horizontal . the work done by applied force is

Answer 1

Hey friend,

◆ Answer-
W = 7.5 kJ

◆ Explanation-
# Given-
d = 10 m
μ = 0.5
F = √3 kN = 1.73×10^3 N
θ = 30°

# Solution-
Force of friction is given by -
Fs = μF
Fs = 0.5 × 1.73×10^3
Fs = 866 N

Work is calculated by -
W = F.d
W = Fdcosθ
W = 866 × 10 × cos30
W = 8660 × 0.866
W = 7500 J
W = 7.5 kJ

Work done by force of friction is 7.5 kJ.

Hope this helps you...