What is the coefficient of friction of the carpet?

Question

You need to move a 53 kg armoire across the room. If you push with a force of 68 N, and the armoire accelerates at 0.10 m/s² squared, what is the coefficient of friction of the carpet?

Answer 1

m = 53 kg

F  = net force

F_a = applied force = 68 N

f = force of friction = μmg

a = acceleration = 0.10 m/s²

 

F = ma

F = F_a - f = F_a - μmg

 

ma = F_a - μmg

 

μ = (F_a - ma) / (mg)

 

μ = [68 - (53)(0.10] / (53*9.81) ≅ 0.121