Question

a force of 35.0 n is required to start a 6.0-kg box moving across a horizontal concrete floor. what is the coefficient of static friction between the box and the floor?

Answer 1

0.595

Step-by-step explanation:

Force of static friction = Fs

μ = coeff. of static friction

N = normal force = W = mg

Fs = μN

μ = Fs / N

Fs has to be less than 35 N, otherwise the box wouldn't move.

μ < (35 N) / (6 kg)(9.8 m/s²) = 0.595