Question

A force of 350 N pushes a 20 kg box in a horizontal plane. The box achieves a velocity of 2.5 m/s in a time of 4 seconds starting from rest. Find the coefficient of friction between the box and the plane

Answer 1

Answer:

0.967

Explanation:

F = ma = m$\frac{\Delta v}{\Delta t}$ = (20×2.5/4) N = 12.5 N

difference of force = 350-12.5=338.5 N

coefficient of friction = 338.5 / 350 = 0.967