Question

A 5kg box slides across the floor with an initial velocity of 5m/s. If the coefficient of kinetic friction between the box and the floor is 0.1,how much time will it take for the box to come to a stop?​

Answer 1

The net force on the block perpendicular to the floor is

∑ F[perp] = F[normal] - mg = 0

so that

F[normal] = (5 kg) g = 49 N

Then

F[friction] = 0.1 F[normal] = 4.9 N

so that the net force parallel to the floor is

∑ F[para] = -4.9 N = (5 kg) a

Solve for the acceleration a :

a = (-4.9 N) / (5 kg) = -0.98 m/s²

Starting with an initial velocity of 5 m/s, the box comes to a stop after time t such that

0 = 5 m/s - (0.98 m/s²) t

⇒   t ≈ 5.1 s