Question

A box of mass m is pushed horizontally on a rough floor with an initial speed of 2 m/s. The coefficient of kinetic friction between the surface and the box is µ_k = 0.1. Calculate the distance the box will move before stopping.

Answer 1

The box is pushed initially.
F = ma;
m - mass of box
a - acceleration. 

ma = -mu_k mg
a = -mu_k g

Equation of motion: v^2 = u^2 + 2as

The box comes to a rest, v = 0 m/s

Here, u = 2 m/s and a = -`mu_k g = - 0.1 xx 9.81 = - 0.981 m/s^2
Thus, u^2 = -2as
s = u^2/(-2a) = (2^2) / (-2 x -0.981) = 2.04 m. 

So the box will move a distance of 2.04 m over the rough surface.