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.

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.