A box of mass m is pushed horizontally on a rough floor with an initial speed of 2 ms−1. The coefficient of kinetic friction...

Since the box is initially pushed, some force has been imparted on it. This force has a magnitude of ma, where 'm' is the mass of box and 'a' is its acceleration. This force is equal to the force of friction, since the applied force will be dissipated to counter the friction. Hence,

`ma = -mu_k mg`

`or, a = -mu_k g`

Using the equation of motion: v^2 = u^2 + 2as

Since the box...

Since the box is initially pushed, some force has been imparted on it. This force has a magnitude of ma, where 'm' is the mass of box and 'a' is its acceleration. This force is equal to the force of friction, since the applied force will be dissipated to counter the friction. Hence,

`ma = -mu_k mg`

`or, a = -mu_k g`

Using the equation of motion: v^2 = u^2 + 2as

Since 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

or, s = u^2/(-2a) = (2^2) / (-2 x -0.981) = 2.04 m. 

Thus, the box will move a distance of 2.04 m over the rough surface before coming to a full stop.

Hope this helps.