# A crate with a mass of 12kg is sliding across the floor with a velocity of 8.8m/s. If the coefficient of kinetic friction is 0.16, how far will it slide before coming to rest?

Sep 22, 2016

Given

$v \to \text{Velocity of the crate"=8.8"m/s}$

$m \to \text{Mass of the crate} = 12 k g$

$\mu \to \text{Coefficient of kinetic frixtion} = 0.16$

$\text{Let the crate slides x m before coming to rest}$

By consudering conservation of energy we can write

$\mu m g x = \frac{1}{2} m {v}^{2}$

$\implies x = {v}^{2} / \left(2 \mu g\right) = {8.8}^{2} / \left(2 \times 0.16 \times 9.8\right) m \approx 24.7 m$