# If a 11 kg object moving at 45 m/s slows down to a halt after moving 900 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

${\mu}_{k} \times m \times g \times d = \frac{1}{2} \times m {v}^{2}$
${\mu}_{k} = {v}^{2} / \left(2 g d\right) = {45}^{2} / \left(2 \times 10 \times 900\right) = 0.1125$