# If a 10kg object moving at 7 m/s slows down to a halt after moving 5/2 m, what is the friction coefficient of the surface that the object was moving over?

$\mu m g \times s = \frac{1}{2} m \times {v}^{2}$
$\mu = {v}^{2} / \left(2 g s\right) = {7}^{2} / \left(2 \times 9.8 \times \frac{5}{2}\right) = 1$