# If an object is moving at 11 m/s over a surface with a kinetic friction coefficient of u_k=22 /g, how far will the object continue to move?

${\mu}_{k} m g \times d = \frac{1}{2} m \times {v}^{2}$ ,where d is the distance traversed.
$\implies d = {v}^{2} / \left(2 {\mu}_{k} g\right) = {11}^{2} / \left(2 \cdot \frac{22}{g} \cdot g\right) = \frac{11}{4} = 2.75 m$