## Essential University Physics: Volume 1 (3rd Edition)

We know that $a=\frac{v^2}{2d}$ $a=\frac{(14)^2}{(2)(56)}$ $a=1.75\frac{m}{s^2}$ Now we can find the coefficient of the kinetic friction as $\mu_k=\frac{ma}{mg}=\frac{a}{g}$ We plug in the known values to obtain: $\mu_k=\frac{1.75}{9.8}=0.18$