Answer
Please see the work below.
Work Step by Step
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:
$\mu_k=\frac{1.75}{9.8}=0.18$
