Answer
(a) $0.39$
(b) please see the work below.
Work Step by Step
(a) We know that
$a=\frac{v^2-v_{\circ}^2}{2\Delta x}$
We plug in the known values to obtain:
$a=\frac{(13)^2}{2(22)}$
$a=3.8m/s^2$
We also know that
$f_s=ma$
$\implies \mu_s N=ma$
This simplifies to:
$\mu_s=\frac{a}{g}$
We plug in the known values to obtain:
$\mu_s=\frac{3.84m/s^2}{9.81m/s^2}$
$\mu_s=0.39$
(b) We first find the runner's acceleration. We equate the force associated with the acceleration and the static friction between the runner's shoe and the track. Finally, we find $\mu_s$.
