Answer
(a) $0.315$
(b) $0.127$
Work Step by Step
(a) We can find the coefficient of static friction as
$\Sigma F_x=\mu_smgcos\theta-mgsin\theta=0$
$\mu_s=tan\theta$
$\implies \mu_s=tan17.5^{\circ}$
$\implies \mu_s=0.315$
(b) We can find the coefficient of kinetic friction as
$a=g(sin\theta-\mu_kcos\theta)$
$\implies a=(9.81)(sin17.5^{\circ}-\mu_k cos 17.5^{\circ})$
$\implies a=(9.81)(0.30-0.9537\mu_k)$
We also know that
$v_f^2=v_{\circ}^2+2ax$
We plug in the known values to obtain:
$(3.11)^2=(0)^2+2(9.81)(0.3-0.957\mu_k)(2.75)$
$\implies 0.3-0.957\mu_k=\frac{9.6721}{53.955}$
$\implies 0.9537\mu_k=0.3-0.179$
$\implies \mu_k=\frac{0.121}{0.9537}$
$\implies \mu_k=0.127$
