Answer
a) $d=0.39m$
b) $\mu_s<0.53$
c) $1.38\frac{m}{s}$
Work Step by Step
a) $E_{Ki}=E_{Pf}+W_f$
$\frac{mv^2}{2}=mgd\sin(\phi)+0.25d\times F_N$
$F_N=mg\cos(\phi)$
$d=\frac{v^2}{2g(\mu_k\cos(\phi)+\sin(\phi))}=\frac{(2.3\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(0.25\cos(28.0^o)+\sin(28.0^o))}$
$=0.39m$
b) $F_{sf}