Answer
a)$\mu_{s}$=1.76
b)$v_{1}$ =0.67m/s
Work Step by Step
a)We proceed to use $F_{net}=ma$
From the free body diagram,
$f_{s}-F_{spring}=0$
$\mu_{s}mg-kd=0$
$\mu_{s}$=$\frac{kd}{mg}$= $\frac{(20)(0.086)}{(0.1)(9.8)}$=1.76
b)We first apply $F_{net}=ma$ to find the maximum compression
$\mu_{s}mg=kd$
d=$\frac{(0.6)(0.1)(9.8)}{20}$=0.0294m
Now we apply work-energy theorem,
$W_{tot}=K_{final}-K_{initial}$
$K_{initial} = \frac{1}{2}mv^{2}, K_{final}=0$
$W_{tot}=W_{spring}+W_{friction}$=-0.5(20)(0.0294)-0.47(0.10)(9.8)(0.294)=-0.02218J
So, -0.02218= -$\frac{1}{2}mv^{2}$
$v=\sqrt \frac{(2)(0.2218)}{0.1} = 0.67m/s$
