Answer
$V=0.0883\text{ m/s}$
Work Step by Step
$W s\mathrm{in}(20^{\circ})=\tau\mathrm{A}$
$\tau=\mu{\frac{V}{b}}$
$\mathrm{Wsin~20^{\circ}=\mu\frac{V}{b}A}$
$W=m g$
$\begin{array}{r l}{V}&{{}={\frac{6W\sin\left(20^{\circ}\right)}{\mu A}}={\frac{{(0.0001~)}{(10)(9.81)(0.1)}}{(0.38)(0.1)}}=0.0883{\frac{\mathrm{m}}{s}}}\end{array}$
