Determine the time in which the smaller block reaches other end of bigger block in figure.

Friction between 2-kg and 8-kg blocks is kinetic in nature, so

        $\mathrm{F}=\mathrm{m}\times 2\mathrm{g}=0.3\times 2\times 10=6\mathrm{N}$

$\begin{array}{l}\mathrm{For} -2 \mathrm{kg} \mathrm{block}, 10-6=2{\mathrm{a}}_1\\ \mathrm{For} 8-\mathrm{kg} \mathrm{block}, 6-8{\mathrm{a}}_2\Rightarrow {\mathrm{a}}_1=2{\mathrm{ms}}^{-2},{\mathrm{a}}_2=\frac{3}{4}{\mathrm{ms}}^{-2}\end{array}$

Acceleration of 2-kg block relative to 8-kg block is

        ${\mathrm{a}}_{\mathrm{rel}}={\mathrm{a}}_1-{\mathrm{a}}_2=\frac{5}{4}{\mathrm{ms}}^{-2}$

Using the equation of motion, $3=\frac{1}{2}\times \frac{5}{4}{\mathrm{t}}^2\Rightarrow \mathrm{t}=2.19\mathrm{s}$