A block is kept on the rough horizontal surface and force acting on it as a function of time is shown by a graph. Find velocity (in m/s) of the particle at the end of 10s .
 

$\begin{array}{l}{\mathrm{f}}_{\mathrm{L}}=0.2\left(20\right)=4\mathrm{N}\\ {\mathrm{f}}_{\mathrm{k}}=0.15\left(20\right)=3\mathrm{N}\\ \mathrm{F}=\mathrm{t}\end{array}$
Block will start to move when $\mathrm{F}={\mathrm{f}}_{\mathrm{L}}$,
$\Rightarrow \mathrm{t}=4\mathrm{s}$
From $\mathrm{t}=4\text{ to }\mathrm{t}=6$
$\sum \mathrm{F}=\mathrm{ma}$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{t}-3=2\frac{\mathrm{dv}}{\mathrm{dt}}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\int }_0^{\mathrm{v}} \mathrm{dv}={\int }_4^6 \frac{1}{2}(\mathrm{t}-3)\mathrm{dt}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}[\mathrm{v}{]}_0^{\mathrm{v}}=\frac{1}{2}{\left[\frac{{\mathrm{t}}^2}{2}-3\mathrm{t}\right]}_4^6\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{v}=2\mathrm{m}/\mathrm{s}\end{array}$
From $\mathrm{t}=6\text{ to }\mathrm{t}=10$
$\begin{array}{l}\sum \mathrm{F}=\mathrm{ma}\Rightarrow 6-3=\mathrm{ma}\\ \Rightarrow \mathrm{a}=1.5\mathrm{m}/{\mathrm{s}}^2\\ \mathrm{v}=\mathrm{u}+\mathrm{at}\\ \Rightarrow \mathrm{v}=2+(1.5)\left(4\right)=8\mathrm{m}/\mathrm{s}\end{array}$