A block of mass m = 4.4 kg lies on a horizontal  rough surface. The coefficient of friction between  the block and the surface is $\mathrm{\mu }=0.5$  A force F
starts acting on the block making an angle $\mathrm{\theta }={37}^{\circ }$ to the horizontal. The force changes with time as 
shown in the graph. I at   time  $t_0$ ,the block begins to move  and the maximum speed attained by the  block is $V_{max}$ , then
$\left[\tan {37}^{\circ }=\frac{3}{4};\mathrm{g}=10{\mathrm{ms}}^{-2}\right]$

 For 0 < t < 10 s
F = 4t
If the block begins to move at time t₀, then  
$$\begin{gathered} \mathrm{Fcos}\mathrm{\theta }=\mathrm{\mu }(\mathrm{mg}-\mathrm{Fsin}\mathrm{\theta }) \\ 4{\mathrm{t}}_0\times \frac{4}{5}=\frac{1}{2}\left(4.4\times 10-4{\mathrm{t}}_{\mathrm{o}}\times \frac{3}{5}\right) \\ \Rightarrow {\mathrm{t}}_0=5\mathrm{s} \end{gathered}$$
 The block accelerates in the interval 5 < t < 15 s
Hence speed is maximum at t = 15 s
$\mathrm{\Delta p}=$Impulse 
$$\begin{gathered} \therefore {\mathrm{mv}}_{max}={\int }_5^{15} \mathrm{Fcos}\mathrm{\theta dt}-{\int }_5^{15} \mathrm{\mu }(\mathrm{mg}-\mathrm{Fsin}\mathrm{\theta })\mathrm{dt} \\ \therefore {\mathrm{mv}}_{max}=\cos \mathrm{\theta }{\int }_5^{15} \mathrm{Fdt}-\mathrm{\mu mg}{\int }_5^{15} \mathrm{dt}+\mathrm{\mu sin}\mathrm{\theta }{\int }_5^{15} \mathrm{Fdt} \\ =(\cos \mathrm{\theta }+\mathrm{\mu sin}\mathrm{\theta }){\int }_5^{15} \mathrm{Fdt}-\mathrm{\mu mg}\times 10 \\ =\left(\frac{4}{5}+\frac{1}{2}\times \frac{3}{5}\right)\left(\text{ shaded area }\right)-\frac{1}{2}\times 4.4\times 10\times 10 \\ =\frac{11}{10}\times \left[20\times 10+\frac{1}{2}\times 20\times 10\right]-220 \\ \Rightarrow 4.4{\mathrm{v}}_{max}=110 \\ \Rightarrow {\mathrm{v}}_{max}=\frac{110}{4.4}=25{\mathrm{ms}}^{-1} \end{gathered}$$