The 10kg block is resting on the horizontal surface when the force F is applied to it for 7s. The variation of F with time is shown. Calculate the maximum velocity reached by the block and the total time t during which the block is in motion. The coefficients of static and kinetic friction are both 0.50. (Take, $g=9.8 \mathrm{m}/{\mathrm{s}}^2$ )

Block will start moving at,$F=\mu mg$ or 

$25t=(0.5)\left(10\right)(9.8)=49 \mathrm{N}$ 

$\therefore t=1.96 \mathrm{s}$

Velocity is maximum at the end of 4 second.

$$\begin{gathered} \frac{dv}{dt}=\frac{25t-49}{10}=2.5t-4.9 \\ {\int }_0^{v_{max}}dv={\int }_{1.96}^4(2.5t-4.9)dt \\ {v}_{\max }=5.2 \mathrm{m}/\mathrm{s} \end{gathered}$$

For $4 \mathrm{s}<t<7 \mathrm{s}$

Net retardation $a_1=\frac{49-40}{10}=0.9 \mathrm{m}/{\mathrm{s}}^2$

$\therefore$$$\begin{gathered} v=v_{\max }-a_1{t}_1 \\ =5.2-0.9\times 3 \\ =2.5 \mathrm{m}/\mathrm{s} \end{gathered}$$

For$t>7 \mathrm{s}$

$$\begin{gathered} \text{ Retardation }{a}_2=\frac{49}{10}=4.9 \mathrm{m}/{\mathrm{s}}^2 \\ t=\frac{v}{a_2}=\frac{2.5}{4.9}=0.51 \mathrm{s} \\ \therefore \text{ Total time }=(4-1.96)+(7-4)+(0.51) \\ =5.55 \mathrm{s} \end{gathered}$$