Two blocks A and B of masses 2 kg and 4 kg are placed one over the other as shown in figure, A time varying horizontal force F = 2t is applied on the upper block as shown in figure. Here t is in second and F is in newton. Draw a graph showing accelerations of A and B on y-axis and time on x-axis. Coefficient of friction between A and B is $\mu =\frac{1}{2}$ and the horizontal surface over which B is placed is smooth. $\left(g=10 \mathrm{m}/{\mathrm{s}}^2\right)$

 Limiting friction between A and B is

$f_L=\mu m_{A}g=\left(\frac{1}{2}\right)\left(2\right)\left(10\right)=10 \mathrm{N}$

Block B moves due to friction only. Therefore, maximum acceleration of B can be

$a_{\max }=\frac{f_L}{m_B}=\frac{10}{4}=2.5 \mathrm{m}/{\mathrm{s}}^2$

Thus, both the blocks move together with same acceleration till the common acceleration becomes $2.5 \mathrm{m}/{\mathrm{s}}^2$, after that acceleration of B will become constant $f_L\leftrightarrows 10N$

while that of A will go on increasing. To For $t\ge 7.5 \mathrm{s}$

find the time when the acceleration of both the blocks becomes $2.5 \mathrm{m}/{\mathrm{s}}^2$ (or

when slipping will start between A and B we will write

$$\begin{gathered} 2.5=\frac{F}{\left(m_A+m_B\right)}=\frac{2t}{6} \\ t=7.5 \mathrm{s} \\ t \le 7.5 \mathrm{s} \end{gathered}$$

Hence, for

$a_A=a_B=\frac{F}{m_A+m_B}=\frac{2t}{6}=\frac{t}{3}$

Thus, $a_A$versus t or $a_B$ versus t graph is a straight line passing through origin of slope$\frac{1}{3}$.

For,

$t\ge 7.5 \mathrm{s}$

and

$a_B=2.5 \mathrm{m}/{\mathrm{s}}^2=$ constant

$a_A=\frac{F-f_L}{m_A}$

$a_A=\frac{2t-10}{2}$or $a_A=t-5$