Question

In the figure, force $F=\alpha t$. Which of the following represents the acceleration-time graph of both the blocks? ($a_1$ and $a_2$ are the accelerations of blocks $m_1$ and $m_2$ respectively)

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

The FBDs of the blocks are as shown



From the FBDs we have
$N_1=m_{2}g$
$f_{\text{max}}=\mu N_1=\mu m_{2}g$
Let us consider the two blocks move together, then the common acceleration will be
$a_c=\frac{F}{m_1+m_2}$
The two blocks will move together till $f=\mu m_{2}g$ and will have acceleration $a_c$.
Hence, at limiting condition $f_{\text{max}}=m_1{a}_c$
$\Rightarrow \mu m_{2}g=\frac{m_{1}F}{m_1+m_2}=\frac{m_1\alpha t}{m_1+m_2}$
$\Rightarrow t=\frac{\mu m_{2}g(m_1+m_2)}{\alpha m_1}$

After the time $t=t_0$, the blocks will separate, where $t_0$ is the time to reach the limiting condition.

So, from the FBDs the acceleration of both the blocks after $t=t_0$ are given as
For block of mass $m_2$:
$F-f=m_2{a}_2\Rightarrow a_2=\frac{\alpha t-\mu m_{2}g}{m_2}$
$\Rightarrow a_2=(\frac{\alpha t}{m_2}-\mu g)$
It will vary linearly increasing with time
For block of mass $m_1$:
$f=m_1{a}_1\Rightarrow a_1=\frac{f}{m_1}=\frac{\mu m_{2}g}{m_1}$, which is constant.

Hence, the graph will be as shown