In the Fig. Force $F = α t$ is applied on the block of mass $m_{2}$ . Here $α$ is a constant and t is the time. Find the acceleration of the block. Also draw the acceleration-time graph of both the blocks.

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Solution

If $F \leq μ \frac{m_{2}}{m_{1}} (m_{1} + m_{2}) g$ , then both will move together.
Here, $t \leq μ \frac{m_{2}}{α m_{1}} (m_{1} + m_{2}) g$
During this time, acceleration of both blocks
$a_{1} = a_{2} = \frac{F}{(m_{1} + m_{2})} = \frac{α t}{(m_{1} + m_{2})}$
If $t \geq \frac{μ m_{2}}{α m_{1}} (m_{1} + m_{2}) g$ , friction will be of kinetic nature.
Free-body diagrams:
Acceleration of $m_{1} : a_{1} = \frac{μ N}{m_{1}} = \frac{μ m_{2} g}{m_{1}}$
Acceleration of $m_{2} : a_{2} = \frac{α t - μ m_{2} g}{m_{2}}$
Acceleration-time graph for $m_{1}$ :
$t_{0} = \frac{μ m_{2}}{α m_{1}} (m_{a} + m_{2}) g$
$a_{t_{0}} = \frac{α t_{o}}{(m_{1} + m_{2})} = \frac{μ m_{2} g}{m_{1}}$
Acceleration-time graph for m_2:
Before time $t_{0}$ . slope of the graph is $\frac{α}{m_{1} + m_{2}}$ and after $t_{0}$ slope of the graph becomes $α / m_{2}$ . Slope increases after $t_{0}$ , hence the graph.

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