Question

Two blocks A and B of mass ${{m}_{A}}$ and ${{m}_{B}}$respectively are kept in contact on a frictionless table. The experimenter pushes block A from behind so that the blocks get acceleration. If the block A exerts a force $f$ on block B. What is the force exerted by the experimenter on A?

Answer 1

Hint: Since the two blocks are kept in contact, we can consider both the blocks as one system in whole. Also, the net acceleration of the system will be equal to the acceleration of any one of the blocks and we shall use this fact in our solution to get the required force applied by the experimenter.

Complete answer:
Let the force exerted by the person on the first block be given by ${{F}_{P}}$ .
Since this force is the only force in the horizontal direction and the two blocks are in contact with each other, therefore the net acceleration (say $A$ ) of the system will be equal to:
$\Rightarrow A=\dfrac{{{F}_{P}}}{{{m}_{A}}+{{m}_{B}}}$ [Let this expression be equation number (1)]
Now, this is the acceleration by which both the body is moving on the table.
Also, we have the net force on block B as $f$ .
Therefore, the acceleration of the second block (say ${{a}_{B}}$ ) can be written as:
$\Rightarrow {{a}_{B}}=\dfrac{f}{{{m}_{B}}}$ [Let this expression be equation number (2)]
Since, the acceleration as a whole is acceleration of any single block, therefore equation number (1) will be equal to equation number (2). Thus, on equating we get:
 $\begin{align}
  & \Rightarrow \dfrac{{{F}_{P}}}{{{m}_{A}}+{{m}_{B}}}=\dfrac{f}{{{m}_{B}}} \\
 & \Rightarrow {{F}_{P}}=f\left( \dfrac{{{m}_{A}}+{{m}_{B}}}{{{m}_{B}}} \right) \\
 & \Rightarrow {{F}_{P}}=f\left( 1+\dfrac{{{m}_{A}}}{{{m}_{B}}} \right) \\
\end{align}$
Hence, the force exerted by the experimenter on the system or block A is equal to $f\left( 1+\dfrac{{{m}_{A}}}{{{m}_{B}}} \right)$.

Note:
The net external force applied on the blocks in the horizontal direction is by the experimenter only. As the surfaces of contact with each other and the table are smooth and frictionless, therefore no frictional force acts on the block from the table that can oppose its motion and bring an extra term in our calculation.