Question

A block of mass $\mathrm{M}$ is pulled along a horizontal frictionless surface by a rope of mass $\mathrm{m}$. If a force $\mathrm{P}$ is applied at the free end of the rope, the force exerted by the rope on the block is

• A. $\frac{\mathrm{Pm}}{\mathrm{M}+\mathrm{m}}$
• B. $\frac{\mathrm{Pm}}{\mathrm{M}-\mathrm{m}}$
• C. $\mathrm{P}$
• D. $\frac{\mathrm{PM}}{\mathrm{M}+\mathrm{m}}$

Answer 1

The correct option is D

$\frac{\mathrm{PM}}{\mathrm{M}+\mathrm{m}}$

Step 1: Given data and diagram

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{a}\mathrm{block}=\mathrm{M} \\ \mathrm{Mass}\mathrm{of}\mathrm{a}\mathrm{rope}=\mathrm{m} \\ \mathrm{Total}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{system}=\mathrm{M}+\mathrm{m} \\ \mathrm{Force}\mathrm{applied}\mathrm{at}\mathrm{free}\mathrm{end}\mathrm{of}\mathrm{rope}=\mathrm{P} \end{gathered}$$

Also the horizontal surface is frictionless.

The diagram for the given situation is shown below:

Step 2: Calculating the force exerted by rope on the block

Total mass of the system$=\mathrm{M}+\mathrm{m}$

Horizontal force$=\mathrm{P}$

Therefore the acceleration of the whole system is given by:

$$\begin{gathered} \mathrm{Force}=\mathrm{mass}\times \mathrm{acceleration} \\ \Rightarrow \mathrm{P}=\left(\mathrm{M}+\mathrm{m}\right)\times \mathrm{a} \\ \Rightarrow \mathrm{a}=\frac{\mathrm{P}}{\mathrm{M}+\mathrm{m}} \end{gathered}$$

Now the force exerted by the rope on the block will be given by:

$$\begin{gathered} \mathrm{Force}=(\mathrm{mass}\mathrm{of}\mathrm{block})\times \mathrm{acceleration} \\ \mathrm{Force}=\mathrm{M}\times \left(\frac{\mathrm{P}}{\mathrm{M}+\mathrm{m}}\right) \\ \mathrm{Force}=\frac{\mathrm{MP}}{\mathrm{M}+\mathrm{m}} \end{gathered}$$

Hence, option(D) is the correct answer.