Question 67 of 210

A block of mass $M$ 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$P$D$\frac{\mathrm{PM}}{\mathrm{M}+\mathrm{m}}$

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Answer: (D) $\frac{\mathrm{PM}}{\mathrm{M}+\mathrm{m}}$

Since the displacement for both block and rope is same so, the acceleration must be same for both



$\Rightarrow \mathrm{p}=(\mathrm{m}+\mathrm{M}) \mathrm{a} \quad \Rightarrow \mathrm{a}=\frac{\mathrm{P}}{\mathrm{m}+\mathrm{M}}$
$\mathrm{T}=\mathrm{M} \cdot \mathrm{a}=\frac{\mathrm{PM}}{\mathrm{m}+\mathrm{M}}$