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Question

(c)

Acceleration of the system $=\frac{P}{m+M}$

The force exerted by rope on the mass $=\frac{M P}{m+M}$

$=\frac{(M)(2Mg)}{(M/2)+M}=\frac{4 M

Vedclass · Accepted Answer

$\frac{4 Mg}{3}$

Vedclass · Answer

(c)

Acceleration of the system $=\frac{P}{m+M}$

The force exerted by rope on the mass $=\frac{M P}{m+M}$

$=\frac{(M)(2Mg)}{(M/2)+M}=\frac{4 Mg}{3}$

A block of mass $M$ is pulled along a horizontal frictionless surface by a rope of mass $M / 2$. If a force $2 Mg$ is applied at one end of the rope, the force which the rope exerts on the block is -

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