Question

Block $A$ of mass $\frac{m}{2}$ is connected to one end of a light rope which passes over a light frictionless pulley as shown in figure.


A man of mass $2m$ climbs the other end of the rope with a relative acceleration of $\frac{g}{2}$ with respect to rope. Find the acceleration ($a_0$) of block $A$ with respect to ground.

• A. $a_0=\frac{g}{2}$
• B. $a_0=\frac{2}{3}g$
• C. $a_0=\frac{3}{2}g$
• D. $a_0=g$

Answer 1

The correct option is D $a_0=g$

Let block A move up with acceleration $a_0$ w.r.t the ground. Then rope which the man is holding will move downward with acceleration $a_0$
Given : Acceleration of man w.r.t rope
$a_{mr}=\frac{g}{2}$ (upwards)

We know ${\overrightarrow{a}}_{mg}={\overrightarrow{a}}_{mr}+{\overrightarrow{a}}_{rg}$
Taking upwards +ve,
$a_{mg}=a=\frac{g}{2}-a_0....\left(i\right)$

From FBD of block:


$T-\frac{mg}{2}=\frac{m{a}_0}{2}.......\left(ii\right)$

From FBD of man:


$T-2mg=2ma.......\left(iii\right)$

Substituting $\left(i\right)$ in $\left(iii\right)$:
$T-2mg=mg-2m{a}_0$
$\Rightarrow T-3mg=-2m{a}_0....\left(iv\right)$

Subtracting $\left(ii\right)-\left(iv\right)$
$\frac{5mg}{2}=\frac{5m{a}_0}{2}$
$\therefore a_0=g$