Question

A man of mass $\text{2m}$ is pulling up a block of mass $\text{m}$ with constant velocity. The acceleration of man is- (Neglect any friction)-

(A). $\text{g}$
(B). $\text{2g}$
(C). $\text{3g}$
(D). ${\text{g}}/{2}\;$

Answer 1

The figure has two separate systems and different forces are acting on each system. Resolve the systems. There are two main forces- tension in string and weight of block. Analyse each system and make equations. Solve the equations to find acceleration of man.

Formula used:
$\Rightarrow \text{ T = mg}$
$\text{mg = 2ma }\\!\\!'\\!\\!\text{ }$

Complete step-by-step answer:
The block, pulley and man form an isolated system.

Let us divide the fig into two systems- the block and pulley system and the man and pulley system and consider the upward direction as $(\text{+}\text{)}$ .
Forces acting on block and pulley system are- $\text{T}$ and $\text{mg}$ , where $\text{T}$ is the tension in the string and $\text{mg}$ is the weight acting on the block. The block is moving with uniform velocity in the upward direction which means there are no external forces acting on block, so,
$\text{a= 0}$ Here $\text{a}$ is acceleration of block
Therefore-
$\text{T - mg = 0}$
$\Rightarrow \text{ T = mg}$ - (1)
Force acting on the man pulley system is $\text{T}$ and the system is moving opposite the direction of $\text{T}$ with some acceleration. Therefore,
$\text{T = (2m)a }\\!\\!'\\!\\!\text{ }$ - (2)
$\text{2m}$ is mass of man, $\text{a }\\!\\!'\\!\\!\text{ }$ is acceleration of man.
From eq (1) and eq (2), we get,

& \text{mg = 2ma }\\!\\!'\\!\\!\text{ } \\\ & \Rightarrow \text{ a }\\!\\!'\\!\\!\text{ = }\dfrac{g}{2} \\\ \end{aligned}$$ Therefore the acceleration of man is $$\dfrac{\text{g}}{2}$$ . **So, the correct answer is “Option D”.** **Note:** It is important to resolve the two systems because in the block- pulley system, there are no external forces hence their motion is uniform while in the man- pulley system, external forces are acting on man due to which it has non-uniform motion. The pulley is considered to be at rest.