A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

Work Power and Energy PYQ — JEE Advanced 2009 · Shift 2 · Q50 with Solution

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Correct answer: 8

Let

Since $m_2 > m_1$ , the $0.72\,\text{kg}$ block moves downward and the $0.36\,\text{kg}$ block moves upward.

For a light string over a smooth pulley, the tension is same on both sides.

For an Atwood machine,

a = \frac{m_2 - m_1}{m_1 + m_2}g

Substitute values:

a = \frac{0.72 - 0.36}{0.72 + 0.36}\cdot 10 = \frac{0.36}{1.08}\cdot 10 = \frac{10}{3}\,\text{m/s}^2

For the lighter block $m_1 = 0.36\,\text{kg}$ moving upward,

T - m_1 g = m_1 a

So,

T = m_1(g+a) = 0.36\left(10 + \frac{10}{3}\right)

T = 0.36\cdot \frac{40}{3} = 4.8\,\text{N}

The system starts from rest, so for the $0.36\,\text{kg}$ block,

s = ut + \frac{1}{2}at^2

with $u=0$ , $t=1\,\text{s}$ :

s = \frac{1}{2}\cdot \frac{10}{3}\cdot (1)^2 = \frac{5}{3}\,\text{m}

The string exerts tension upward, and the block also moves upward, so angle between force and displacement is $0^\circ$ .

Hence,

W = Ts\cos 0^\circ = Ts

W = 4.8 \times \frac{5}{3} = 8\,\text{J}

The work done by the string on the $0.36\,\text{kg}$ block during the first second is

\boxed{8}

This matches the stored correct answer.