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 $g=10 \mathrm{~m} / \mathrm{s}^{2}$, 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.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Laws of Motion

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 / s^{2}

, 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.

Answer: 8

Solution

Let

m_{1} = 0.36

kg and

m_{2} = 0.72

kg. The acceleration is

a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}} = \frac{0.36 \times 10}{1.08} = \frac{10}{3}

m/s². The tension in the string is

T = m_{1} (g + a) = 0.36 \times \frac{40}{3} = 4.8

N. The displacement of

m_{1}

in 1 second is

s = \frac{1}{2} a t^{2} = \frac{1}{2} \times \frac{10}{3} \times 1 = \frac{5}{3}

m. Work done by tension is

W = T \times s = 4.8 \times \frac{5}{3} = 8

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer

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