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QJEEAdv09Phb2e6: JEE Advanced 2009 Physics - Class 11 JEEAdv09Phb2e6 Solution: A light inextensible string that goes over a smooth fixed…

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.

IIT-JEE 2009 Paper 2 Offline Physics - Work Power & Energy Question 3 English

Step-by-Step Verified Solution

We have,

a=(m1−m2m1+m2)g=(0.72−0.360.72−0.36)×10=g3=103a = \left( {{{{m_1} - {m_2}} \over {{m_1} + {m_2}}}} \right)g = \left( {{{0.72 - 0.36} \over {0.72 - 0.36}}} \right) \times 10 = {g \over 3} = {{10} \over 3}

T=2m1m2gm1+m2=2×0.72×0.36×100.72+0.36=4.8T = {{2{m_1}{m_2}g} \over {{m_1} + {m_2}}} = {{2 \times 0.72 \times 0.36 \times 10} \over {0.72 + 0.36}} = 4.8 N

s=12at2=12×103×12=53s = {1 \over 2}a{t^2} = {1 \over 2} \times {{10} \over 3} \times {1^2} = {5 \over 3} m

The work done by the rope on 0.36 kg is

W=Tscos⁡0∘=4.8×53=+8W = Ts\cos 0^\circ = 4.8 \times {5 \over 3} = + 8 J

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