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.

For 2m block , 2mg-T=(2 m)aFor m block, T-mg=maUsing both equation we get, a=g/3 and T=4mg3Work done by tension force on 'm' block isW=T×s=T×12at2=4mg3×12×g3×12=2009×0.36=8 Joules

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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.

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Detailed Solution

$For 2 m block, 2 mg - T = (2 m) a$

$For m block, T - mg = ma$

Using both equation we get,

$a = g / 3 and T = \frac{4 mg}{3}$

Work done by tension force on 'm' block is

$W = T \times s = T \times \frac{1}{2} {at}^{2} = \frac{4 mg}{3} \times \frac{1}{2} \times \frac{g}{3} \times 1^{2} = \frac{200}{9} \times 0.36 = 8 Joules$