A light inextensible string that goes over a smooth fixed pulley connects two blocks of masses 0.36 kg and 0.72 kg. Taking g=10m/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.

Let m1 = 0.36 kg and m2 = 0.72 kg. Let T be the tension in the string and 'a' be the acceleration of the system.
For mass m2 (0.72 kg):
2mg - T = (2m)a
For mass m1 (0.36 kg):
T - mg = ma
Adding the two equations:
mg = 3ma
=> a = g/3 = 10/3 m/s²
Substituting this value of 'a' in the equation T - mg = ma:
T = m(g + a) = 0.36(10 + 10/3) = 0.36 * 40/3 = 4.8 N
The distance 's' covered by the block of mass 0.36 kg in the first second is given by:
s = ut + (1/2)at² = 0 + (1/2)(10/3)(1)² = 5/3 m
Work done (W) by the string on the block of mass 0.36 kg is:
W = T * s = 4.8 * (5/3) = 8 Joules