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.

Step-by-step solution

Given m=0.36 ~kg , M=0.72 ~kg . The figure shows the forces on m and M . When the system is released, let the acceleration be a . Then T-m g=m aM g-T=M a a= (M-m) g M+m =g / 3 and T=4 mg / 3 For block m: u=0, a=g / 3, t=1, s=? s=u t+ 1 2 a t²=0+ 1 2 g 3 1²=g / 6 Work done by the string on m is T s =T s=4 m g 3 g 6 = 4 0.36 10 10 3 6 =8 ~J