A light inextensible string that gas over a smoth fixed polley as shown in the figure connect two blocks of mases it `0.36` kg and `0.72` kg Taking `g = 10 ms^(-2)` , find the work done by the string on the block of mass ` 0.36` kg doring the first second after the system is refosed from rest ,

Explanation

Given `m = 0.36 kg , M = 0.72 kg ` The figre shown the force on `m` and `M` when the system is released , let the acceleration be a . Then

`T - mg = ma `
Mg - T = Ma
` :. 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 = ut + (1)/(2) at^(2) = 0 + (1)/(2) xx (g)/(3) xx 1^(2) = g//6`
:. work done by the string on `m` is
`T^(rarr) .s^(rarr) = Ts = 4(mg)/(3) xx (g)/(6) = (4 xx 0.36 xx 10 xx 10 )/(3 xx 6 ) = 8 J`