Two block of masses `m_(1)` and `m_(2)` are in equilibrium. The block `m_(2)` hangs from a fixed smooth pulley by an inextensible string that is fitted with a light spring of stiffness k as shown in fig. Neglecting friction and mass of the string, find the acceleration of the bodies just after the string S is cut.

Explanation

FBD: Let the spring forces be `F=kx` just after cutting the spring. Hence, at that instant, the forces acting on `m_(1)` are `T=kx rarr m_(1)g darr and N uarr`, on `m_(2)` the force are `m_(2)g darr` and `T uarr`.
Force equation: Initially, all the particles are stationary, hence, `a_(1)=a_(2)=0`. applying Newton's second law.
For `m_(1)`: `Sigma F=T'-kx=0`
For `m_(2): Sigma F=T-m_(2)g=0`
for spring: `Sigma F=T-kx=0`
Solving eqs. (i), (ii) and (iii), we have
`T'=T=kx=m_(2)g`
Just after the string S is cut, tension T' vanishes immediately but the spring cannot regain its shape and size instantanously. Therefore, the spring force remains as it is . Then, just after cutting the string, the net force acting on `m_(1)` is equal to kx, whereas the net force acting on `m_(2)` is equal to `T=m_(2)g`. Hence the acceleration of `m_(1)` and `m_(2)` just after cutting the string is given by
`SigmaF=m_(1)a_(1)=kx=m_(2)g, Sigma F=m_(2)a_(2)=T-m_(2)g=0`
This gives `a_(1)=((m_2)/(m_1)) g and a_(2)=0`.