In (b), position A of the upper disc shows undeformed spring. Let the external force applied be F. If on applying the force F, the upper disc of mass `m_(1)` is pressed downwards by `x_(1)` (as at b)
`F+m_(1)g=kx_(1)`…(i)
Now if on releasing the upper disc the extension of the spring is `x_(2)` (as at C) then by conservation of mechanical energy
`1/2kx_(1)^(2)=1/2kx_(2)^(2)+m_(1)g(x_(1)+x_(2))`
i.e., `1/2k(x_(1)^(2)-x_(2)^(2))=m_(1)g(x_(1)+x_(2))`
i.e., `x_(1)=(2m_(1)g)/k+x_(2)` ...(ii)
Now, the lower disc will leave the table only and only if
`kx_(2) gtm_(2)g " i.e., " x_(2) gt m_(2)g//k` ....(iii)
Substituting the values of `x_(1)` and `x_(2)` from Eqns. (i) and (iii) in (ii),
`(F+m_(1)g)/k=(2m_(1)g)/k+(gt(m_(2)g)/k)`
or `F gt(m_(1)+m_(2))g`
i.e., `F_(min)=(m_(1)+m_(2))g`
So, the lower disc will spring back and rise off the table if the spring is pressed by a force greater than the weight of the system.
