A disc of mass `m_(2)` is placed on a table. A stiff spring is attached to it and is vertical. The other end of the spring is attached to a disc of mass `m_(1)` What minimum force should be applied to the upper disc to press the spring such that the lower disc is lifted off the table when the external force is suddenly removed?

Explanation

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.