An object is attached to a vertical springs and lowered slowly to its equilibrium position. This stretches the spring by a distance `d`. If the same object is attached to the same vertical spring, but permitted to fall freely, through what distance does it stretch the spring ?

Explanation

As is known, in equilibrium position,
`F=mg=Kd`
or` K=(mg)/(d)` …(i)
When the object falls freely , its gravitation `P.E.` is converted into elastic `P.E.` of the spring
`i.e., mgx=(1)/(2)Kx^(2)=(1)/(2)((mg)/(d))x^(2)`
or `1=(x)/(2d)` or `x=2d`