One end of an un-stretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lowered to its equilibrium position. If S is the gain in spring energy and G is the loss in gravitational potential energy in the process, then :
Answer & explanation
Correct answer: option 3
As : \(kx = mg\)
At equilibrium position : \(x = \frac{mg}{k}\)
\(U_{spring} = \frac{1}{2} k x^2\)
= \(\frac{1}{2} k (\frac{mg}{k})^2\)
= \(\frac{mgx}{2}\)
= \(\frac{1}{2} \text{loss in GPE}\)
\(S = \frac{1}{2}G\)