91Ó°ÊÓ

Because the force applied by the spring is always in the opposite direction of the displacement, this force is known as a restoring force.

Let the mass of fish is \(m\).

Now when fish is slowly lowered the spring will extend till the gravitational force on the fish is balanced by the restoring force of spring. Such that,

\(\begin{aligned}{}mg = kd\\ \Rightarrow k = \frac{{mg}}{d}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 1 \right)\end{aligned}\)

When the same fish is allowed to fall vertically, the gravitational potential energy of the block is converted into elastic potential energy of spring.

Let\(d'\)be the extension of spring.

Then we have,

\(\begin{aligned}{}mgd' = \frac{1}{2}kd{'^2}\\ \Rightarrow d' = \frac{{2mg}}{k}\\ \Rightarrow d' = \frac{{2mg}}{{\frac{{mg}}{d}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{\rm{by }}\left( 1 \right)} \right)\\ \Rightarrow d' = 2d\end{aligned}\)

So, maximum distance is \(2d\).