Here we have, the mass of fish is \(m = 3\,{\rm{kg}}\).
The force constant of spring is \(k = 900\,{\rm{N/m}}\).
Let\(y = 0\)at the initial position.
We take point 1 at the initial position. So we have
\({y_1} = 0,\;\;{x_1} = 0,\;\;{v_1} = 0\)
Now we take point 2 \(0.05m\) below the initial position, and since the spring stretches the same distance the fish descends, so we have
\({y_2} = - 0.05\,{\rm{m}},\;\;{x_2} = 0.05\,{\rm{m}}\)
Since the fish starts from rest and from the zero gravitational potential energy, we get:
\({K_1} = 0\)
\({U_{grav,1}} = 0\)
And since the spring is initially either stretched or compressed, so we get:
\({U_{el,1}} = 0\)
Now, put value of \({y_2}\;{\rm{and }}m\) in equation (3),
\(\begin{aligned}{} \Rightarrow {U_{grav,2}} & = \left( {3.00} \right)\left( {9.8} \right)\left( { - 0.05} \right)\\ \Rightarrow {U_{grav,2}} & = - 1.47\,{\rm{J}}\end{aligned}\)
Now, put value of \({x_2}\;{\rm{and k}}\) in equation (4),
\(\begin{aligned}{} \Rightarrow {U_{el,2}} & = \frac{1}{2}\left( {900} \right){\left( {0.05} \right)^2}\\ \Rightarrow {U_{el,2}} & = 1.125\,{\rm{J}}\end{aligned}\)
Now, put all these values in equation (1), we get
\(\begin{aligned}{}{K_1} + {U_{grav,1}} + {U_{el,1}} & = {K_2} + {U_{grav,2}} + {U_{el,2}}\\ \Rightarrow 0 + 0 + 0& = {K_2} - 1.47 + 1.125\\ \Rightarrow {K_2} & = 0.345\,{\rm{J}}\end{aligned}\)
Substitute into equation (2) with our value for \(m\) and solving for \({v_2}\) we get:
\(\begin{aligned}{} \Rightarrow 0.345 & = \frac{1}{2}\left( {3.00} \right){v_2}^2\\ \Rightarrow {v_2}& = \sqrt {\frac{{2 \times 0.345}}{3}} \\ \Rightarrow {v_2} & = 0.48\,{\rm{m/s}}\end{aligned}\)
Hence after the speed \(0.48\,{\rm{m/s}}\)it has descended \(0.0500\,{\rm{m}}\) from its initial position.
