91Ó°ÊÓ

Work energy theorem is given by:

\({K_1} + {U_{grav,1}} + {U_{el,1}} + {W_{other}} = {K_2} + {U_{grav,2}} + {U_{el,2}}\)

Since there are no other forces, so the equation become:

\({K_1} + {U_{grav,1}} + {U_{el,1}} = {K_2} + {U_{grav,2}} + {U_{el,2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 1 \right)\)

Where kinetic energy is given by:

\(K = \frac{1}{2}m{v^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 2 \right)\)

The gravitational potential energy is given by:

\({U_{grav}} = mgy\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 3 \right)\)

And the elastic potential energy is given by:

\({U_{el}} = \frac{1}{2}k{x^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 4 \right)\)

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.

Let’s take point 2 at the maximum speed point.

Since the stretched the same distance the fish descends, so we have:

\({y_2} = - y,\;\;{x_2} = y,\;\;{v_2} = {v_{\max }}\)

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 \({y_2}\) in equation (3),

\( \Rightarrow {U_{grav,2}} = - mgy\)

Now, put value of \({x_2}\) in equation (4),

\( \Rightarrow {U_{el,2}} = \frac{1}{2}k{y^2}\)

Now, put all these values in equation (1), we get

\(\begin{aligned}{}{K_1} + {U_{grav,1}} + {U_{el,1}} & = {K_2} + {U_{grav,2}} + {U_{el,2}}\\ \Rightarrow 0 + 0 + 0 & = {K_2} - mgy + \frac{1}{2}k{y^2}\\ \Rightarrow {K_2} & = mgy - \frac{1}{2}k{y^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 5 \right)\end{aligned}\)

Now, we know that the maximum speed occurs when the kinetic energy is maximum.

We also know that, at the point of the maximum kinetic energy in \(K - y\) graph, the slope of the tangent is zero, so we get:

\(\begin{aligned}{} \Rightarrow \frac{{d{K_2}}}{{dy}} & = mg - ky & = 0\\ \Rightarrow ky & = mg\\ \Rightarrow y & = \frac{{mg}}{k}\\ \Rightarrow y & = \frac{{3.00 \times 9.8}}{{900}} & = 0.032\,{\rm{m}}\end{aligned}\)

Now put the value of \(y\) in equation (5)

\(\begin{aligned}{} \Rightarrow {K_2} & = 3.00 \times 9.8 \times 0.032 - \frac{1}{2} \times 900 \times {\left( {0.032} \right)^2}\\ \Rightarrow {K_2} & = 0.48\,{\rm{J}}\end{aligned}\)

Now by equation (2)

\(\begin{aligned}{} \Rightarrow 0.48& = \frac{1}{2} \times 3.00 \times {\left( {{v_{\max }}} \right)^2}\\ \Rightarrow {v_{\max }} & = 0.57\,{\rm{m/s}}\end{aligned}\)

Hence \(0.57\,{\rm{m/s}}\) is maximum speed of the fish as it descends.