91Ó°ÊÓ

Work energy theorem is given by:

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

Here we have, the mass of brick is\(m = 3\,{\rm{kg}}\).

The force constant of spring is \(k = 1500\,{\rm{N/m}}\).

Here we have both gravitational and elastic potential energy.

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:

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

Let \(y = 0\) at the initial position of the brick.

We take point 1 at the initial position and the point 2 at the final position; and since the spring stretches the same distance the brick descends, so we have

\({y_1} = 0,\;\;{x_1} = 0,\;\;{v_1} = 0\)

\({y_2} = - {d_1},\;\;{x_2} = {d_1},\;\;{v_2} = 0\)

Since the brick starts from rest and eventually comes to rest, so we get:

\({K_1} = {K_2} = 0\)

Also it descends from the zero gravitational potential energy level \(y = 0\),

So we get, \({U_{grav,1}} = 0\)

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),

\(\begin{aligned}{} \Rightarrow {U_{grav,2}} = \left( {3.00} \right)\left( {9.8} \right)\left( { - {d_1}} \right)\\ \Rightarrow {U_{grav,2}} = - 29.4{d_1}\end{aligned}\)

Now, put value of \({x_2}\;{\rm{and k}}\) in equation (4),

\(\begin{aligned}{} \Rightarrow {U_{el,2}} &= \frac{1}{2}\left( {1500} \right){\left( {{d_1}} \right)^2}\\ \Rightarrow {U_{el,2}} & = 750{d_1}^2\end{aligned}\)

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

\(\begin{aligned}{}{K_1} + {U_{grav,1}} + {U_{el,1}} & = {K_2} + {U_{grav,2}} + {U_{el,2}}\\ \Rightarrow 0 + 0 + 0 & = 0 - 29.4{d_1} + 750{d_1}^2\\ \Rightarrow {d_1} & = \frac{{29.4}}{{750}}\\ \Rightarrow {d_1}& = 0.04\,{\rm{m}}\end{aligned}\)

Hence at \(0.04m\)the spring will stretch whenIf we suddenly put a\(3.0 - {\rm{kg}}\)adobe brick in the basket.

Let \(y = 0\) at the initial position of the brick.

We take point 1 at the initial position of the brick, so we have

\({y_1} = 0,\;\;{x_1} = 0,\;\;{v_1} = 0\)

When brick is released, it falls \(1.0m\) then it settles in the basket, then it descends with the basket a distance \({d_2}\) until it stops, making the spring to stretch the same distance, so we have,

\({y_2} = - \left( {{d_2} + 1} \right),\;\;{x_2} = {d_2},\;\;{v_2} = 0\).

Since the brick starts from rest and eventually comes to rest, so we get:

\({K_1} = {K_2} = 0\)

Also it descends from the zero gravitational potential energy level \(y = 0\),

So we get, \({U_{grav,1}} = 0\)

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),

\(\begin{aligned}{} \Rightarrow {U_{grav,2}} & = \left( {3.00} \right)\left( {9.8} \right)\left( - \right)\left( {{d_2} + 1} \right)\\ \Rightarrow {U_{grav,2}}& = - 29.4\left( {{d_2} + 1} \right)\end{aligned}\)

Now, put value of \({x_2}\;{\rm{and k}}\) in equation (4),

\(\begin{aligned}{} \Rightarrow {U_{el,2}} & = \frac{1}{2}\left( {1500} \right){\left( {{d_2}} \right)^2}\\ \Rightarrow {U_{el,2}} & = 750{d_2}^2\end{aligned}\)

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

\(\begin{aligned}{}{K_1} + {U_{grav,1}} + {U_{el,1}} & = {K_2} + {U_{grav,2}} + {U_{el,2}}\\ \Rightarrow 0 + 0 + 0 & = 0 - 29.4\left( {{d_2} + 1} \right) + 750{d_2}^2\\ \Rightarrow 750{d_2}^2 - 29.4{d_2} - 29.4 & = 0\end{aligned}\)

\(\begin{aligned}{} \Rightarrow {d_2} & = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ \Rightarrow {d_2} & = \frac{{29.4 \pm \sqrt {{{\left( { - 29.4} \right)}^2} - 4\left( {750} \right)\left( { - 29.4} \right)} }}{{2\left( {750} \right)}}\\ \Rightarrow {d_2} & = 0.22\,{\rm{m}},\; - 0.18\,{\rm{m}}\;\end{aligned}\)

Distance cannot be negative.

Hence at\(0.22\,{\rm{m}}\)the spring stretch at its maximum elongation if, you release the brick from\(1.0\,{\rm{m}}\)above the basket.