91Ó°ÊÓ

Elastic potential energy is stored by deforming an elastic item, such a spring. It's equivalent to the spring's stretching effort, which depends on k and the distance extended.

\(U = \frac{1}{2}k{x^2}\)

Here we have, the mass of block is \(m = 0.50\,{\rm{kg}}\).

The force constant of spring is \(k = 40\,{\rm{N/m}}\).

The length of the spring is \(l = 0.6m\) and applied force is \(F = 20.0\,{\rm{N}}\)

Here we have only elastic potential energy.

So, work energy theorem is given by:

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

Where kinetic energy is given by:

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

And the elastic potential energy is given by:

\({U_{el}} = \frac{1}{2}k{x^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 3 \right)\)

We also know that the work done by a constant force is given by

\(W = F \cdot s\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 4 \right)\)

Where \(s\) is displacement.

Let point 1 at point A and point 2 at point B,

So we have,

\(\begin{aligned}{}{x_1} = 0,\;\;{v_1} = 0\\{x_2} = 0.25\,{\rm{m}},\;\;{v_2} = {v_B}\end{aligned}\)

Since the force is applied the distance from point A to B, so we have,

\(s = {x_2} = 0.25\,{\rm{m}}\)

Now, put value of \(F\) and \(s\) in equation (4),

\(W = \left( {20.0} \right) \cdot \left( {0.25} \right) = 5\,{\rm{J}}\)

Now, the block starts from the rest.

So,\({K_1} = 0\)

Also, it starts from the equilibrium point.

So,\({U_1} = 0\)

Now, put value of\({x_2}\;{\rm{and }}k\)in equation (3),

\( \Rightarrow {U_2} = \frac{1}{2} \times 40 \times {\left( {0.25} \right)^2} = 1.25\,{\rm{J}}\)

Now, put all these values in equation (1)’

\(\begin{aligned}{} \Rightarrow 0 + 0 + 5 = {K_2} + 1.25\\ \Rightarrow {K_2} = 3.75\,{\rm{J}}\end{aligned}\)

Now, put value of\({K_2}\;{\rm{and m}}\) in equation (2),

\(\begin{aligned}{}3.75 = \frac{1}{2}\left( {0.50} \right){v_2}^2\\ \Rightarrow {v_2} = 3.87\,{\rm{m/s}}\end{aligned}\)

Hence the block’s speed when the back of the block reaches point B, which is \(0.25m\) to the right of point A is \({v_2} = 3.87\,{\rm{m/s}}\).

Let point 1 at point B and point 2 at the closest point from the wall.

So we have,

\(\begin{aligned}{}{x_1} = 0.25\,{\rm{m}},\;\;{v_1} = 3.87\,{\rm{m/s}}\\{x_2} = ?,\;\;{v_2} = 0\end{aligned}\)

Since the force is more applied.

We get \({W_{other}} = 0\)

Since the block eventually comes to rest.

We get\({K_2} = 0\)

Since also above we calculate the energy quantities at point B.

We have,\({U_1} = 1.25\,{\rm{J}}\)and \({K_1} = 3.75\,{\rm{J}}\)

Now, put all these values in equation (1)’

\(\begin{aligned}{} \Rightarrow 3.75 + 1.25\_0 = 0 + {U_2}\\ \Rightarrow {U_2} = 5\,{\rm{J}}\end{aligned}\)

Now, put value of\({U_2}\;{\rm{and }}k\)in equation (3),

\(\begin{aligned}{} \Rightarrow 5 = \frac{1}{2} \times 40 \times {x_2}^2\\ \Rightarrow {x_2} = 0.5\,{\rm{m}}\end{aligned}\)

Now, closest distance from the wall is

\(\begin{aligned}{}d = l - {x_2}\\ \Rightarrow d = 0.6 - 0.5\\ \Rightarrow d = 0.1\,{\rm{m}}\end{aligned}\)

Hence the block get\(0.1\,{\rm{m}}\)closeto the wall where the left end of the spring is attached.