A \(500 - {\rm{kg}}\)block, attached to a spring with length\(0.60\,{\rm{m}}\)and force constant, is at rest with the back of the block at point A on a frictionless, horizontal air table (Fig P7.69). The mass of the spring is negligible. You move the block to the right along the surface by pulling with a constant\(20.0 - {\rm{N}}\)horizontal force.

(a) What is the block’s speed when the back of the block reaches point B, which is \(0.25\,{\rm{m}}\)to the right of point A?

(b) When the back of the block reaches point B, you let go of the block. In the subsequent motion, how close does the block get to the wall where the left end of the spring is attached?

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.