This question aims to find the speed of the block when it gets **released** from its **compressed state**. The spring of the block is compressed by length delta x from its initial length $x_o$.

The tension and compression present in the spring obey **Hooke’s law** which states that the minor **displacements** in the object are **directly proportional** to the **displacing force** acting on it. The displacing force can be twisting, bending, stretching and compressing, etc.

It can be mathematically written as:

\[F \propto x \]

\[F = k x \]

Where **F** is the **force applied** on the block which displaces the block as **x**. **k** is the **spring constant** that determines the **stiffness** of the spring.

## Expert Answer

The “**to and fro” motion** of the block exhibits both kinetic and potential energy. When the block is at rest, it exhibits **potential energy** and it shows **kinetic energy** in motion. This energy is conserved when a block moves from its mean position to the extreme position and vice versa.

\[ \text { Total energy (E) }= \text { Kinetic energy (K) } + \text{ Potential energy (U) } \]

\[\frac{ 1 }{ 2 }k A^2= \frac { 1 }{ 2 }m v^2 + \frac { 1 }{ 2 }k x^2\]

The **mechanical energy** is **conserved** when the sum of the kinetic and potential energy is constant.

The energy stored in the spring must be equal to the kinetic energy of the released block.

\[K.E = \frac{ 1 }{ 2 } m v_o ^ {2}\]

The potential energy of the spring is:

\[ K.E = \frac { 1 } { 2 } k \Delta x ^ 2\]

\[\frac { 1 } { 2 } m v_o ^ {2} = \frac { 1 } { 2 } k \Delta x ^ 2 \]

\[ v_o = \Delta x \times x \sqrt { \frac { 2 k } { m }}\]

By keeping mass and change in length constant, we get:

\[ v_o = \sqrt { 2 } \]

## Numerical Results

**The velocity of the released block attached to the spring is $ \sqrt { 2 } $**.

## Example

To find the change in length of the same block, rearrange the equation as:

The mechanical energy is conserved when the sum of kinetic and potential energy is constant.

The energy stored in the spring must be equal to the kinetic energy of the released block.

\[ K.E = \frac { 1 }{ 2 } m v_o ^ {2} \]

The potential energy of the spring is:

\[ K.E = \frac { 1 }{ 2 } k \Delta x ^ 2 \]

\[ \frac { 1 }{ 2 } m v_o ^ {2} = \frac { 1 }{ 2 } k \Delta x ^ 2 \]

\[ \Delta x = v_o \sqrt { \frac{ m }{ 2 k }} \]

**The change in length is equal to $\dfrac{ 1 }{ \sqrt {2} }$.**

*Image/Mathematical drawings are created in Geogebra**.*