 # What is the block’s speed now? 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.

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.