# Deriving the energy conservation law and solving an example problem

We will derive the energy conservation law here, as it is used in this article.
-> Let’s think about a pendulum with air resistance

The explanation is as follows.

## Integrating the equation of motion with the velocity multiplied

Let us consider the motion of a mass point with mass $$m$$.
We will take a 1 dimensional motion for simplicity.
The equation of motion can be written as

$$\begin{eqnarray} m\frac{{\rm d}v}{{\rm d}t} = F, \end{eqnarray}$$

with $$v$$ the velocity component.
Here, note that acceleration is written as $$\frac{{\rm d}v}{{\rm d}t}$$.

Multiplying the velocity $$v = \frac{{\rm d}x}{{\rm d}t}$$ and integrating with $$t$$,

$$\begin{eqnarray} \int m\frac{{\rm d}v}{{\rm d}t} v{\rm d}t &=& \int F v{\rm d}t \\ \int \frac{1}{2}m\frac{{\rm d}}{{\rm d}t}v^2 {\rm d}t &=& \int F \frac{{\rm d}x}{{\rm d}t}{\rm d}t \\ \frac{1}{2}mV^2-\frac{1}{2}mv_{0}^{2} &=& \int F {\rm d}x, \end{eqnarray}$$

we denoted the velocity of the final state as $$V$$, and the initial state as $$v_{0}$$.

The last equation

$$\begin{eqnarray} \frac{1}{2}mV^2-\frac{1}{2}mv_{0}^{2} = \int F {\rm d}x \end{eqnarray}$$

is the equation of the energy conservation law.

### When the force $$F$$ is a conservative force

Here, we assume the force $$F$$ a conservative force.
The conservative force is a force that the integral $$\int F{\rm d}x$$ does NOT depend on the path.
We can define a potential energy

$$\begin{eqnarray} U = \int_{x_{0}}^{x}F{\rm d}x, \end{eqnarray}$$

and the energy conservation equation can be written as

$$\begin{eqnarray} \frac{1}{2}mV^2-\frac{1}{2}mv_{0}^{2} = U\left(x\right)-U\left(x_{0}\right) \end{eqnarray}$$

For example, the energy of a spring or gravity can be provided in this form.

## (Example problem) a spring A spring with the equilibrium length $$l_{0}$$ stretched to $$l$$.

Let’s derive the energy conservation law of a spring to get used to it.

We think about stretching the spring equilibrium length $$l_{0}$$ to $$l$$ and letting it go with zero initial velocity.
A mass point with mass $$m$$ is on the edge.

If we take the origin at the equilibrium position, the displacement is $$x = l-l_{0}$$ and the equation of energy conservation is

$$\begin{eqnarray} \int F {\rm d}x = \int_{l-l_{0}}^{0}\left(-kx\right){\rm d}x, \end{eqnarray}$$

here, the resilience is given by $$F = -kx$$ by the Hooke’s law.
Integrating this, we get

$$\begin{eqnarray} \int_{l-l_{0}}^{0}\left(-kx\right){\rm d}x &=& -\left[\frac{1}{2}kx^{2}\right]_{l-l_{0}}^{0} \\ &=& \frac{1}{2}k\left(l-l_{0}\right)^{2} \end{eqnarray}$$

On the other hand, as for the left hand side, the initial velocity is 0, and if we write the velocity $$v$$ at $$x = 0$$, it is given by $$\frac{1}{2}mv^2$$.
Using the energy conservation law,

$$\begin{eqnarray} \frac{1}{2}mv^2 &=& \frac{1}{2}k\left(l-l_{0}\right)^{2} \end{eqnarray}$$

This is the common form of the energy conservation equation.

## Summary

Now, that’s all what I wanted to write.

This time, we derived the energy conservation equation and solved an example problem.