Physics: Calculus-based:Conservation of energy

Within a closed system, the total amount of energy is always conserved. This translates as the sum of the n changes in energy totalling to 0.

$\sum _{k=1}^n\mathrm{\Delta }{𝐄}_k=0$

An example of such a change in energy is dropping a ball from a distance above the ground. The energy of the ball changes from potential energy to kinetic energy as it falls.

$\begin{array}{ccc}{U}_g& =& m𝐠h\\ K& =& \frac{1}{2}m{𝐯}^2\end{array}$

Because this is the only change in energy within our system, we will take a simple physical problem and model it in order to demonstrate.

An object of mass 10kg is dropped from a height of 3m. What is its velocity when it is 1m above the ground?

We start by evaluating the Potential Energy when the object is at its initial state.

$\begin{array}{ccc}{U}_g& =& m𝐠h\\ U_g& =& 30𝐠\\ 𝐠& =& 9.807m{s}^{-2}\\ U_g& =& 30\cdot 9.807\\ U_{g3}& =& 294.21J\end{array}$

The Potential Energy of the object at a height of 1m above the ground is given in a similar fashion.

$\begin{array}{ccc}{U}_g& =& m𝐠h\\ U_g& =& 10𝐠\\ U_g& =& 10\cdot 9.807\\ U_{g1}& =& 98.07J\end{array}$

Hence the change in Potential Energy is given

$\begin{array}{ccc}\mathrm{\Delta }{U}_g& =& 294.21-98.07=196.14J\end{array}$

By definition, the change in Potential Energy is equivalent to the change in Kinetic Energy. The initial KE of the object is 0, because it is at rest. Hence the final Kinetic Energy is equal to the change in KE.

$\begin{array}{ccc}\mathrm{\Delta }{U}_g& =& \mathrm{\Delta }K\\ 196.14J& =& \frac{1}{2}m{𝐯}^2\\ 196.14& =& 5{𝐯}^2\end{array}$

Rearranging for v

$\begin{array}{ccc}\frac{196.14}{5}& =& {𝐯}^2\\ \sqrt{\frac{196.14}{5}}& =& 𝐯\\ 𝐯& \approx & 6.263m{s}^{-1}\end{array}$

We can check our work using the following kinematic equation.

$\begin{array}{ccc}{𝐯}^2& =& {𝐮}^2+2𝐚𝐬\\ {𝐯}^2& =& 0^2+2𝐠𝐬\\ 𝐯& =& \sqrt{2𝐠𝐬}\\ 𝐯& =& \sqrt{2\cdot 9.807\cdot 2}\\ 𝐯& \approx & 6.263m{s}^{-1}\end{array}$

This follows because we can actually use the equations for energy to generate the above kinematic equation.

$\begin{array}{ccc}\mathrm{\Delta }{U}_g& =& \mathrm{\Delta }K\\ m𝐠h& =& \frac{1}{2}m{𝐯}^2\\ m𝐠\mathrm{\Delta }h& =& \frac{1}{2}m{\mathrm{\Delta }(𝐯}^2)\\ 𝐬& =& \mathrm{\Delta }h\\ 𝐠𝐬& =& \frac{1}{2}\mathrm{\Delta }({𝐯}^2)\\ 2𝐠𝐬& =& {𝐯}^2-{{𝐯}_0}^2\\ 2𝐚𝐬& =& {𝐯}^2-{𝐮}^2\\ {𝐯}^2& =& {𝐮}^2+2𝐚𝐬\end{array}$