# Mechanics: Chapter 13 – Conservation of Mechanical Energy

When a force is conservative, the work done by the force on an object equals the change in the potential energy of the object. If the work done is positive, there is a decrease in the potential energy of the object. Similarly, if the work done is negative, there is an increase in the potential energy of the object. Hence,

• $W = - \Delta U$

Energy is always conserved. Essentially the sum of the kinetic energy $K$ and the potential energy $U$ of a system is constant.

• $K + U = c$ where $c$ is a constant

This also implies that if no external energy is added or removed from the system, the initial energy $E_{i}$ of a system is equal to the final energy $E_{f}$ of a system, and hence

• $K_{i} + U_{i} = K_{f} + U_{f}$

This is easy to prove in the case of a conservative force like gravity. Let’s say an object is dropped from rest in free fall, now the work done, $W$ by gravity on the object is, as I’ve already established,

• $W = - \Delta U$

and since the force of gravity accelerates the object from rest to a speed $v$, the work done by the force of gravity on the object increases the Kinetic energy of the object. In other words,

• $W = \Delta K$

Now

• $\Delta K = - \Delta U \Rightarrow \Delta K + \Delta U = 0$

If the initial Kinetic energy and potential energy of the system were $K_{i}$ and $U_{i}$ and

• $K_{i} + U_{i} = c$

Then the final kinetic and potential energies, $K_{f}$ and $U_{f}$ of the system can be written as the sum of the initial energies and the change in energies. Hence,

• $K_{f} + U_{f} = (K_{i} + \Delta K) + (U_{i} + \Delta U) = (K_{i} + U_{i}) + (\Delta K + \Delta U)$

We know that $\Delta K + \Delta U = 0$,  hence

• $K_{i} + U_{i} = K_{f} + U_{f}$

Quod erat demonstrandum.