General definition of potential energy
Consider a flower pot falling from a balcony. As we have seen, there are two ways of looking at the process: the work approach and the potential energy approach.
Work approach
|
|
As the pot falls, gravity does work on the pot, increasing its kinetic energy. The pot started from rest: \(v_0 = 0, \Rightarrow E_{\mathrm{K}0} = 0.\) The work–kinetic energy theorem allows us to find the change in the kinetic energy: \[\sum{W} = \Delta E_{\mathrm{K}} = \frac12 m v^2 - 0 = \frac12 m v^2.\] This work is done by gravity: \[\sum{W} = m g h \cos0°{} = m g h.\] So overall, we can write \(m g h = \frac12 m v^2.\) |
Potential energy approach
|
|
We can also say that the pot-Earth system had energy when the pot sat at the balcony due to the pot’s position in the Earth’s gravitational field. This potential energy gets converted to kinetic energy during the fall. The sum of the kinetic and the potential energy stays constant in the process. Potential energy is energy due to position, so first we need to define a reference level to which this position is compared. Let us place it to the final position. To stay consistent with other laws of physics, such as the work–kinetic energy principle, the value of the potential energy at height \(h\) above the reference level must be equal to the work the force (in this case the force of gravity) does as the object moves from its current position to the reference position. We have seen that it is \(m g h\): \[E_{\mathrm{P}0} = m g h.\] The pot is at rest here, so its initial kinetic energy is zero. In the final position, the potential energy is zero, as the reference level is there: \(E_{\mathrm{P}1} = 0.\) Here the kinetic energy is \(E_{\mathrm{K}1} = \frac12 m v^2.\) Let us compare the sum of the kinetic and the potential energy in the two positions: \[E_{\mathrm{P}0} + E_{\mathrm{K}0} = m g h + 0 = m g h;\] \[E_{\mathrm{P}1} + E_{\mathrm{K}1} = 0 + \frac12 m v^2 = \frac12 m v^2.\] We have seen in the work approach that \(m g h = \frac12 m v^2,\) which means that if we introduce the potential energy due to gravity as we did (\(E_{\mathrm{P}}(h) = m g h\)), the total energy of the system, that is, the sum of its kinetic energy and potential energy, will be conserved (will stay the same): \[E_{\mathrm{P}0} + E_{\mathrm{K}0} = E_{\mathrm{P}1} + E_{\mathrm{K}1} = \mathrm{const}.\] This allows us to describe the motion of the object between any two positions. |
How did we introduce the potential energy due to gravity in the example above? As the work done by gravity as the object moves from its current position to a reference position. This definition can be generalised to any conservative force, not just gravity.
Potential energy
The potential energy of a system associated with a conservative force acting in the system is always equal to the work done by the conservative force as the object moves from its current position to a reference position:
\[E_{\mathrm{P}}(\mathbf{r}) := W_{\mathbf{r} \rightarrow 0},\]
where \(0\) denotes the reference state and \(E_{\mathrm{P}}(\mathbf{r})\) is the potential energy in position \(\mathbf{r}.\)
Using integration, the potential energy can be written as
\[E_{\mathrm{P}}(\mathbf{r}) = \int_{\mathbf{r}}^{\mathbf{r}_0} \mathbf{F}(\mathbf{r}) \mathrm{d}\mathbf{r},\]
where \(\mathbf{r}_0\) is the position vector in the reference state.
Potential energy is only defined for conservative forces.
The choice of the reference level is arbitrary, thus the value of the potential energy in any given position is never defined unambiguously – it can assume any value, depending on the reference level.