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Skill 12 of 13
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# Potential energy vs force

Gravitational potential energy $U_g$ is stored as the boat reaches the water fall. When it tips over, the gravitational force – the weight $w$ – makes it fall and the stored energy is released metre by metre during the fall.

• Had gravity been larger, so a larger force (weight $w$), then more energy $U_g$ would have been stored to begin with. During the fall more energy would then have to be released per metre.
• Similarly in a stretched spring, if the spring is stiffer1 so the elastic force $F_\text{elastic}$ is larger, then more energy $U_\text{elastic}$ is stored even for the same elongation. More energy must be released per metre when it springs back.

It seems to be a pattern that the force represents the potential-energy change per distance. A ‘potential-energy change‘ would be written $\Delta U$ and the ‘distance‘ $\Delta s$, so $F=\frac{\Delta U}{\Delta s}$. Since the force might change during the energy release (for instance, the elastic force changes), it represents a tiny, tiny change over a tiny, tiny distance before its a new force, so in general we’ll write it with $\mathrm d$ instead of $\Delta$ as usual: $F=\frac{\mathrm d U}{\mathrm d s}$.2

Note: This is a derivative not to time as we are used to, but to distance or actually: position. In 3D, when a ball is rolling in a landscape, the 3D version of “derivative” is of what we call the gradient with the symbol $\nabla$. So, we can write this general recipe for this ‘force-and-potential-energy‘ relationship:

$$\vec F=\begin{pmatrix}\frac{\mathrm dU}{\mathrm dx}\\\frac{\mathrm dU}{\mathrm dy}\\\frac{\mathrm dU}{\mathrm dz}\end{pmatrix}=\nabla U$$

Note that this usually becomes simpler when used. For instance, weight $\vec w$ only pulls vertically down, so along the $z$ axis, thus the gradient of the gravitational potential energy only has a $z$ component:3

$$\vec w=\nabla U_g=\begin{pmatrix}0\\0\\\frac{\mathrm dU_g}{\mathrm dz}\end{pmatrix}$$