# Is force the derivative of energy?

In my lecture today my professor briefly mentioned that force is the derivative of energy but I did not really get what he meant by that. I tried to express it mathematically:

$$\frac{d}{dt}K_E=\frac{d}{dt}\left(\frac{1}{2}mv^2\right)=mv\frac{dv}{dt}$$

This looks really close to Newton's second law $F=ma$ but there is an extra "$v$" in there. Am I missing something here?

• Good start! See what happens if you also have a potential energy term $V(x)$. – Holographer Jan 24 '15 at 12:53
• there is also a very similar discussion on physicsforum.com – glS Jan 24 '15 at 12:58
• Search term: "Work-energy theorem". But note that you are taking a derivative with respect to a particular independent quantity. What happens when you chose a different one. You are also examining only one kind of energy. – dmckee Jan 24 '15 at 16:29

## 2 Answers

It is important to understand to which derivative you are referring to, i.e. derivative with respect to what?.

For conservative systems, it is true that the force can be expressed as minus the gradient of the potential energy: $$\tag{1} \textbf F(\textbf x) = -\nabla V( \textbf x),$$ which can be though of as the defining property of a conservative system.

The gradient $\nabla$ reduces for one-dimensional systems to the derivative with respect to the space coordinate, i.e. you have in this simple case $$\tag{2} F = -\frac{dV}{dx}.$$

Taking as an example the case of a mass $m$ in the gravitational field of the earth, you have the potential energy $$\tag{3} V(z) = mgz,$$ where $z$ is the distance from the ground. The force in the $z$ direction is then given by $$F_g = - \frac{dV(z)}{dz} = -mg,$$ which is what you would expect.

• Makes sense now. Thank you for taking the time to write a response. – qmd Jan 24 '15 at 13:08

I would like to emphasize on a point in the previous answer - you get different quantities depending on what you differentiate energy with respect to.

In other words, if you differentiate energy with respect to time, you get Power (as you have derived in your question - $$m \times a \times v$$ would equal $$F \times v$$, which is power; also power is the rate of change of energy, so it makes sense.)

On the other hand, if you differentiate energy with respect to position, you get the variable force at that position, as explained above by another user.