# Conservative Forces

What’s this, and why is it different from any other Force?

It’s not a far-right force, nor is it a frugal force. No, it’s merely a type of force for which one can describe the Work done by the force in terms of some kind of Kinetic or Potential Energy trade-off.

Let me explain.

It’s just a type of force, k? It’s a categorization of forces – they can either be conservative or non-conservative.  Making sense?

What makes a conservative force special is the point I made above – if a force is conservative, we can say something about it. Namely, that the Work-Kinetic Energy relationship is completely reversible.

So the work done on an object by a conservative force gets completely turned into kinetic energy (a la The Work-Energy Theorem) – and you can get all of that back via the same force (stored as Potential Energy). At least, you can conceivably do that.

The easiest-to-point-to conservative force out there is Gravity. You’ve got some chemical energy stored in your biceps (hopefully) that you got from some PBJs your mom made (optional), and you wanna throw a pop-fly (baseball) on the moon for some reason (no atmosphere = no worries). At first, all that energy is in you – you have the power(energy)! – and the total energy of the ball is merely the sum of it’s kinetic energy (currently nil) and it’s gravitational potential energy (how high ya holding that ball?): $E_{total}=K+U$

If you were to drop that ball, it’ll fall down (doy), and it’s potential energy decreases at the same rate as it’s kinetic energy increases: ${\Delta}KE = -{\Delta}U$

Now, if you threw it, you do work to the ball, and it’ll go flying off – but what goes up must come down, and that energy you transferred to it will work against the force of gravity for as long as it can, until at the peak of it’s ascent, when it will have it’s maximum potential energy, and it’s vertical kinetic energy is zero.

The point I wanna make with this example however, is that it’s the component of the work you did that’s in opposition to gravity that really matters. If you threw it straight up, then at some point it only has potential energy, and no kinetic energy, and as it hits the ground, it’ll have only kinetic energy and no potential energy – hence the whole thing about gravity being conservative.
(horizontal velocity vectors need not apply)

The consequence of this is that conservative forces have the property that the work done by that force in moving an object from point A to point B is independent to the path taken. The same cannot be said of non-conservative forces, like Friction for example. It totally matters whether you take the scenic route, cuz thems what makes your summer tires look tired!

To make the point about the path independence clearer, realize that it’s only a particular parameter that decides the potential in space – in Newton’s Universal Law of Gravity, it ‘s the displacement from the point mass $U=-mgh \vec{g}$ and in Coulomb’s Law, for the electric potential off a point charge Q (Coulomb force is also conservative!), it’s the inverse displacement $U=\frac{1}{4{\pi}{\epsilon}_0}\frac{Q}{r}\vec{r}$

More on this over at Work-Energy Theorem, Energy, and probably other places too. Google it folks!