# Does work only equal the change in energy for non-conservative forces? If so, is power measuring work of non-conservative forces?

I have these formulas: work(net forces) = change in kinetic energy, and then work (non-conservative forces)= change in energy, and power=w/t=change in energy/time. So im confused about what the work in the power formula stands for? Is it only for non-conservative forces? Because it can’t be for net forces since net forces equal kinetic energy, right?

• Your question is rather broad. If you could give a specific example and ask questions of that example, it would be easier to provide an answer to your question. Commented Feb 14, 2022 at 22:46
• I find the definition of work unclear, for example here: en.wikipedia.org/wiki/Work_(physics)#Work_and_energy. I will stick to energy. Commented Feb 14, 2022 at 23:33

then work(nonconservative forces)= change in energy,

This is where you are going wrong. A change in kinetic energy can be due to conservative forces as well as nonconservative force.

Consider the work done by gravity, which is a conservative force, on an object falling from rest at a height $$h$$ from the ground in the absence of air resistance. Per the work energy theorem

$$W_{net}=mgh=\frac{1}{2}mv_{final}^{2}-\frac{1}{2}mv_{initial}^{2}$$

For $$v_{initial}=0$$

$$mgh=\frac{1}{2}mv_{final}^2$$

Now the time it takes to fall a distance $$h$$ is based on

$$h=\frac{1}{2}gt^2$$

so

$$t=\sqrt\frac{2h}{g}$$

Given that

$$P=\frac{W}{t}$$

You can calculate the power delivered to the falling object by gravity.

Hope this helps.

• It is true, however, that the work done by non-conservative forces is equal to the change in kinetic energy plus the change in potential energy. From the way the OP phrased things, I suspect that's what they meant. Commented Feb 14, 2022 at 23:30
• @MichaelSeifert Is it not true that If work results in a change in potential energy, then a conservative force has to be involved? Can you give me an example where there is a change in potential energy when no conservative force is involved. Commented Feb 17, 2022 at 21:45
• You're correct there, but I wasn't trying to say otherwise. I was trying to point out that if by "energy" the OP means mean "kinetic energy plus potential energy", then the OP's statement that "work done by non-conservative forces = change in energy" is actually correct (contrary to your first sentence.) Commented Feb 17, 2022 at 21:48
• @Michael Seifert Perhaps. But it was my impression the OP thought that only non conservative forces can cause a change in KE. I think we can agree that it’s not really clear what is being asked. Commented Feb 17, 2022 at 21:57

From $$\mathbf F = m\mathbf a$$, making the dot product with a infinitesimal displacement $$\mathbf {dr}$$, we have $$dw = \mathbf{F.dr} = m\mathbf{a.dr}$$. The right side of the equation can be proved to be $$d\left(\frac{1}{2}mv^2\right)$$.

So, the work of the net force is always equal to the change in kinetic energy. The force doesn't need to be conservative.

Power is the derivative of the work with respect to time.

The definition of work done being equal to the change in kinetic (or potential) energy is specific to conservative systems. Power, the timed rate in change of a systems energy usage, is not specific to conservative systems.

These are two separate concepts. For example, if a conservative force $$F$$ changes an object's velocity, then we say that the work done by $$F$$ is $$W=\Delta E$$ where $$E$$ is the kinetic energy.

But the power required to do so is $$P=\frac{\Delta E}{T}$$ where $$T$$ is the time taken to change the object's velocity. But at the same time, our object could have been made out of a metal, and its velocity changed by a magnetic force, which is not conservative. Nevertheless, the amount of power is still defined by $$P=\frac{\Delta E}{T}$$

Power is therefore a general term and not specific to conservative systems, and always describes the rate of energy used by a system. It is not only defined in mechanical systems, but others, like electrical systems.

It sound like you have a decent handle on conservative forces. Let's compare an instance of conservative forces with non-conservative forces. Conservative: An object falling in a vacuum in constant gravity. Bob D has already spelled out that example. Now, how about an object falling in constant gravity, but inside a thick atmosphere, or even water, instead? In this case the change in potential energy will still be $$mgh$$. But the change in kinetic energy, $$(mv_{final}^2-mv_{init}^2)/2$$, will be something smaller than $$mgh$$, perhaps much smaller. And this raises the question of where the energy went.

Well, as the object was passing through the atmosphere, or water, it collided with millions (billions, whatever ...) of molecules and transferred energy to them. Gravity gave energy to the object, but the object, instead of turning all of that energy into kinetic energy, "gave" some to the atmosphere.

If we had the ability to measure, we would see that the atmosphere was heated somewhat by the object, and the energy of the heat would be equal to the "missing" kinetic energy of the object at the end of its fall. So energy is always conserved, as far as we know (and that's what it's looked like for about 400 years now), if the energy of the entire system is taken into account. In this case, the energy of the entire system is the falling object plus the atmosphere. So in this case, the work done by gravity, which is still equal to $$mgh$$, does not equal the change in kinetic energy. But the work done by gravity does equal the total change in energy of the system.

As for power, that is still the work done by gravity on the whole system divided by the time over which the work is done. But it is no longer the change in kinetic energy divided by the time. So power is still $$mgh/T$$, but not change in kinetic energy divided by $$T$$.