# Conservative and non-conservative forces problem

I know that conservative forces conserve mechanical energy and non-conservative forces do not conserve mechanical energy and that for conservative forces a potential energy $PE$ can be defined while for non conservative force a potential energy cannot be defined since the work done is independent of the path taken in the first case and in the second case the work done is dependent on the path taken and hence the energy transfer in both the cases, i.e., for conservative forces,

$$W=-\Delta PE$$

Now i am really messing up conservative forces, the equation $W=-\Delta PE$, and non-conservative forces.

Does this equation mean that conservative forces can change only the potential energy of a body and that non-conservative forces can change only the kinetic of the body or the opposite or that both conservative and non-conservative forces change both the kinetic and potential energy of the body?

A better way to look at your equation, in my opinion, is $$\Delta PE = -W_{\mathrm{cons}}.$$

That's closer to a definition of potential energy. But, in either order, another concept is that you can substitute a change in potential energy for the work by a conservative force when you analyze the motion of a system. In other words, you use either the work done by a conservative force or the potential energy contributions of that conservative interaction, but not both.

For example, in a system involving gravity and air resistance one could write (using $K$ for kinetic energy, $W$ for work, $U$ for potential energy) $$K_{\mathrm{initial}}+W_{\mathrm{grav}}+W_{\mathrm{air}}=K_{\mathrm{final}}$$ using the work-energy principle, $$\Delta K = \Sigma_{\mathrm{all}} W.$$ Or$$K_{\mathrm{initial}}+U_{\mathrm{g,initial}}+W_{\mathrm{air}}=K_{\mathrm{final}}+U_{\mathrm{g,final}}.$$

You can see these are equivalent by subtracting $U_{\mathrm{g,final}}$ from both sides of the last equation and applying $\Delta U_{\mathrm{grav}}=-W_{\mathrm{grav}}$.

So the answer to your "or" question is neither of those is correct:

1. Conservative forces can change kinetic energy and can be accounted either by the work they do or the change in potential energy of the system, and
2. Non-conservative forces can change kinetic energy and must be accounted by the work they do. There isn't a potential energy function which helps us.
• So conservative forces can change both the kinetic and potential energy while non-conservative forces can change only the kinetic energy(and potential energy too but this potential energy is not useful to us.Right?
– MrAP
Oct 14, 2016 at 15:19
• Looking at the definition of potential energy, potential energy changes directly with the work done by the conservative force associated with it. Non-conservative forces do not directly change the potential energy function because they have no potential energy associated with them. Non-conservative forces transfer energy in or out of the system and that affects the total energy within the system. Oct 14, 2016 at 16:23

Both conservative and non-conservative forces can change both kinetic and potential energy. I'll try to clear your confusion packing it all in one equation:

$\Delta KE + \Delta PE = W_{ncf}$

Where $W_{ncf}$ is the work done by non-conservative forces. But as you said, $\Delta PE=-W_{cf}$ so

$\Delta KE = W_{ncf} + W_{cf}$

Both conservative and non-conservative forces can change kinetic energy (for example gravity or you pushing a block, respectively). Conservative forces, by default, change $PE$ (potential energy naturally tends to its minimum) while increasing $KE$. However (as Bill pointed out) non-conservative forces can't change potential energy because they don't have a potential associated with them, they can only change kinetic energy. This change in KE can now change potential energy.

DERIVATION OF THE ABOVE EQUATION

We know (from Newton's law) that

$m\frac{dv}{dt}=\sum F_i(x)$

Multiplying the above by $dx$

$m\frac{dv}{dt}dx=m\frac{dx}{dt}dv=mvdv=\sum F_i(x)dx$

integrating the equation above we have

$m \int_{v_1}^{v_2} vdv=\frac{1}{2}m(v_2^2-v_1^2)= \Delta KE= \int_{x_1}^{x_2}\sum F_i(x)dx=W_{ncf} + W_{cf}$

But as $\Delta PE=-W_{cf}$ we have that

$\Delta KE + \Delta PE = W_{ncf}$