Work done by a non-conservative force and change in potential energy I know that the work done by a non-conservative force is equal to the change in total mechanical energy (from Work-Energy Theorem). But I read in a place that "Non-conservative forces don't affect PE".
So I am confused. How does the work done by a non-conservative force affect the potential energy?
 A: Your question seems to arise from a problem in which there is both a conservative and a non-conservative force. When you say "PE" you must be referring to the PE of the conservative force (by definition there is no PE of a non-conservative force).
The work done by the conservative force does not depend on the path. Therefore you can define the potential as
$$\phi(x_0) - \phi(x) \equiv W_{x_0\to x}$$
Notice that:


*

*The potential is defined up to a global offset: you can arbitrarily choose the value $\phi(x_0)$ but afterwards any value of $\phi(x)$ is defined.

*This is a well posed definition just because $W_{x_0\to x}$ is a well defined quantity (depends only on $x_0$ and $x$, by definition of conservative force). This is not the case for the non conservative force.

A: In Newtonian Mechanics, the work-energy theorem states that net work = change in KE (not in PE). 
Let's consider an example. Imagine a block of wood falling down on a rough ramp.  The non-conservative force here is friction, and the conservative force is gravity. On a frictionless surface, gravity converts PE into KE while conserving the total energy. However, in the rough ramp that we are considering, the non-conservative friction does work on the block and thus decreases KE to 0 eventually(dissipation of energy). 
The interesting point about work-energy theorem is that it doesn't rely on the force being conservative, while conservation of energy clearly does. So there is really no need to think about the concept of PE while using work energy theorem because PE is defined only for conservative force fields. 
