I will give this question a little context. Firstly, as I understand it, as soon as I "close the switch" on a circuit, electric current pretty quickly establishes a steady state where, at any given cross section along a wire, the average kinetic energy of that slice is both constant and equal anywhere along the wire. (Please correct me if I am misunderstanding). 

If the average kinetic energy is constant anywhere throughout the wire in the circuit (which confuses me, as I would think that the electrons should all accelerate after exiting the resistor since they are no longer being impeded *as much*), that means that the average velocity is constant. So, in the example of a simple circuit with a battery of v Volts and a resistor of o Ohms, my question is the following:

Because a 'voltage drop' is known to occur across the resistor, what type energy is being "traded" to generate the heat that radiates from the resistor? I would think that the reflexive answer is "electrical energy"...hence the VOLTAGE drop (the sacrificed energy is clearly not kinetic, as the velocity is constant everywhere). However, I find this confusing. Does this mean that if I had a positive test charge, it would be *easier* for me to bring it to the beginning of the resistor as compared to bringing it to the end of the resister? 

Further, if the correct answer IS "electrical energy", why exactly DOES electrical energy get "dissipated" as the electrons pass through the resistor? I always here the comment **Oh! It's because all the electrons are running into densely packed crap which impedes their flow** but that to me sounds like a reason for their KINETIC energy to be reduced. However, clearly that's not the case. So, ultimately, I guess the real question is:

\*\*What goes on in the resistor that is literally removing electrical energy \*\*. I am sure that this is a quantum mechanics question, but my quantum mechanics knowledge is not terribly strong. If I could get an answer that is devoid of crazy wave function equations, I would greatly appreciate it. Thanks!