 For our purposes, we define work as the integral of force with respect to displacement. In a past physics class or something you may have described work as the force times displacement itself, but that's only true if force is constant. If it's constant, it comes out of the integral and we're left with force times delta x. For our purposes, we have to account for the fact that the force might be changing, and our work will be changing as a result. We don't just use force times displacement. But we can approach it from this perspective in a couple of different directions. We could apply our definition of work to a problem, or we could use a couple of shorthand simplifications for specific purposes. Those shorthands include electrical work. In electrical work, what we are describing for our purposes here is actually electrical power. And electrical power we are saying is the voltage multiplied by current. Now yes, there's a conversation to be had here about power factor, but for right now we are saying the voltage times the current gives us electrical power. And then because we can relate voltage current and resistance to each other with Ohm's law, we could write that in a couple of different ways as well. We could say, since voltage is IR, we can write this as current squared times resistance, or we could say voltage divided by resistance squared. And if we really wanted to write this as a work, we can always remember that power is going to be work divided by time. Across an entire process that would be work divided by duration, in which case the electrical work could be written as electrical power times duration. And we can come up with this equation by looking at the actual forces involved, considering coulombs and whatnot. But for our purposes here we are just going to use this shorthand, which comes from our definition. The next shorthand simplification we'll be encountering is shaft work. For shaft work we start with our definitional work. And we recognize that for displacement what we are really talking about is a rotating shaft crossing a system boundary. This might be attached to say a propeller, something like that, maybe a fan. And instead of displacement, we really would rather talk about revolutions. And as this thing is revolving, the displacement that is effective here is going to be the perimeter multiplied by how many times it's gone around. So we could describe that displacement in terms of revolutions per second and the radius of the effective shaft. Furthermore, the force is a little bit more useful if I actually talk about torque. We describe torque as force times our radius, therefore I can write force as torque, which I will just shorten to t divided by radius. And if the torque is constant and the radius is constant, then force would be constant. It would come out of our integral and we'd be left with force times displacement. And I could substitute in torque over radius for force and instead of displacement I could describe the perimeter of that shaft, which I will describe as pi times the diameter, but instead of diameter I want radius, so pi times two times radius. That's the perimeter and then I will call number of revolutions N. Now my shaft work simplifies because the radius cancels to two times pi times torque times number of revolutions completed. And since again I'm probably more conveniently able to write shaft power, that's going to be two times pi times time times number of revolutions divided by time. And dt I could just wrap into N and write this as two times pi times torque times number of revolutions per unit time, which I will write as N dot. Does that make sense? So this is just a simplification, a shorthand that we use for describing shaft power, and this is the torque applied multiplied by the number of revolutions per second. So if we're talking about shaft work coming off of an engine, if it has a torque of 200 foot-pounds and I'm describing 2,000 revolutions per minute, I could figure out how much shaft work is actually being applied here. Our next shorthand simplification for work is spring work. And spring work is useful in describing springs, imagine that. And for our situation here we are only describing one type of spring, a linear spring, where x is measured as some sort of displacement from its initial resting position and the force associated with expanding or contracting the spring is written in terms of a coefficient that we call k times that displacement. That k value is the spring coefficient, and then our spring work is going to be the integral of that force, k times x, with respect to displacement. k comes out, I'm left with the integral of x dx, which is going to be 1 over x squared evaluated from x1 to x2. So I can write this as 1 half times the spring coefficient, multiplied by x2 squared minus x1 squared. And note that these two x values are relative to the initial, the resting position of the spring. Then our last shorthand simplification for work is called boundary work. And boundary work is the form of these that we are going to encounter the most often in our class here. Boundary work is the work associated with a moving boundary. Like for example if we had a piston cylinder arrangement and the piston moved down, expanding our volume, there is a work associated with that. That expansion process is described as occurring with a work and we call that associated work the boundary work. And like with shaft work, this is really just a matter of simplifying the definition of work so as to be more convenient for the sorts of parameters that we're going to have access to in this sort of problem. So I'll start with the integral of force with respect to displacement. And then I will consider what I'm likely to have in a piston cylinder problem. If I probably don't actually care about the force itself, I will probably more conveniently have access to say pressure. If this was a gas, I would probably be able to describe the pressure of the gas. And I know that the pressure of the gas is going to be the effective force applied by that gas divided by the area of effect. Therefore I could write the force applied by a pressure as being that pressure multiplied by area. And this boundary work, pressure times area times displacement, can be simplified a little bit further by writing area times displacement as d volume. And that volume differential represents a tiny little step. In this case, a little slice of cylinder that is infinitesimally thin going all the way down from X1 to X2. Even in a situation where our area was changing, say instead of a piston going down, we were blowing up a balloon, that work associated with the expansion process. That's actually kind of a bad example because there's also going to be spring work associated with the resistance of the elastic. But depending on how you define your perspective on the problem, you could consider that as either boundary work or spring work. But the work associated for the expansion process would be written out as an area times displacement, or rather pressure times the area times displacement integrated, and the area would change with each step. So for each value of dx, as you're going out in the expansion process, you're going to have a different area. But your volume, d volume, is encompassing that. So even if area isn't constant, grouping them together as the volume differential is still the most convenient way to represent that expansion process. Therefore we write boundary work as the integral of pressure with respect to volume, and that gives us our fourth shorthand simplification for work.