 Okay, so, I don't know if you remember or if you watched. I did a video where we looked at the momentum principle. And so in that I had a little robot space probe that I could fly around. And so let me redraw that. And it's got a little rocket on it. Like that. And let's just say it has a mass of, I'm going to make this and it has a mass of two kilograms. And then this thruster exerts a force of, let's say, three newtons, making stuff up. Now, before I did several cases where I looked at if the rocket fired for a certain amount of time, then what would be the new momentum and what would be the new position? What if I change the problem? What if I said, okay, this robot is going to, let's call that M. Let's call that FT. And let's say the robot wants to fire the thruster for a distance S. Let's just call that one meter. Okay, so it's going to fire the thruster. So it starts at rest, fire the thruster, and then turn off the thruster right here. Now I know how far it's going to go. But how fast is it going to go? What if I want to use a momentum principle? What if I want to say F equals delta P over delta T? Can I do that? Well, I know F and I could call this X, Y. So I know F, I know the initial momentum. So this is equal to, I could write this as P2 equals P1 plus F delta T. If I write that as P2 minus P1, I could do that. And so I know P1 is the zero vector. I know F, but I don't know T. There is a way you could find T and we've done things like that before when we dropped objects. You'd have to actually use the average velocity and things like that. Okay, but there's another way. If I don't know the time, I'm not going to use the momentum principle. The momentum principle uses time. So in this case I use the work energy principle. So this says the work is change in energy. So why is this better? Because what is work? What's the definition of work? If I only have one force, the work done by that force is going to be F dot, let's call that delta S. Notice it's called S. Don't you hate when I change things? I will. Where this is, you can't multiply two vectors. It's the dot product. Okay, and I explained that other places. But in this course, pretty much you could probably get away with always doing, writing it as the magnitude of F, the magnitude of S times the cosine of the angle between those two. So in this case, you're moving that way, the force is that way. So theta would be zero and the cosine is zero is one. But here you see this is a product of force and displacement, not force and time. So when we deal with things that move over a certain distance, then the work energy principle is much easier to use. Now, let me go ahead and point out one other huge difference between the work energy and the momentum principle. Vectors. This deals with vectors. So the momentum principle deals with the momentum as a vector and the force as a vector. These are vectors, but when we take the scalar product or the dot product, you get a scalar answer. So there's really, if you write it like this, that's a scalar, scalar, scalar. So it really doesn't tell you anything about the direction. Okay, so let's go ahead and set this problem up. So I can, what about the energy? That's work. What about delta energy? Now when you do delta in the change in energy, it's very important. First, it's a change, just like here we're dealing with a change momentum. Second, what is my system? What am I talking about? It's changing energy. In this case, I think you would probably just pick the space probe, the robot thing. That's the best choice. But in other cases, you could pick other things. But the key is whatever, if you have a system, this is really the work done on it by the outside. So in this case, I'm going to have work done by the rocket. I can't have the rocket and all the gases as part of my system. If I did that, I wouldn't have work done by that. Okay, so here I have a spacecraft, a little robot. What kind of changes in energy can it even have? It can just have a change in kinetic energy. So let's put change in k. And kinetic energy is one-half m v squared. So where v is the magnitude of velocity squared, because you can't square a vector. Okay, so let's put this all together. So the work is going to be fs. And then the change in kinetic, that's going to be equal to the change in kinetic energy. I'll put that as k2 minus k1. The initial kinetic energy is zero. So fs equals one-half m v2 squared. So I can solve that for v2. v2 equals two fs over m square root. Now let's just put in some values because that makes everyone happy. Square root of two. And then I had the force of three newtons. The displacement was one meter over two kilograms. Okay, so let's check our units to make sure units work out right. A newton is a kilogram meter per second squared times meters, it's going to be squared, divided by kilograms, and then square root. So the kilograms cancel. I get meters squared per second squared, take the square root, I get meters per second. Okay, so it's six, so it's square root of three meters per second. Let me see how much time I have because I want to... Okay, really quick. Really, really quick, let me show you that this is the exact same thing we got before for how fast is something moving when you drop it. If I drop this ball from a height h, then I have gravity mg, and there's my height h, and it starts from rest. So if you turn that sideways, it's exactly the same as this. So I'm going to get the work is going to be mg times h. It's still going to be positive. I'm doing work done by gravity. I'm not doing gravitational potential energy. So I get the work is mg h, and then I get this initial kinetic energy zero, so I get one-half mv2 squared. Solve this for v2, and I get the square root, the masses cancel, and I get 2gh. And that's exactly, if you look back what we did before, that's exactly the same thing we got before. Okay, so that's a simple work energy example.