Now we'll calculate the work in a few cases. And we're going to use the force of gravity rather than me pushing so that we can actually calculate something. So first we're going to do drop and shoot. So you remember dropping shoot, two blocks and one fell and the other got shot, and they landed at about the same time, right? So we'll look at drop, look at the work that gravity did, and we'll look at shoot. And for all of these since we're doing now vector dot products, we need to be careful with our coordinate system. So let's put positive x this way, positive y this way, which means positive z has to be out. x cross y is how you find the z Direction. Let's see, in drop we had the case that we had a mass here, mass m, and it basically just falls through a displacement d, all right? Or let's call it delta r, just to match our previous notation. And we could say, what's the work by gravity? Well, the work is f dot delta r. So the force of gravity f is what? It only has a y component, right? So it's 0 i hat minus mg j hat plus 0 k hat, all right? And what is delta r if we want to write it as a vector? Is let's see, in the x direction is zero i hat, and it is we dropped a distance d. I meant to say that's the displacement vector, but we dropped a distance d, right? So the displacement was negative d in the j hat direction. We didn't move in the k hat direction. So to find the work using our Cartesian definition of the dot product, then it would be just this times this, 0 times 0 is 0, it would be minus mg times minus d, right? That's mgd, and then the two minuses turn into a plus. That's just a scalar. Plus this one times this one is also 0. So the work is mgd, the work that the gravitational force did on the object. Now, although, let's look at shoot, all right? So shoot was the one that we shot it with a little bit of an initial horizontal velocity. It also fell through d, but it kind of went like that, all right? And it made it through some distance of the range. So we'll call that r, but we'll still keep this as the origin, just like we did before, okay? So let's calculate the work. In this case, let's see, the force, again, is only minus mg in the j hat direction the whole time. There was no force in the x, it just had an initial velocity in the x. And the delta r, the displacement was, let's see, everything's negative here. It was no, it's positive r. r i hat minus d j hat plus 0 k hat, all right? So there's your two vectors. And if we want to do this one Cartesian, why not? We would say the i hat components, well, it was 0 here is 0 times r is 0, so the fact that it moved that direction doesn't really help us any. The work 0 for the i hat contribution minus mg times minus d mgd for the y contribution. And then 0 times 0 is 0. So you see that gravity did the same amount of work whether you dropped it or whether you shot. I'll talk about what that means in a minute. Let's do one more. Let's do a ramp. So here I have a one-dimensional ramp set up. And Frank is going to roll down the ramp for us, okay? So Frank is here, he's feeling a gravitational force, and it's going to push him down the ramp. So we're going to figure out how much work is being done down the ramp. Here we go. Let me draw that real quick. So it looks a little bit like this. Let's imagine it went down the same distance d, and let's say the ramp is at an angle theta, okay? And it went down like there he was, [COUGH] like that, lose down the ramp. Let's see, so this is a case where we wanted to maybe think about doing it in polar coordinates just for fun, right? So if we think about this, then the displacement would be this vector, all the way from here all the way down to there. There's delta r. And the force, well, the force, of course, mg is down. There's a normal force up, in the end you get an unbalanced component that pulls it to the side, you get mg sine theta. So the force pulling you down is mg sine theta. So let's think about doing the work the polar way. Work is the magnitude of the force times the magnitude of the displacement times the cosine of the angle between them, okay? Well, let's see, look at it. Let's see, the magnitude of the force is mg sine theta, okay? We have a sine theta, even though it's not this theta, it's because it's just a component of the force. The displacement is what? [COUGH] It's not just d. You gotta go with the actual displacement. It did a force this way. We have to go this way. We have to know what this distance is. Well, sine theta is the opposite d over the hypotenuse. And this is a case where we actually want the hypotenuse. Now, usually you're given the hypotenuse and you break it into components. This is a little different, we're given one of the sides. So actually the hypotenuse we want is d over sine theta. d over sine theta, that's the magnitude of the displacement. And then we get the cosine of the angle between them. Cosine, what's the angle between them? 0, I'm doing this down the plane, I'm not breaking things into components. Cosine is 0, and that's just 1, right? So that's 1, and those two cancel, and you get the same thing, mgd every time, all right? So the point is what you're supposed to notice is if the mass, if the force is really down, okay, and if the mass moves a distance d down, it doesn't really matter what else it does. In this case, it moved a distance d down, mgd. In this case, it also had some horizontal component, doesn't matter, there's no force that way, mgd. In this case, we didn't even treat the forces down. We calculated the component this way and it still came out mgd. So that's why sometimes you'll see this written in books as, I don't know, the work is like the force parallel times the distance that it goes. And what that means is just the force or is the distance along the force direction or something. But what they're really trying to find a way to say is it's the dot product. The dot product gives you the component of one vector along another. And all that really matters in work is, how far did it travel? What was this displacement along the actual force? And that's what we got here, three different ways.