 Okay, a more complicated work energy example, I still didn't do one with angles and stuff in it, but that's okay. Let's proceed on anyway. This is real. Okay, so there's this guy called Professor Splash, and what he does, other than being crazy, is jump from a very high distance into a tiny pool that's like one foot deep with one foot of water. This is like, let's say, let's just make it 0.3 meters depth, call it D, and this will be a height. I think he jumps from like 30 feet, so that's about, let's say, 10 meters. I'm just making up stuff. I do have the exact numbers somewhere. Okay, so the question is, what's going to happen to this guy when he hits the water? Well, he's going to stop. Let's see if we can find the force, the water exerts on him while he's stopping, and from that we can find his acceleration while he's stopping, because really, one of the measures of how dangerous something is or how much, how likely you already be injured is the acceleration. Everyone has different masses, so the force isn't really a good measure there. So we'll calculate the force, we'll get an expression for that, and then find the acceleration. Okay, so what else do we need to know? That's his mass. I think that's the depth of the water is D, and the height is H. Right away, you should say, I have two things that I know. I know the momentum principle, and I know the work energy principle. And I need to say, momentum principle or work energy, which gun do you draw, right? In this case, I mean, either will work, really, but which one's going to be a little bit easier? The key here is we're giving distances. If you know something how it changes over distance, then work energy's going to be better, because it deals with distance. So in this case, let me first do a couple things. Let me pick my axis, here's x and y, so y equals zero is down here. Then let me pick my system. It's going to be the professor's splash, I'll put that as doctor's splash. I don't know if he actually has a Ph.D. in splashology or not, or maybe it's just an informal term, plus the earth. Why plus the earth? Why do I include that? You don't have to. But in this case, I want to. If I have my system as both of these things, then the gravitational force between those two objects, professor's splash and the earth, wouldn't be in the work term. But we have to take into account that somehow, if we have a system like that, then we can have two things. Our energy at any given time would be the kinetic energy plus gravitational potential energy, where close to the surface of the earth, this is going to be MGY. And I derived that in the book, so you can check that out. So it's still kind of complicated because then what's going to do work on the guy? Work is going to be done while he's slowing down during this part right here. So let me blow that part up. Here's the pool with water, and then he lands in this position. So during that time, there is a gravitational force on him, but really I'm not thinking about that so much. I'm not worried about the gravitational force because it's part of my system. I am worried about this force of the water, I'll call it FW, and this is acting over distance D. Okay, so let's set up our work energy expression. So work equals change in kinetic plus change in potential, and he's going from here to there. This is position one, position two. So what does work? Well, the water does work, and that's it. So we're going to have F water, magnitude, D water, the displacement in the water, times cosine theta, and that's the work. So here, although he falls all the way down a height H, this force is only doing work over this distance D, and the force is that way, it's pushing up on him to slow him down, and he's moving that way. So what's theta? Theta would be 180 degrees, so in this case the work would be negative FWD, I'll just leave it as D. I want to emphasize it's the distance the force is exerted while he's moving in the water, not the whole thing. Okay, what about the change in kinetic energy? What does kinetic energy appear? If he jumps from rest, it's zero. What about down here? If he stops, it's zero. Okay, so the change in kinetic energy is zero. What about the change in potential? It's going to be final potential, which is going to be zero, because he's at y equals zero, minus initial MGH. So that's it. Do I have enough to solve for F? I do. So I got those zeros in there, so I get FW equals MGH over D, yeah. And so the positive thing on it means that yes, this is doing negative work, but you can do that, it means it decreases the energy of the system, because right before he hits right here he's moving pretty fast, he has kinetic energy, and then he stops. He also decreases in potential. But from this I can find out that that is the direction that he's going to be pushing. It does have the right units. MGH has units of force, distance over distance cancel, so that is a force. And also what about the higher I start, the greater that force? The shorter that I stop, the greater that force. Okay, so this is the force exerted on the jumper. Now I don't want to calculate that, because I don't really know his mass. It would be rude to ask someone's mass, right? It's rude. So let's just calculate the acceleration during that time. So I could say Fnet in the y direction equals MAy. Okay, so now I do have to include that gravitational force, if I want to calculate the net force. So here I have, I want to calculate the acceleration. Okay, so the Fnet is going to be this going up in the y direction, MGH over D minus MG, and that's going to be MAy. MAy should be positive, right? He is accelerating that way, even though he's moving down. So the masses cancel, I get GH over D minus G equals Ay. So this little term isn't going to do very much compared to these, but let's just put in some values and see what we get. Ay is going to be G, 9.8, I'll leave off the units, H I said was 10, D was 0.3 minus 9.8. Okay, so it's 98 over 0.3, I don't want to make a mistake, let me just do it on the calculator real quick, otherwise I'd feel foolish. So I get 316 meters per second squared. One of the common terms is to get that in terms of G's, so it's 32 G's, where 1 G equals 9.8 meters per second squared. Now there's, that's not really the best thing for G's, the G's actually wouldn't have that G in there, it's G, because if I'm standing here what's my acceleration, zero. But I'm considered 1 G, so it's a little confusing. But that's pretty high, but apparently it's on the, it's on the survivable point, okay. So I think the important thing to see here is that, I mean you can make this a pretty complicated problem. You can break it into two parts, this part where he's free falling and this part where he's slowing down, but you don't have to do that. You can just put it all together in one big thing and it's not that bad.