 Oh, yeah, that helps, too. All right, yesterday we started working with the work energy equation. Rather than laboriously developing each little part one at a time, I gave you the whole thing. I gave you the whole work energy equation, but we didn't get a chance to actually do anything with it. Each part is developed in the book, read through it. You don't need to study it in great laborious detail. There are homework problems that help you work through each little piece of it as you go, so you know how to calculate each little piece. But most of them are pretty straightforward, and most of them you're familiar with already. If you took high school physics, especially in New York State, I know what's in that curriculum, and most of this business was in there. So let's get the work energy equation back up on the board. Remember what it was? Jolly W for work, but I had something extra on there. One, two. I don't, I can't remember. I don't think our book does that. That's just a reminder of what fact. That this is work, as we look at it, is the action of a force moving from one place to another. That's the full solution right there. But for most of our problems, remember what the dot product does for you here, why the dot product was so important to this? Yeah, we need the part of the force that's in the same direction as the movement, and that's what that does for us. So if the forces are constant, let's say we only have one, this comes down to be the F in the x direction, the force in the x direction times the movement in the x direction. The y component of the force, if any, isn't contributing to the work being done. It's pushing sideways, not pushing along with the way the object is to be going. It's that simple picture. In fact, you have a couple problems that will address exactly this. The force in the direction we're moving times the distance we move that is our definition of work. We'll go through a couple practice parts of that. There's several problems in the book where they give you a setup, a force, and some movement there, and you just post-calculate how much work is being done. It's fairly straightforward for most of the things we're doing. We want to get the rest of the work energy equation in there, and then we're going to talk about it a little bit more and then do a couple sample problems and show you how to work through it. Next part, we're doing some work on some object, whatever that crate was or anything else in these problems. But we want that work to do something for us. One of the possibilities is it can speed things up. If we're pushing in the same direction as motion, it's moving. That generally speeds things up. That's what happens when you push a box across the floor, speeds it up, it wasn't moving. Then it is moving. You spin it up. If you're pushing against the way it's moving, you can slow it down. That's what you do when you put on the brakes on your car. The friction is pushing against the direction the car is moving until it finally brings the car to the stop. So we see that in terms of our change in kinetic energy. Remember, each of the terms on this right-hand side are change in terms, and that's very important. This business doesn't work right. It starts to fall apart if you neglect these change in terms. So I want to see those deltas there every time I see this equation. What is the next term? Doesn't matter really what order they are. But what did I? We can do work on something to change its speed. We can also do some work on something to raise or lower it. If we lift something up, we're doing some work on it. We're pushing in the direction it's moving. We change its gravitational potential energy and put it up on the shelf. If we bring it down, we're pushing up still. We have to hold it up in the air, but we're bringing it down. It's moving in the opposite direction. We're doing negative work, and we're decreasing its potential gravitational energy. And then the last term was, Patrick, do you like that one? The change in the potential energy of an elastic medium. For our purposes here, it's simply a linear spring. What about work done by friction? What about work done by friction? Let's say there's some parameters. Let me, I need to do one more thing with the stuff that's on this side and the stuff that's on that side that may explain exactly where the friction falls. It may help you remember exactly where it falls. Obviously, this work term involves some forces. And I'll leave a little bit of a space above it for now, because I'm going to put something in there in a second. But it's also obvious, I hope, that these two termed over here involve forces. This just really involves speed. But these two very clearly involve forces. The force of gravity and the force of a spring are very much tied up in here. So I'm going to leave another little space because these are two different types of forces on either side of this equal sign. The forces over here are forces that we call non-conservative. Doesn't mean they're in the Democratic Party. What that means is we put forces over here that if we turn the problem around and go back to where we started, those forces get even worse in terms of how much work is done. For example, I have to push something across the floor. I push on it. I push it all the way across the floor. I turn around. I push it back to the first place where I stopped. I'm twice as tired because I had to come back. I didn't do some work going out and then get that work coming back and come all rested here. It took me twice as much work to bring it back. That's the nature of non-conservative forces. When you turn the problem around, you can't return the situation to where it was before. That's not true with gravity. If I have something on the ground here and I put it up on the shelf and then I bring it back down from the shelf, yeah, I've done a lot of work, but gravity is exactly the same. The box is exactly the same as it was before. The gravitational potential energy is exactly the same as it was before. In terms of the gravity, the box isn't exactly the same situation it was before. That's the nature of conservative forces. The spring, too. If I stretch the spring out and then I let it stretch back to where it was before, the spring is exactly the same. Assuming I didn't stretch it so far that I bent it, that's why we don't actually bring springs out in the lab. They all get mysteriously bent because all the boys try to show us how much they can stretch our springs and then they are ruined. But if I stretch the spring out and then I let it come back to rest, it's exactly the same spring. Nothing's changed. So that's why these ones are over here. These are considered conservative forces. So the question was friction. Where does friction fall? Here or there? Well, let's see. If I'm pushing something across the floor and it's scraping over the floor because of friction, it's gouging up the linoleum and it's heating up the bottom of the box and I turn around and I push it back. Does that repair all the gouges and cool off the bottom of the box? Not only does it not do that, it makes it even worse. So where should friction go? Here or here? Friction's gotta be over here because we can't get it back. We can't do anything in our problem to recover what we already lost in terms of the friction. That's damage done. Just like the effort our muscles are making. That's damage done. I can't get that back. That's the nature of the world. That's the nature of those forces. We can return the gravitational situation to exactly the same thing it was before. We can return the spring to exactly the same. We could unhook it, stick it back in the box, take it back if we had the receipt and get our money back. If he's not gonna look at it and say, where does I got it? You use this, you can't tell. And then this doesn't really involve forces either but these are all energy terms. Energy we do that we can actually only spend in one direction, whether it's work we're doing or work that's being done against the object, we can't turn around and get that back. This is kinetic energy, the energy involved in an object moving at a certain speed. We can return that to where it was before. If something's not moving and it speeds up and then later it's not moving again, the kinetic energy is just exactly what it was before. And these ones, we can return those to exactly the same situation they were before. But they're all energy terms. So that helped a little bit with where friction goes. Friction is always over here and it always does negative work, doesn't it? Why is that? Why does friction always do negative work? If the crate moving that way over the floor, which way is friction pointing? Always in the opposite direction. There's no way we can push something across the floor in that direction and have friction also point in that direction. Well, we could, I guess, put it on a conveyor belt, but that's a whole different situation. It's still, in terms of us, slighting that crate across the floor, the friction's always gonna do negative work. When we look at the work done by friction, it's always gonna be with a minus sign in there because the friction's one direction, the movement's in the other direction. We can't change that, folks. Okay, should we do a couple sample problems starting from easy and working towards beautiful? How about I decide? I like that idea. All right, in fact, what I'll do is we'll leave a look at the chapter seven homework problem. So that'll lighten up the spring break a little bit. Maybe you can get one more six pack of Miller light in there now. Yeah, Mark, you like that idea? Yes, sir, engineering students gone wild. That video's coming up. I wanna see if you guys can get on with it. All right, so we'll look at one problem here. We'll look at a couple. We got a chance to do a couple here. All right, problem seven seven. How much work is done by a 70 kilogram man racing up the Empire State Building? 1,600 steps in 320 meters in 10.9. So we got the Empire State Building there. That's a scale drawing. It's 320 meters high. The man is 75 kilograms. How much work he does racing to the top. So here's what I want you to do on these work energy problems. Well, first, I think I told you this yesterday. I hope I did. If not, I'll tell you now. The work energy equation is very useful for doing a particular type of kinetics problem. Or kinetics problems are those that involve the forces and the acceleration and the position and the distance. We've got the forces in there. Did I tell you what type of problems? Work energy problems of the work energy equation is particularly good for? Yeah, position dependent problems, which this is. This problem depends upon his position going up the Empire State Building. He has a distance to cover. Here's what I'd like you to do with the work energy equations when you need. Now some of the problems just say find the work. You know, you just give them the force and the distance you're supposed to calculate for work. But this one, you're not given the forces involved here you're to find whatever parts of it that you can. So here's what I want you to do with the work energy equation. Write it down and then go look at any parts of it that are zero because automatically the problem becomes smaller. And the smaller you can make the problem, the easier it is to complete it. So go look at any part of the problem and see if you can make it smaller as you go through it. Well, we know there's work being done because that's what we're supposed to find. Plus, if he runs to the top of the building, he's gonna be pretty tired. And if he runs back down, he's not gonna get less tired to be completely rested at the bottom. He's gonna be even a little bit more tired. He might have recovered his heart rate some but he's still gotta make some effort to do that. So we're looking for that work. This is just to the top or to the top and back down again? I think it says just to the top. They do have an annual race up the Empire State Building and I think they only go up. I don't think they come back. They're all still up there. They don't come back down. The Delta K term, what do we look at here? This is velocity. Now, wait a second, wait a second. Do we look at the velocity or do we look at the speed? Remember, kinetic energy, one half MV squared. Do we put in here the speed or the velocity? There's no such thing as the square of a vector so it can't be the velocity. It's gotta be the speed. Notice that means then, kinetic energy doesn't care anything about direction. It just cares about speed. That's it. The matter if you're going up, down, sideways, circles doesn't matter. Any mass with some velocity has got kinetic energy. So the question is, is his speed changing from the start at the bottom to the top at the finish? He's standing down there at the starting lines. He says go. He gets up here. The instant he crosses the finish line, he stops. He has to stop because he's gonna throw up. That's what they do up there too. That's where they can come down into the floors all steep. So there's no change in kinetic energy in this problem. So already it's getting smaller. The problem's already 25% smaller. Is there any change in gravitational potential energy? Yes. Yeah, of course. All you look for is, is there a change in height? Is there any, is there a spring in the problem? No, there is no spring in the problem unless he has those really expensive nikes. Now the problem's 50% smaller than when we started. And so it becomes pretty easy to find the work done. We merely have to calculate the change in the gravitational potential energy. And you're given all those numbers. M is his mass because that's the mass that we're going up through the gravitational field. We're given the height of the building. G, I guess you have to look up. Simple as that problem becomes. Oh, except there's a second part to that problem. It then says at what rate is the work being done? We find this work, just plug in these numbers that we have and calculate that. In fact, it's such an easy calculation. I'll show you. It comes out to be 235 kilojoules. So you can check that if you haven't already. Then the problem says at what rate is the work being done? All they're asking here is how long did it take to do this? That would be the rate at which the work is being done. So we've got that term, we know how much work is being done and then they also told you 10 minutes, 59 seconds, just under 11 minutes. So that's the time it took to do that work. In fact, the rate at which work is being done is a term you're very familiar with. The rate at which work is being done. Because obviously anybody who's 75 kilograms goes to the top doing the same amount of work. That's all that was involved here. But if he does it in 11 minutes and my brother-in-law does it in two and a half hours, there's something very, very different about the two of those. They both did the same work, but one of them did it very, very quickly. In fact, it's hard to do it very, very quickly. That's why one particular person's the winner and nobody else is. Because he's the only one that could do it that fast. So it should be pretty obvious that doing work in a very short time is something different than doing the same work in a real long time. And it's a term you're all very familiar with. Anybody know what it is? Anybody have the power to figure this kind of thing out? No? Bill? What? It's power. The rate at which work is being done is power. I depend to put a double thing on it there because a lot of times I use P for a force that's being applied and so on. So I put that on it. Power. The rate at which work is being done. What are its units? Well, what are the units on work? Joules. Joules or Newton meters, either one. So we put both down and the units on time, seconds. Joules per second. What should we call that? Do we have a dead white male German physicist we need to honor? What? What? I've been waiting all year for that one joke. There's only three or four physics jokes and that's one of the best right there. That's a watt. That's how we define the watt. Another W, we've got three Ws now so be careful. We've got weight, we've got work, we've got watts. The rate at which work is being done. All right, so that problem is very easy to do with the work energy equation. Recognize that because a big part of that problem was, had to do with position, which is definitely a part of that. The position and the gravitational field. Actually the change in position for most interested in. Can you have a W1 of force? No, work is not a force, but it involves force. If there is no force, remember non-conservative forces, forces we can't turn around and get back to where we were before. If there is no force in a problem, there is no work being done. At least no non-conservative work. These are actually work terms, but they're conservative work terms. We can get that work back. So yes, this is not a force, but it involves forces. Remember, work is defined as F dot ds. So if we don't have any forces, we don't have any F dot ds, we don't have any work. Actually that's what we're gonna do in the next chapter. Those situations where there is no work being done, but there can still be changes in these. In fact, you're familiar with one of them. If this guy then steps off the top of the building and goes into free fall, he'll lose all of his gravitational potential energy. Where will all that energy go? Into kinetic energy with which he'll hit the ground. At quite a high rate of speed. That's a free fall problem we looked at. When there is no work being done and energy just goes between these three terms somehow, that's called conservation of energy. That term also you've probably heard. And we'll study that in chapter eight. This stuff is all chapter seven. Why was the space constant? No, he starts with no speed and he finishes with no speed. As soon as he crosses that finish line, he stops. They don't keep going, they stop immediately. And then his mass decreases because he throws up. And then he must do the Corning Tower run and throw up on that one. You know, the Corning Tower, the big, real tall building in Albany, they do a run up bet every year too. I think for MS Society Fundraiser thing. Yeah, you guys need to throw up more often, I think. Mark's going on the other one. That was good play. Take it easy, Mark. I want you to live through the term. No, actually, your tuition check is clear. So, all right. Other chapter seven problems. A lot of them you're just given a situation, say calculate the kinetic energy, calculate the potential energy, not necessarily looking at the change in those values. But that is very important to us in general, is the change in those terms. All right, so here's another problem. Also, clearly dependent upon position during the problem. It's number 67. You don't have to pull it out. I'll give it to you, it's very straightforward. Ignoring friction, it shows that a skateboarder goes down something like that. This part is 5.5 meters above the bottom. And that tail part over there is 2.5 meters above. And we're asked to find the final speed at this point here. So it starts here, finishes here. We're supposed to find this speed right there. I think it's 67, unless they, no, no, I'm still using the same addition. So the numbers shouldn't have changed. Isn't that 67? 7 to 67. So we're defined as speed at this point. Obviously, it's a position dependent upon a problem. It depends on where he is in the problem. That's what the basis of the problem is. Is this freefall, well, obviously gravity's involved because otherwise he wouldn't slide down the hill and then slide up this little part of the hill. Sorry, sorry, Mike. You probably want to see it as a Colorado mountain. Sorry, sorry, something like, because Mike's going to Colorado next week. Not that I'm jealous, but if I went to Colorado next week, I'd have to see my sister and I don't want to do that. Hey, there's Mike. What's Mike saying? Mike saying, whee! All right, there's Mike. Did I ruin the trick for you? You're really hippy laughing at us when he's at the top of the Bale Mountain and we're not chipping ice off of our driveways for another month. Go ahead, Mike, go ahead, go to Bale. We don't care. Anyway, what was there? Do you have any? Oh, a question? It's not freefall. What forces? Skateboard are going out. What other forces are there? Well, weight is, weight's a part of freefall. What's normal force? What normal force? Where? Yeah, anytime he's touching the rap, especially down here, when he goes across the bottom, there's quite a bit of normal force from the bottom. So it's not necessarily, it's not a freefall problem. Plus, we don't normally have freefall problems where the thing goes down and then comes back up. It's not going to come back up, but something doesn't help it do that. So, it does seem to be position dependent, so we'll look at it as a, write down the work energy equation. Don't forget the deltas. Doesn't mean anything if those are missing. Is there any work being done by non-conservative forces? If you remember, it may have slipped by an introduction to a problem, it said ignoring friction. So there's no friction being done. That's always, and maybe even our number one non-conservative force. Any other non-conservative forces? What about, what about that normal force right there? It's balanced with the weight, but that doesn't necessarily mean it's, how much work is that normal force doing? Let's blow this little part of the picture up. So, there's the surface there, there's the skateboarder right there. Now I'm talking about the normal force. I'm not asking how big it is, I'm just asking how much work it's doing. Why zero? You're absolutely right. Why is the normal force doing zero work? Here's the normal force. The normal force is not pointed in the direction he's moving. He's moving in that direction. What's the dot product of two vectors that are perpendicular? Remember? It's zero. Dot product of perpendicular vectors is zero. So the normal force doesn't do any work. And there's no friction. I can't remember if it says, Joe, you have it open there? Does it say that he pulls or pushes off with his foot or anything or it just says he goes from there down to there? Starts from rest. Starts from rest. So if he was doing some pushing or something, we'd have to count that as some work being done. So what work is being done? Non-conservative work being done. What work is being done? Phil? Is there a centripetal force? Well, the centripetal force at any instance is gonna look exactly like this, isn't it? So the centripetal force isn't doing any work either. In fact, the normal force is the centripetal force for that curved part of the course right there. So what non-conservative work is being done? Well, coming up with any, there isn't any, there isn't any non-conservative work being done. This happens to be a situation of conservation of energy. All the energy we start with, we finish with because all it does is transfer from one section to the other here, is there a change in kinetic energy? What do we do? What do we do about the fact that at point one, he's not moving, he's gonna be going real fast right here, and he's gonna slow down a little bit and go, what do we do about the fact that we've got no speed, a whole bunch of speed, and then some intermediate amount of speed? How do we handle that? Under that curve, the area under this curve is how much dirt there is for that ramp. That's all that is. But that's not a bad try, except it's even easier than that. Remember what this kinetic energy term is? It's what? The kinetic energy, oh, my ride's here. All right, I got it, we hear it, my ride's here. The governor's coming to get me. What is delta K? Remember what our definition of delta is? Well, yeah, but what do I write now for an equation and something to do? It's K2 minus K1. Notice, I don't care what K is down here, I don't care what K is at one, what K is at two. It's all I care about, nothing else. I don't care about any other speed in between. In physics and engineering, we call that kind of thing that only depends upon the points. We call that a point function. Only cares about the end points. Nothing that goes on in between has anything to do with the problem, at least in terms of the kinetic energy there. Yeah, Len, you're right. That's not true with the work. Remember, the work is defined as the integral of f dot ds between two points, one and two. And that definitely does depend upon what goes on in between. Which one of, here's two points. Here's a, we gotta get from here to here. From S1 to S2, we can do it. We can do it with a force that does that. Or maybe we can do it with a force that does that. Do those two require the same work, not at all. Because it's the area under the curve. This one, the top one, has a lot more area than the bottom one does. Well, you know that. You need to push something from here to there. You push it straight to there. You don't push it out into the hall, down the hall, back again, out into the parking lot, cross the parking lot, bring it in, and then stop here. That's a whole lot more work, but you still only went from one place to the other in the same. Work is what we call a path function. Its value depends upon how we get from one point to the other. The path is very important. Kinetic energy, it's a point function. It doesn't depend upon anything else. Start and finish. Take the rest of the time off in between. Are either of these two things zero? Yeah, k1 is zero. There's no speed at the start, so the problem's getting smaller and smaller as we go. Is there any change in potential gravitational potential energy? Of course there is. We start at one height. We finish at a different height. Notice if the problem had been that kind of thing where that's point one, that's point two, now what's the change in potential energy? Zero, because we only care what it is at one and what it is at two. We don't care about what it does in between. But that wasn't the problem given. Could have been, but it wasn't. So that problem's still in there. And this? Zero. There's no string in this problem. That's all you need to look at and we're done with that one. So delta k is one half m v2 squared minus v1 squared, but v1 is zero. And in fact, we're looking for v2. Do we have this mass? We don't have this mass. Delta u, that's the same thing every single time. That's mg delta h and we don't have. G we know, oh, what do you put in for g? Let's see, yeah, the problem's in meters. So what are you gonna put in there? Yeah, not going to have one. But I think we need to vote. Who says we should, everybody agrees that the numerals are 9.81 and the units are meters per second squared. But a couple people said put a negative in there. A couple, I'm seeing a couple heads here. We need to vote. Do I put a negative on here before I put it in for g? Who says yes, put a negative. Everybody has to vote. Who says no, don't put a negative. It's about even. You say no, don't put a negative. Why not? Don't change the value of gravity. You can't leave it in front of g. You don't put a negative on it, I'm saying. Half credit, yes. The value of g. You're right. g is always just that number. No negative sign. Where's the negative sign come from? Yeah, we put it on there. So there's got to be a negative sign in here, doesn't there? Where's it going to be? On the delta H. That needs to be negative because we decrease in height in that gravitational field. G itself never has a negative. But this g, you think it stands for gravity. It doesn't stand for God. God said that this is what gravity is. We can't change that. Republicans will probably try, but they can't. So we've got mg and delta H. What's delta H here? It's what? We start here. We finish here. That's a delta H of negative three meters. So let's see. Let's put everything back into here now. Delta K. Let's see. The work is zero equals delta K, which is one half mv2 squared because v1 squared was zero plus delta ug, which is mg delta H. And we've got all those terms. Ah, what happens to m? Chances. That's why we didn't give an m. We didn't need it. And we've got all those things just solved for v2. How small that problem is. Start out with this equation. It's not all that big itself, but we know it has a lot of parts to it, whether it reduces down to something incredibly simple, which I like. I don't know about you. Well, John doesn't. John likes it. It's left complicated, challenging. Just keep them up nice. But I like simple why I am simple. We got all those parts. No negative on G. That's negative three. So we're losing gravitational potential energy that's going over here and becoming kinetic energy. That's where your speed is coming from. All right, any questions before I give you an exciting problem? Alan? A bar is going horizontal in your name on the wall. All we were asked was how fast is he moving at point two? So we don't care how far he's traveling. We would if we had to have friction in there. Who said that? Who said ignoring friction? I heard somebody say it. You're actually right. If we had friction in here, friction has a lot to do with the distance over which it applies. In fact, if this was, if there was a lot of friction, he might not even make it up to here. He might lose so much of that energy to that negative work. I guess what I'm seeing is that when those of you go over some of that motion that's going up into the air and it goes around, or who's to say it's all in a horizontal direction? That's not a vector. I don't care what direction he's going. So maybe it's right at that point that he is actually in that direction. I don't know. That's way more detail than we have on this problem. Okay. Let's up the stakes some. Here's a new problem for you. All right. Oh, in fact, I think I showed you this problem, this type of problem, the UPS delivery problem box. We'll put some numbers to it and do it. Package delivered of eight kilograms goes down a ramp that's 3.2 meters long where it meets a bumper to protect it from smacking into the bottom of the wall there. That bumper is 3.2 meters from where the crate is let go. Just to speed things up a little bit instead of giving you the friction I'll give you the coefficient. We'll go ahead and figure the friction. We'll have to find the force friction and then find out how much negative work that force is doing. Kinetic friction or static friction? Kinetic. The box is sliding over the surface. The two surfaces are in motion with respect to each other. Kinetic friction. The strength of the spring is 121 Newton meters to either stretch that spring a meter or compress that spring a meter. It's gonna take a force of 121 Newton. And the last thing, the slope here, 23 degrees. We wanna find how much spring squish there is because if they don't build enough space back here the thing's gonna hit the wall anyway. It'll compress the string completely and smack into the wall then. So we wanna prevent that. We wanna have a little bit of extra space in there. Need to find, with that spring, that ramp, that box, how much spring squish is there gonna be? Clearly I hope a position-dependent problem. We have this distance here. We have a spring, depends on how much that's squished. We have gravitational height changes, a very position-dependent problem. So you should suspect, and we're working with it anyway, that the work energy equation might apply. Right down the whole work energy equation, easy place to start, then go to the four terms and see if any of them are zero and we're done with them automatically. Any work being done by non-conservative forces, any kind of friction or problem, you're gonna have a work term. There may be other forces doing work. For instance, we could have somebody coming in here and push it down the ramp. Any change in speed? Well, there is a change in speed because the guy sets it there on the ramp, then it starts to slide down the ramp, then it hits the bumper and squishes down the spring away. What speed did it start with? Zero, he just sets it down on the ramp, let's go with it. What speed does it finish with? Zero as well, because that's how we can tell how much the spring is squished. When it squishes the spring all the way, it comes to a stop again. If it was still moving, there'd be more spring squish to come. So it's gotta be that point where it comes back to a stop. So there's no kinetic energy change in this problem. Yeah, well, there is a kinetic energy change, but start to finish, point one to point two here and where it's squished both speeds are zero. Starts from rest, finishes at rest. Problem's already 25% smaller. Any change in gravitational potential energy? Yeah, of course. Because starting high, finish low. Any change in spring energy? Of course, there's a spring just sitting there doing nothing and then in the end, it's squished up some so it's got some potential energy in it. So that term's in there as well. All right, so let's do the two term, the three terms independently. Because then we're working on little tiny problems. One thing that's absolutely crucial here is one, we gotta get the units right. I talked about that yesterday. All have to have the same units or they're not gonna go together. Also though, if you're not picking it up, we need to have the negative signs correct or this is gonna be screwed up. We gotta get the negative signs right. If we have a negative sign on one of these where we shouldn't, it's all wrong. You're doing a different problem. So what's doing work? No, well, gravity's doing work, but that's over here because that's conservative work. We can get that work back by just taking it back up to the top of the ramp. What? Yeah, friction. Friction's doing work, so that's negative F, negative because it's against the direction it's moving and the distance it's moving. What distance is it moving? Almost, it's moving 3.2 meters plus del because it slides a little bit farther than that. So it's gonna be 3.2 meters plus del. How big is the friction force? It's what? Let's work towards the number. So if you just throw out a number, half the class, I know when I was in the class, we'd go where'd that come from? Yeah, it has to do with the coefficient of friction times the normal force. Coefficient of friction we have, what's the normal force? Is it W? Box of any one time. There's its weight. That's not the same W as over here. That's why I like the 1 and 2 on here because it reminds me, this isn't weight, this is work. There's no weight. What other forces? The friction going up the ramp, that's why there's a negative here because it's moving down the ramp, friction is up the ramp. Other forces. Normal force. How's the normal, what's the normal force look like? The perpendicular surface. Perpendicular to the surface. So that looks something like that. In fact, in this case, the normal force is equal to W or a G. That's the same angle as the ramp. That's the 23 degrees. So it's W cosine theta because we're adjacent to that 23 degree angle. The angle between the weight and the normal is always equal to the angle of the ramp even if it's zero. So normal, since it's non-accelerating, perpendicular to the ramp, then the normal force must equal W cosine theta. UK, W cosine theta, and we have all those numbers. We have mu, we have W, well, we have M, but MG, and we have theta is the 23 degree angle. I just don't happen to have that number separate. Who's got a calculator handy? Go. 9.8 times 0.24 times eight times cosine 23. I want you to have it there. 17.3 newtons. All right, so this is minus 17.3, 3.2 plus del, and it's gonna be newtons times meters, newton meters. Notice how compact that term looks. It's not very confusing now when we took care of it all by itself. Delta UG, we'll go to the next little piece. Just do them one at a time. That way we can watch the units. That way we can watch any minus signs so we don't screw this up. Let's see, MG, delta H, and we've got G, we've got no negative sign on it. It's just G, what's delta H? Here's delta H, and it's 3.2 plus del sine 23, plus del sine 23, 3.2 meters plus del sine 23. We're looking for del. We can't find del until we have the whole thing put together because we have a del in here. We have a del in here. We have a del in there. You watch, it's gonna become very easy. So I already have this one. Minus 98.1 plus 30.7 del, and it's kilogram meter per second square, which is Newton meter, it's Newton meters. So checking my units. Oh, sorry. Sorry, this is also a minus in here. Minus on both of those terms because it's going down from both of them. Oops, I didn't have my minus in here for the delta H. Make sure that minus sign's in there because we're dropping down. That's a pretty simple term too. It's gonna be very easy to put it in the equation with those terms. It's gonna be very easy. Delta Ue, oh shoot, we're out of time. Okay, we'll, oh, we have to wait a full week for the answer. What all, here, here's what I wanna give you. Just so you can see it, and then you can go. Wolf vision, that's cool. Here's why I like you to do each of these little pieces one at a time. Here's the equation, the work energy equation, when you start with it. If you replace all those terms, it starts to get that big. If you then take all the terms you know and replace them with the values you've got, it now gets that big. And you have to solve that for del, which is here, here, here. You can't miss one single minus sign. You can't drop one single square and you gotta get your units right in that thing. Does that look like that's a, there it is, there's the whole equation. That's what'll happen if you go from this right to this and you're gonna screw up. I don't know anybody could solve that without screwing up. This, we're doing little tiny pieces one at a time. I'll post this whole solution online when it's all done. Oh, I do have it, I do have, there was a typo in it, but you can see what the solution looks like. When you put all these together, that's all you're left with. It's quadratic in del. Is that something you could solve? Yeah, quadratic equations are pretty easy to solve. Look at that compared to that. My way, here's your choice. The old fashioned, bang your head against the wall of Joey. Joe's not gonna move that way. All right, oh, you're gonna go ahead and turn it in, huh? All right, Mike, have a good, be careful, don't break anything else, just yield.