 Nope so go ahead where are we left off? Do you wait until the tape is rolling? What? Let's do Wednesday. I don't know. I don't know. I don't know. It's your job to look. I just accept it when it comes to me. All right, we finished a week and a half ago, two weeks ago, with the work energy equation. So we're going to take it from there. A huge chance to do much with it, so we'll do a little bit more with it today. And also then there's some special cases, just like there were some special constant acceleration type cases that changed some of the way we approach the things. It's the same truth here with the work energy equation. Do you remember what the work energy equation is? Joe got the easy part. Joe's got the work part. Changing kinetic energy. If we do some work on something, we might change its speed. We can either speed it up if we're working in the same direction it's already moving anyway. Or if we work against the direction it's moving, we can decrease its kinetic energy, slow it down as it breaks too. But we can do other things if we do some work on it. What else? What symbols I use for potential energy? Some folks use a V. Some use a P for potential. I forbid we do something simple and direct like that. That just means that we can raise or lower something, too, by doing some work on it. And one last piece. Patrick, do you want to go for that piece? That's it? That's very disappointing. What did you do over a spring break that puts you in such a downward mode? Do you want to try again? Change in? I'm just going to do it. I see. Even Len's smiling. I kind of wanted to. Yeah. Okay, we'll come back to you then. Samantha is a pillow way up. Smile on the bed. Well, let's go over each of the pieces first. Make sure we've got those. How do we calculate the work done? What? I heard something. It has to do with some of the forces or the net forces acting on something. Because each force could do its own little bit of work. There might even be forces working against each other. That's what we have to do against friction to get things to move. Sort of. Not times. What does that do for us? Do you remember? We don't actually have to calculate the dot product. Because there's probably more direct ways to come up with what we need from it. But that does work then for any vectors you've got. And it certainly gives us the result we need to have in here. What does that do for us? That dot product? It makes sure that the amount of force that we're applying. So let's say the resultant force is something simple like that. Or it might be that there's only one force. And so by default it's the net force that's acting on the object. It makes sure that only the component in the same direction that we be moving is contributing to the work that we calculate. The other component of that force, in this case the vertical component, is doing no work. It's contributing nothing to the calculation of the work. It doesn't mean it doesn't affect the problem. It most certainly can't. For example, this force down would increase the normal force up. And a greater normal force is going to increase the friction force. If the friction force goes up, then this net force is going to change because it's part of the calculation, part of the contribution to the work. This isn't quite complete. What else? Integral. That's what this DS business is. So we'll integrate from point one to point two. That's what this subscript here reminds us. And this allows us then to handle the fact that forces can change with position. Maybe you push for a little while. You're starting to get tired. Your force drops a little bit. Buddy comes in to help. That force goes up then. Could be you're pushing across two different types of floors with a friction would change in certain positions. It handles any of those. In effect, we'll do a problem with that just after this. For most of our problems, though, that force is constant. And then it comes out of the integral. And the dot product could probably reduce to something like that. For most of the problems. So like most of the things we're doing, there are some simplifications we can make to keep this stuff going as we learn our way through it. And then each step as you go through it gets a little bit more complicated as you mature through the subject. If you do happen to have a graph of force with position, it could change continuously. That integral then is the area under the force time graph if you happen to have that. I think we didn't have a couple of problems with just that. That's the left-hand side of the equation. What forces go in here? Any and every force in the problem? Or just certain forces? I need maybe even a little bit different word. Remember to get to moving it from S1. Of course, but that is to be taken care of automatically with that. So I don't need to say. Well, the thing I'm getting at is there's force involved here. The force of gravity is involved here. But I don't put that calculation in here. In here we have springs involved in this problem. So they exert forces. But I don't put that calculation in here. What is it about these two forces that means that they don't contribute to this side of the equation? And there was a specific word I used to label those, if you remember. Over here, these are work being done by non-conservative forces. And these two involve conservative forces. How do you tell the difference? How do you go home and tell your mom and your dad the difference between conservative and non-conservative forces when every time you talk to your dad about politics, his eyes look out and the veins in his neck start to leave. You can't even talk to him about it anymore. Is it like that at all? There's a very easy way to tell the difference between the two as it would mean to you in either applying those forces or having to work within some of those forces. Philip, you remember? All forces are vectors. These don't have quite the vector quality to them because, for one, this only acts up and down because that's the direction of the gravitational field. And this acts just merely, it has to do with how long the spring is, but it still comes from the forces the springs exert. Joey, you remember? He said something like, non-conservative is like pushing a box. That's exactly it. Non-conservative forces. If you turn the problem around and bring it back to the starting place, you don't get back to the same place you were with non-conservative forces. That usually things are even worse. If you push against friction across the floor, you turn around, you want to push back. If you double the friction that you've had to work against, you went twice as far now. The friction didn't care which way you were going. It's bad no matter which way you're going. You can't get that kind of force back to where it was. Gravitational force, if something's low, we raise it and lower it back down. Gravity is exactly the same as it was before. There's no change in the earth. There's no change in the gravitational pull. There's nothing different. Same with springs. We stretch them out a certain distance in a problem. We stretch back to the original length. They're exactly the same as they were before for all our intensive purposes. These things we can recover. These things are conserved. These things are non-conservative. No matter what we do, we lose what we do there. And it gets even worse as we try to get back. So that's the difference there. Be careful. Don't count the weight over here. The weight's being counted for over there. Remember the K? Give it to me real quick. Remember K? That's change in. So it's K2 minus K1. Remember the Ks? You'll remember when we say them. What's kinetic energy? V squared. So this is 1 half MV2 minus 1 half MV1 squared. Be careful. Students very commonly don't write that out. And if V2 happens to be zero, if it's a problem where we bring the object to a stop, if V2 happens to be zero, you've got to make sure this minus sign is still there. So be careful with this. Students a lot of times just write 1 half MV squared because one or the other of them is zero and they lose that minus sign. So be careful with that. Delta UG. Change in gravitational potential energy. Remember the form of that one? Almost. MG delta H. It's kind of like vectors. If you have vectors in one place, they're not on the other than they can't be equal. This is the same thing with these delta signs. We have something changing here. We have to have a change here. Students very often leave off that H. It doesn't lead to the kind of problem that it does in the other one necessarily. But leave it in because it is a change in the height that's important, not the height itself. How do we find delta UE? Change in, of course. And either of those, or even both, could be zero. Remember how to find that? How's it getting in it? Because that tells us how strong this spring is. It has to do with that, too. One half. FK times del. Yeah, one half K. Del 2 squared minus del 1 squared. What was that definition of del? The book uses X. A lot of books do. I don't like to use X there because that X could be different than the X used for position in the problem as a whole. Remember what was the definition of that del? Turn them up a foot there. The change in length of the spring limits equilibrium length from its rest length, the very length it had when you took it out of the box. Notice it doesn't care what angle the spring is pointing. It doesn't matter if it's pointing diagonally or not. It just matters whether or not it's been stretched or compressed. Either one of those situations can make del non-zero. All these terms are either equal or added together so they all have the same units. Remember what the units were in the SI system for this equation? For each one of these terms individually? Newton meters. Most easily seen from this work term because that's force times distance. But each of the other ones, and it's not as easy to see them, but we did work through it, each of the other ones are also going to have units of Newton meters. It's a good chance in these problems when you're working on the individual pieces here to check the units and the minus signs to make sure that they make sense in the problem. Another name for a Newton meter jewel, just so you recognize it, capital J. Though if you don't put that, I'm not going to take off. Because I usually stop at different meters myself. Okay, so let's do, I told you there were a couple of special classes. They don't really change how you work with the equation, but it is a different situation in terms of really what's going on. So we'll take a problem here that we've done before, and we'll do it with the work energy equation. So imagine we've got a 3.5 kilogram ball. Given some initial velocity, they're just bounced off of the table and is now starting on its way back up, or whether it's been thrown or shot or something, I don't care, that doesn't matter to me. Rises to a height of 120 meters. Using the work energy equation, find out what V1 is. All right, so we're kind of warm enough from coming back from spring break. So let's do this one together, because there's a couple things in here we need to know about. But the work energy equation, all four parts of it. First thing you do after you've written it down is go and look at those things and see if there's any one of them you can get rid of. If you can, the problem's already getting smaller and smaller. Smaller problems are easier to do than big problems. Any of these problems, zero, is work, the work term, zero. It either is or it isn't. It's as simple as that. However, it's not subject to a democratic vote. But it either is or it isn't. It doesn't mean we can't vote to see what the opinions are. But I heard somebody say, no, it's not zero. Anybody say yes, it is zero. Remember, this is work done by, by what? Unconservative forces. Those forces where somebody reaches in, pushes, pulls, certainly where there's friction. So if the work term is not zero, that means there are forces you can tell me about that are doing work. What are they? You guys who said, yeah, the work term, no, the work term is not zero, tell me what the forces are. Why isn't gravity? Gravity is a conservative force. We take care of it over here. We don't want to count for it twice. We only want to count for it once. So what forces are there doing non-conservative work? No wind resistance. Wow, there will be tomorrow in lab when we do realistic refall. Kinetic force? Want to take that back? Remember, all forces in these problems that we do in this class are caused by something you can point to. I can go and touch. I can see it. I can't see kinetic. So there can't be such a thing as kinetic forces. There are no non-conservative forces in this problem. So that term is zero. Our problem is already 25% smaller. Is there a change in speed? No, I had some yeses. Yeah, there is. It starts with some speed. In fact, we need to find that. What's the speed here? Yeah, that's how we know it got up to 120 meters. When it gets up to there, it stops. That's what marks the speed we needed to get up to there. I didn't want it to go farther. I didn't want it to go less. So, not only is this not zero, do we know if it's negative or positive? It's very straightforward to look at. If the kinetic energy increases, this will be positive. The only way the kinetic energy can increase is if the object speeds up or slows down? Speeds up. Is it speeding up? It's slowing down. It's got good kick of speed here. It doesn't have any speed here. So, we know that this is going to be then negative. So, we know that already. We know it's not zero. We also know it's going to be negative. So, we can check this as we go along so we don't screw it up. We lose the negative signs. We lose the squares. We lose the units. Things are going to square up because it's got kinetic energy at the start. It has none at the finish because it actually comes to a stop up here at the 120 meters. I said I wanted to get up here. I don't want it to go farther. I don't want it to go less. I need to go right there. We know that's going to be there because that's where it comes to a stop. And it starts to fall back down. Didn't ask about that. Is this zero? Anytime there's a change in height in a relational field that's not zero and of course there is no, there is a change in height. Is that positive or negative? Positive. If it gets higher, it gets more potential energy and that goes up. So, we know that one's positive. Non-zero and we know it's positive. So, we haven't done how old I get in the problem just getting smaller and smaller and smaller. There's hardly anything to do. I'll come. There's no spring in the problem. Problems half the size it was when we started even less because we already got that minus sign and plus sign business. This is a special situation. It doesn't really affect it in terms of what we do, but it is a special situation in terms of what the physics is going on. In the absence of outside forces by outside I mean nobody's reaching in to give a push or a pull. There's no dragging friction going on. I don't mean there's no weight because we're taking care of that over there. So, it doesn't even matter if there's gravitational pull or not. We take care of that elsewhere. In the absence of outside forces that work term is zero. Like it is in this example then that side was also zero because they're equal, but this is zero and that's all zero. This is known as the conservation of energy. Energy, conservation of whatever energy we started the problem with we finished with. No more energy comes into or out of the problem. In fact, it tells you that only outside forces can bring energy in to a problem or can take energy out of a problem. All that's going to happen here is energy is going to go from one of these to the other. But the total amount of energy is going to stay the same. The change in energy itself the total we're talking about mechanical energy I'm not including thermal energy. I'm not including chemical energy or nuclear energy or nobody here. Oh yeah Alan has a mustache. So whatever energy he stores in is a mustache from lunch. That's all there. Whatever energy we start the problem with we finish with. The only way energy gets inner out of a problem our problem is by the application of forces of some kind. Did you just get back last night Mike? Been back for a while? I'm going to go over in the lab and take a nap there. Drive and fill nuts. I'm afraid you're going to have a head going to drop back and knock them out. So we happen to have a problem where energy is conserved because there's no non-conservative forces in this problem. However in terms of what we do with it nothing is any different. So let's look at these term by term that way. We can make sure the units are right if we have m v2 squared minus v1 squared. Any part of that zero. Yeah v2 is zero. That's the deal up here. When it reaches that point it has no speed. That's zero. And in fact there's the minus sign we need because every other term is positive. That's the only place our minus sign is coming from. We can put it artificially but we do want to make sure it's there. And right there's our minus sign. So we can put in the terms we know. We're looking for v1 and I'll bring the minus sign out in the front there now. And that's how we're going to find the v1 right there. That's what we're asked to find. What are the units on that? What are the units on delta K? What must the new unit be for that term to be Newton meters or joules? If v1 is in meters per second this will be in Newton meters. If v1 is in anything else that wouldn't be Newton meters. And we want Newton meters on each of the terms. Well it doesn't have to be Newton meters but it does have to be the same thing on each of the terms. What's 3.5 divided by 1.75? That was a small little problem to do so it was pretty easy to keep screwing it up. The only other term we have on the problem was delta UG that's MGH sorry, MG delta H let's see M was 3.5 right there. What do we put in for G? 9.81 meters per second squared however G is acting down. G is always acting down that's where the earth is. G is down. That's a picture right there of the earth. Have I said that before? Nobody wants to stumble in for that one and have a negative in here. Nobody? Tyler? Volunteer? Not for that one. Volunteer for something exciting. What's delta H? It's going higher, this height is increasing. Are the units there okay? 3.5 meters so that's going to be okay. Who's got that number for me? 41.20 back together we had 0 on the left hand side minus 1.75 V1 squared we know that V has got to be in meters per second that's the only way all this is going to work out we've already checked it plus 41.20 that's an easy problem to solve. See how small the problem got? What's V1? 48.5 meters per second alright wait a second something wrong down's not positive that's positive because we're going against the gravitational field so it doesn't matter what we call up or down that's against the gravitational field so it's greater in height that's got to be that's considered automatically a plus then why is this 0? What could go in here in a problem like this? If this was a real problem that we were doing over in the lab assuming the lab was 120 meters high if this was a real problem what we were doing would this term really be 0? What would be in there? Okay I said that and I said yeah we are so this is really a free fall problem isn't it? Remember doing free fall problems about a month and a half ago we started them I think back when you were physics students this high how come? Mass did not matter I haven't even mentioned mass yet in class when we did free fall problems and now mass matters how can we do free fall problems where mass matters now before we were doing free fall problems mass didn't matter we need to go back and start the term over from the beginning because everything's been wrong it's still the same problem though it's still a free fall problem where mass didn't matter I've never even mentioned mass by the time we did free fall problems what did you say Phil? I said we've already got final velocity what do you mean? yeah we've already got final velocity at 0 that doesn't mean this is a free fall problem instead of putting the numbers in let's just use the equations work equals 0 change in kinetic energy with one half m v2 squared minus v1 squared don't even have to put any numbers in yet plus mg delta h is 0 so that's our work energy equation isn't it what happens if we divide through by mass divide through by m doesn't matter doesn't matter what the mass is so this is still truly a free fall problem where mass doesn't matter but if we were doing this as a rocket ship problem when maybe we're neglecting air resistance but we're having mass lost because fuel's being burned and spewed out the back we could still handle the fact that there was no air resistance but we could handle the fact that the mass itself was changing it'd be a more complicated problem but we could do it but it still is a free fall problem it doesn't matter it's just we didn't see that in here because we had the mass in and it actually would divide out as we did the problem I guess when you solve for v1 here you don't see it but the mass divides out and you can do that again that's one of our free fall problems we'll get exactly that same answer okay that was just a warm up problem got our first look at conservation of energy but it doesn't really change the problem we still calculated the same way we'll we'll do another look at the conservation of energy in a little bit here let's see let's do this and this I decide do I want to make it hard or easy what is it with you want hard yeah oh easy oh that simple oh that simple alright well maybe I'll even it's a big stretch here if it's hard enough here's your problem up a 13 degree slope this creek I'm being generous but I'm going to say that you can push with 42 newtons 42 alien but you guys remember initial velocity meters per second so it's already moving when we come upon the problem here some time later up here it's now moving two meters per second I labeled that 0.3 I called that v3 because there's something going on in between at a point 8 meters from the fin from the start 7 meters from the finish the uh to a much rougher surface a nice slippery 0.12 would that be kinetic or static kinetic because the box is moving over the surface of the hillside there friction up here is 0.36 alright let's see if I have all the boxes more massive than that then it won't even be going this 2 meters per second up here if it's less massive than that it'll be going over that 2 meters per second so we needed to arrive up there at the top with 2 meters per second speed to it that's your get out of class question now energy equation will work real nice with this because it's a position dependent problem if I were you I'd work the little pieces of the work energy equation one at a time especially since that makes it easier for me to help you know if one of them went wrong if you put them all together in one equation you can do it but it's going to be hard for me to look at it and say which part went wrong because usually what happens is you make one mistake not not many if you want to just for reference if you think you need to this point one that point three this point in the middle we can call point two if you want the numbers don't matter they're just arbitrary as long as you're doing the deltas between the right points at the right time you're okay oh man that's life taking units no stinking units units for sissies good problem here chance to get out the first day spring break you know bad why didn't it snow like this out in colorado it's only snowing now I want to get that did it Mike that's good John none before that though did you have good sunshine yeah I was like good too oh that's when the girls came I think I hear it feel free to talk to each other if you want see if you need to put your heads together because you know what we say if you want to grade your heads half of what we don't want if I stuck need a little help or advice all of them just turned over in their grades am I doing okay doing something I'll give you a discount on my usual consulting rate because it is class time 50 bucks an hour don't ask me first because I charge 50 bucks an hour he's free or if he isn't you're a fool I got it what's that is that one but not in this problem where they keep water see if any terms are zero and go turn by turn by turn we'll let you stuff on the papers okay you stuck need some help what with the friction part that's kind of what I suspected what's Alan doing about it that's kind of what some people like about the work energy equation others hate very flexible many different ways you can do all these little parts to it guess what work one to three because we're starting at point one and we're finishing at point three what do we do about this friction change in between no you don't forget about it friction is going to be negative work here's part of the magic I think or not the flexibility of the work energy equation if you have different things going on if you have different things doing different things in the problem you can just do each one of them separately and add them up to find the work done from point one to point three calculate the work done from one to two add that to the work done from two to three this happens to have one friction this happens to have another friction but within those everything's constant so you can do the sum of the forces from one to two delta x one to two nothing changes in there so the integral of f dot ds becomes that and then recalculate the friction just redo it for the different section delta x for the first part happens to be eight delta x for the second part happens to be seven only thing that changes between the two is the friction force and then just add them together no no that's just this side then you do the other side remember I said do these pieces separately one at a time this one that's just the whole change from one to three I don't care about two in between because that isn't affected at point two nothing changes at point two and here it's just the two velocities you don't even care what speed it's moving at point two here just skip it so only the work term has any change at two the others are unaffected by point two so just ignore point two each of these are functions of m you just solve for it you'll have one equation and one unknown the only unknown will be m is that how you're doing it Alan? just sort of another way you can do it is you know the velocity here you can figure out what the velocity is here take this velocity here and you know what it is here do it as two separate problems where you figure out the intermediate velocity you can do it that way if you want to and either one is correct like I said some students love that flexibility some students just say just tell me which way to do it and I'll do it normal forces are going to depend on m which means the friction force depends on m which means these two terms are going to depend on m you're going to have an m in there an unknown m because that's what you're trying to find trying to find this m but these two terms will have an m in there too and in the end you'll have one equation one unknown it's the one equation with m as the only unknown the a four term equation with m in each term what can I help you with Samantha? the normal force remember what I said was the very best way to find the normal force free body diagram gotta have a free body diagram to find the normal force so there's our thing on a 13 degree slope any forces on it yeah we got this 42 newtons here that's straight up the slope friction because it's dragging along the slope gravity which acts straight down we don't know what that is because we don't know what m is what else? well the normal force you're looking for is the other thing remember it's perpendicular to the surface it might help though if you call that the x direction and that the y direction generally if you line up your coordinate system with the thing you're looking for it just is a little bit simpler that n is a y direction force so let's sum the forces in the y direction what should the y direction forces sum to? zero because it's not accelerating in the y direction it is in the x direction but not the y direction so that means all the up forces equal all the downwards n is up in the y direction w we've got this little piece over here w in the y direction and then w in the x direction probably didn't get too crowded and up has got to equal w y down those are only two ones in the y direction w y is w that's the same angle as the slope and about the normal force now we don't know what m is nothing else is we know g we know cosine 13 is u k times n that when you get to the second section all the changes is u k nothing else changes set that up for the two sections each one goes in there we push with one pull with the a and b if you want one and two, two and three that's meters what's the 1.8 meters oh that shouldn't be u k I thought this was your work okay you can add those together because that's the total work done from one to three and then that that total will equal this it still has n in it then what else oh okay so total work is that minus the work done by this which would be this times that 8 meters and this times the 7 meters yeah I know well you got this part and then you went on to another part and then you went back to the first part but you hadn't done that first part in the first part so you got all the part well almost all the parts you don't have delta k you got almost all the parts but you're kind of scattering them around let's look at that work term just the work term one way to do it let's see you can just start putting pieces in we're pushing with 42 against friction with the first coefficient of friction in here so that's minus because we're working against friction m which we don't know g which is 9.81 u k which is 0.12 u k cosine 13 is simply the friction force in the first section that we're using a lower coefficient of friction the sum of the forces for that first section this force continues all the way through but this friction only applies for the first section and then that sum of the forces times the length of that first section 8 meters maybe because we got a couple m's in here we'll just change the shape of them we'll make that a capital M just to keep from making a mistake that's just the work done for the first section Tyler is that a hand up? put it up to w x w x this part? no because that's a conservative force that's taken care of I don't have it up there in the gravitational part you don't want to count for it twice it's real easy to do this is just the non-conservative forces now that's just the first section there the 1-2 section the 2-3 section looks exactly the same I don't even need the units because I already worked them out but the friction changes the coefficient changes to .36 cosine 13 and now only 7 meters that's 7 meters times the forces Westville did you want me to do that for you? no I just and yes M is an unknown in here but this comes down to be two numbers when you add all those together collect light terms you just have two little numbers left over in fact I think I even have them there once you collect all these it becomes nothing more than 630 minus 17.2 amp for being a new meter because we checked all the units so that whole friction the work done by the friction part and new pushing comes out to be a simple little number when you work through the steps assuming there was a number sorry it's 42 times 8 plus 42 times 7 or I guess 42 times 15 that's positive that's the work you're doing in the same direction as the object moving this minus 17.2 amp is the friction work that you're losing cement the where you stuck now that is put you there together Joey like this where we have friction on a slope the only thing is we have two different slopes that we just have to add together cement the where you stuck so we know you don't put this WX in however we're not putting gravity in this is the normal force that plus that times the friction coefficient which is the friction force if you want to do it later and find this but if you put in all the numbers you know you should get down to just this simple little form for that entire thing that's the whole work term all the way from 1 to 3 if you put all the numbers in it becomes pretty simple I know getting there may not be as simple but once you get through it that's why we're doing this to try to simplify the little pieces as we go no 42 is given nicely no we're moving in that direction and the force is pointed in that direction so this is f.dx already that force times those distances is all done since they're all lined up you can do a different X and Y for different parts of the problem because remember the coordinates are man made they're arbitrary they don't affect the physics they affect the solution alright well work on it a little bit we'll see what you have 13 no it's not what I got no not even close oh wait there's a decimal in there but it's still not 13 I got 40.6 I'm signed I have that number separate