 Alright? Any questions? Where are we? Oh, we're starting to do no questions. We're okay? Yeah? Alright. We, for those of you that have been in physics one, we hit this sum. We're going to do it a little bit differently. This time so it works. Hopefully it works better. Because we're going to do much more involved problems than we did before. Ultimately we're going to have three ways that we solve kinetics problems. We've done one and we're ready now for the second. Each of the three, and this is no different than it was in physics one, each of the three are a little bit better at solving certain types of problems than the others are. The first one we had for solving kinetics problems is simply f equals m a. That works very well for general problems. You have the acceleration. You need to find the forces to ensure that acceleration. You have the forces. You need to figure out what the resultant acceleration is. Those types of things are other general problems. The second one that we're going to get on today is the work energy equation. We're going to hit this one right from the start in its full form and use it right from there. I think that would be a good way for us to be able to get into some fairly complicated looking problems that if we're controlled and disciplined about doing this become very easy problems. We'll see exactly one of those today. This is very good for position dependent type problems wherein something very much depends upon the position and the problem. This is quite true with things like friction that can very easily change at a certain place in a problem. Also has to do with spring since the forces they exert depend very much on the position of the object when in relationship to the spring. It can also handle the general problem where forces themselves change with position as they often do. The third method we'll use is the impulse momentum equation. That works real well for time dependent problems as we'll see when we get there. Same three general solutions for kinetics problems that we looked at in physics one. We're going to then revisit these near the end of the term when we redo all of this stuff we've been doing for rigid body motion. We're going to take that a step farther than we did in physics one as well. Let's lay out the work energy equation. It's a little bit different in the notation it used with this book than the book we used in physics one and for you other guys one you sought. I'm not sure what it looked like. So we'll stick with the notation that's in this book and just be okay with that. I've got a little bit of a change with it in the way I do it just to I think makes it a little bit more rigorous. But the work energy equation itself looks like this. That first term is the work. When we do work on something we can change its potential energy. We'll come back to each of these pieces one by one. Change its potential energy. We can change its, our book uses a V here. Sorry. We can change its height or we can stretch or squish a spring. We'll call that E for elastic media. For our purposes those are the only things we're going to look at. These are all mechanical energy terms of some time and there are certainly other types of changes that can happen but we're going to stick with these. So the first one we'll look at is the term, this is sort of the input to the problem. These are things that can come from the outside. These are, this is work done by applied forces. The best term for this is what we call non-conservative forces. We use a U here because if we used a W for work then it would look like a W for weight and it would be confusing. Defined as, not every book uses this subscript 1, 2. This implies to us it only occurs when there's movement between two points. They could be the same point. It could be that we come back to where we started but there's got to be some kind of change in position. There's no such thing as a definition of work at a single point. It's only moving from one point to another. So that's why I like to use this subscript there that it reminds us that we've got to be actually doing something, doing somewhere with this. No such thing as work at a single point. So the definition if you remember in its full and general form is something like this. As we go from one place to another, so I'll use S for a general position and this can be 1D or 2D, it doesn't matter. The net force being applied to the object as a vector dotted with our position vector, in this case we'll use the differential elemental form because that allows us the possibility that as position changes the force changes as well. That can be a change in the direction or the magnitude or both as we do that. So that's the full and general form of our definition of work. For a special case just to see what it looks like for constant force, then that comes out of the integral. We're just integrating a dot in there. That's just simply an operation. Remember. And the integral of dS becomes delta S. So if the force is constant, it turns out to be simply the force being applied over a certain distance. This kind of makes sense. It has good common everyday sense. You apply a little force for a very long way. You could be doing as much work as a big force for a very short distance. So those make some pretty good sense. It is very important to us that we recognize the utility of this dot product in the setup of the problem and the definition of the work. Easiest way to show that is imagine pushing some object across the floor some certain distance and doing so with a force like that. The dot product gives us the fact that it's only the component of the force in the direction of the motion that's doing any work. The dot product will do that for us automatically. The component of the force perpendicular to the work does no work because there's no motion in that direction delta Y in the direction of the Y force does no work. It does not mean that this has nothing to do with the problem. If you remember the applied downward force can increase the normal force on an object, you increase the normal force on an object, you increase the friction on an object and friction is most definitely one of these non conservative applied forces. So don't think that FY doesn't have any play in it. It does. It just doesn't have any direct contribution to the work. Sound familiar so far? A little bit? Have you seen this before? Maybe not in this precise form but something like it. Can I have you too? This term non conservative work this tells us what forces we can actually put in here. There are other forces in the problem but they don't all go into this part of the calculation. Non conservative forces are those forces that if we turn the problem around we don't get back what we've lost to those forces or what we can't recover the work done by those forces if we turn the problem around. This example is friction. You push something across the floor, you work against friction the entire way. You turn around and push it back, you double the amount of work you had to do against friction. You don't get that back and return to a zero point. There's still all the abrasion that went on, all the heat that was generated and you don't get any of that back. You cannot return to your original situation with non conservative forces. So we'll do some problems applying this work as we go but let's remind ourselves what each of the four terms are and how to deal with them. Of course the units on this, the force times distance, so the units are Newton meters. When talking about work and energy type problems we define one Newton meter of work or energy as a jewel. I don't find any great utility in finishing a problem like this. If you finish it with Newton meters just leave it there. But it is certainly a viable and well used unit for these type of problems. So that's the work term. It works very nicely if there's a position dependence upon force as we'll see in some of the problems. The second term is the kinetic energy term. It simply means if we do work on an object, we could change its speed. If we do positive work, we can increase its speed. If we do negative work, which is what friction does, because friction is always against the direction of motion. If we do negative work we can slow something down. We can have both going on at once so we'll see this term in all kinds of different ways. The units, the masses, kilograms, velocity squared, velocity, there's velocity squared. What are those units in more, perhaps more familiar terms? If we take kilogram meters, just one of the meters, but both seconds and rearrange them to look like that, same units just rearranged. That first term is a Newton and we have exactly the same units as we had for the work term, which we must or they wouldn't be equal. So we're going to find that all four of these have exactly the same units. If we're in English units we have pound-foot, more commonly said foot-pounds, but we have those very same units. Now every year I go over this. I find it necessary to do this even though I will have a few of you who will neglect to pay attention to this fact failure. I'll put it down here. v2 squared minus v1 squared, which we need for the kinetic energy term is not equal to v2 minus v1 quantity squared. Yeah, you're shaking your head. Jay, somebody in here, might as well volunteer now is going to try to do this. Nobody's volunteering? Frank, you're volunteering? No? Pay attention to that. That isn't even a legitimate physics one mistake. You might as well tell us a second year dynamics mistake. Alright, so so far we've got the possibility if we do some work on something we can change its speed. All these pieces are developed in the book and I don't see any great need to to repeat that. So I think it's a lot more instructed if we get to some problems we wouldn't get to otherwise if we'd be able to do all the development of each of these pieces. So the next piece is the delta vg, the gravitational, that's the g, potential energy. Our book uses a v. Why but it does, it's easy enough. Perhaps you're more familiar with this than even magnetic energy term. This just relates to us the possibility that we might have a change in elevation in the presence of a gravitational field. If there's no gravitational field then delta h is meaningless. If there is then it applies as written. Remember here that this is not a distance traveled by the object. It's simply a change in elevation in the direction of the gravitational field. So if we have some mass we raise up to some height delta h has the same change in gravitational potential energy as an object that goes a greater distance but still only goes up the same height in the in the direction of the gravitational field. And this all depends upon the implication that gravity is straight down. So we don't have to worry about any of the distance it travels just simply its increase in height. If we have a decrease in height then that of course that delta vg would be negative. And once again the units the kilograms meters per second squared times meters for the height well this first piece here. So once again we get new meters or joules for the units. As they should be we can't add one thing with one unit to something with different units. All of these must have the same units. So those usually look pretty familiar to most students. The other one I've always had trouble with the way the books do it because I think they set it up for more errors than need be. So we'll revisit it in a much better way which of course is my fault. This is the elastic potential and the fact that we may have in the problem a spring. Springs can store energy. They can accumulate it. They can disperse it and we have to take into account that fact. The slight bit of complication is the fact that springs can be stretched and store energy or they can be compressed and store energy. It's not a one way thing like it is with the gravitational potential energy. It's a two way thing. Sorry. That's why it's a conservative force. If you pull a spring out then let it return back to where it was before your release in terms of the spring in exactly the same place you were. Other terms the fact that you had to put your chin there and do that or there might be friction those kind of terms are always over here. It depends upon the strength of the spring. So this term is known as the spring strength or spring modulus or spring constant. Any of those terms are all fine. Remember that tells you how much strength this spring can exert once it's been either stretched or squished which we take into account with that second term there. The trouble I find with most of the other books and I believe ours is one that does it, they very often put an X in here which I think is confusing because there could be an X in the problem, a position in the problem labeled X that is different than the X that goes here on the spring equation. That term del is the difference in the spring from its rest length. But any time in the problem it could have length L which is greater than say its rest length so del would be positive. The length of the problem could be, the spring could be also squished some so L is less than its rest length. But once we square that it doesn't matter whether it's positive or negative which just shows us the spring can store energy whether stretched or compressed. This rest length that's the length of the spring as it comes out of the box after you went to K's hardware and got it from Earl. Be right there sitting there right in the box. And the K is also something determined by the manufacturer of the spring. It has to do with how tightly wound it is, how thick the wire is and what material it's made of. Remember the units on K are? No units on one half. It's what? Newtons per meter. It tells us to stretch a spring one meter. How many newtons must we exert on it to do that? If we need to squish the spring one meter, how many newtons must we push to do that? And then del squared of course is a length term so this has units of, that has units of one meter squared so again we have units of newtons meters. That's it. That's the work energy equation. I think all of the other stuff that goes into it is interesting. More so the physics students generally than engineering students, engineering students usually like to get to work on things rather than look at all the background theory and development to get to some place. That's why we're engineers. We have work to do. So let's do, let's set up a problem. It might be the type of thing you'd have to do if you're designing warehouses for UPS. Imagine a crate of somebody's shipment being delivered from the back of a truck into the warehouse or something. It needs to slide down a ramp. You don't want it to damage itself when it gets there so you develop some kind of bumper system to cushion the impact of the box when it gets to the bottom. Put some numbers on this. Let's say that's 23 degrees, 8 kilograms, flesh, coefficient of friction of 0.24 frame constant of 121 Newton per meter. And you want to determine by how much the spring will squish. If it's too much then the box might go ahead and hit the wall anyway. If it's too little then maybe the spring's not really doing as much cushioning as it should. Oh yeah that's in there too. 3.2 meters to the initial contact point with the spring. Fairly simple set, fairly simple problem. But if you just start throwing everything into the work energy equation it's going to be prohibitively difficult to. So if you want to do that just start throwing the things in and ignore me. Go ahead. If you rather fall along I'll show you just how easy it can be to use this equation especially on a problem like this. I've seen this type of problem. I remember trying to do it. You get lost in the units, the mathematics, and the amount of minus signs. So here's our first step. Go through any of the four terms and determine if any of them are zero. If so we're already at a smaller problem and things are getting easier. Is there any work being done by non-conservative forces? Is anybody reaching in and pushing or pulling or shoving in any way? We have friction in this problem. That's the classic non-conservative force. There's no way we can let this slide down the hill, push it back up the hill and get back what we lost to friction. So there's definitely some work being done that we have to keep in there. Is there any change in speed? Well of course there is. It's just sitting here at a rest. Then it slides down the ramp and then it comes into the bumper and is brought to a stop. That's not the concern here though. The concern is is there any change in speed from the initial point compared to the last point? We don't care what happened in between. So this is our first point and this down here is our second point. Is there any difference in the speed between those two points? If there's not, that's all we care about. We don't care about anything else. So in this case we can already make the problems 25% smaller by getting rid of that term. Since V1 it started from rest, it finishes at rest. It doesn't matter what else it did in between. We term something like that a point function. Anything that only depends upon the end points, we call a point function. The work term is a path function because it depends on where we go. The farther we go, the different surfaces we go over, all of those increase the work being done. The third term, very simply, is there any change in altitude? Of course there is. Start it up the ramp, finish at the bottom of the ramp. Is there any change in the length of a spring in the problem? Of course there is. It was at rest to start with compressed at the finish. We want to find what happens when the spring, the box brings that spring all the way to its point of compression. Alright. Now that we've made the problem 25% smaller, go back and do each one of these one at a time. Paying attention to the units and the minus signs at this time. Because now it's a very easy thing to do. The problem is really small. It's not too big a deal. Alright. Any work being done by non-conservative forces? Yeah, we have the friction. Non-conservative work. There's no work being done by the normal force. Why not? No movement in the direction of the normal force. Is there any work being done by the weight? Only movement in the direction of weight. However, what, Jake? That term is taken care of over here. That's conservative work. If we let it fall down and then we bring it back up to where it was, the weight is still the same. Gravity is still the same. There's been no change in the problem in that way. We have to do the work to do that. So the only friction, force we have is friction. It's in the opposite direction of the motion. That's why we have a minus sign. If we did the dot product on the vectors of this, it would give us just exactly that. If you remember, friction is the coefficient times the normal force. And the normal force equal to that component of the weight. That component of the weight in the y direction. What was he saying? Signing a cosine of 23. That angle between the normal and the weight is always the same as the incline. So this is cosine 23. Do we have all of those values? Now hang on. A lot of people working real hard here and I don't think you've got the whole thing in mind here. What is delta x? We've got all these other pieces. Coefficient and friction is 0.24. The weight is 8 kilograms times 9.81 meters per second squared. Cosine 23 you're going to have to look up yourself. What is delta x? 3.2 meters? Not quite. No? No? Plus delta. The amount of the spring squishes. That's why I don't like using that for squishing. That's in meters. So if we carry this calculation all the way through, which we will, we've already ascertained that the delta we're looking for will be in meters. We've got to watch this as we go along. So what's that equal to? That's just a quick calculator of business for us. Use it down to its farthest point. We've got kilograms of meters. We've got units of Newton meters as we should. Let's make all of these terms in Newton meters. Get the units right when we've got just a small problem. This is very easy to do so far. Term plus 1 something times del. Jake you've got it 5.5 or minus del. And we know that all of them have units of Newton meters if del is in meters. Everybody not get that? Are we getting different? We'll do the second term. Delta VG, MG, delta H. What's this term? We have 1 meters per second squared. And delta H is, what's delta H? It's, what'd you say? 3.2 plus del. Plus del times the sign of 23 going down. We need to take care of all the units and the minus signs at one time. So this term also has a delta H in it. When we put all those pieces together you should get, and your sign 23. You've got it? Okay. 9.81 minus what? 30.7. Yep. And we know that's in Newton meters and again it's determined that del itself must be in meters. So keep track of each of these pieces. Little problems one at a time. Let's see. Number two was plus 17.3 del. Alex just gave us, take care of all these things as we go. So the last part, del one is zero. Spring was at rest while the box was at the top of the slope. We'll have units of meters that will give us Newton meters. And so what's that plan? 21 is 60.5. You might have? Yep, we did. Those two are on the same side so we can add those directly together and we can bring this over and we get a simple quadratic in del. Good looks something. Let me look at the calculator. 60.5 del squared. That's the only term we have with the del squared in it. Then remember this is a negative but it's on the other side. So it's 98.1 minus 98.1 plus 55.5 which is like 42.6 but it'll be minus. Minus 30.7 del plus because we'll bring it over 17.3, 13.4. That's a pretty easy problem to solve. If you put everything into here you wouldn't have an easy problem to solve. This is a very easy problem to solve. Some of you even have a quadratic equation solver on your calculator and make it really easy. You'll see if this quadratic actually works out. Look around for it online there's a connector that allows you to download stuff to your calculator online which they have in the math lab I think if we don't have them here I think we have a couple of the plugs here. If not, you can use the quadratic equation solver B plus or minus find out where it equals zero same thing. The value equals here. Does anybody else get that? If it turned out that there were no real roots under the square it was negative what would that mean if there were no real roots to this equation? It would mean with this friction the box actually gets to this frame and there's no real solution to it which is certainly a possibility but that would show itself as non real roots. Were there two roots? I would assume so. Actually it's the negative root because we're talking about spring squish where they probably weren't the same. The positive root I think would be if we squish the spring this much and then let it go the spring would extend to that much as the problem was trying to reverse itself. I think I'm not positive that's what that would happen. But the point being if you do each of the sections one at a time watch your units, watch your non-assigns you end up with a very, very easy problem to solve. If you put everything into there all of this stuff into there first thing you're going to stop worrying about is your units you're not going to put those up there anymore so you could screw those up you're also going to lose minus signs we didn't have to worry about this minus sign but the other way around if this was zero and this wasn't it's very easy to lose that minus sign so I very highly recommend you do this problem each little piece at a time as you go through it it makes it an awful lot simpler. That's extremely easy to solve for you guys. Now if you want to, as an exercise put all of them in there and just make sure you can do it accidently can happen I think for many of us doing it. Alright, ready for one of your own? Energy equation so when you start to solve these that's the first thing you should write down is the work energy equation see if any of the parts disappear and we have an easier problem. Here's a smooth shaft spring there down here at the bottom is a smooth collar that can slide on that shaft attached to the middle of that collar and its height is 25 millimeters that's to that is a cable that runs up this right level with the spring 25 millimeters which is 200 newtons spring stiffness how big a spring do you need to go down to Earl and buy so that the amount of spring is squished as you pull on this force it's going to hit the spring we need to limit del to 75 millimeters now I would suspect sounds like a nasty and complicated problem break it into pieces that aren't so nasty and complicated checking your units checking your minus signs as you go looks like a big problem huh first thing you do is go over that if any of those parts disappear if they do the problem is already smaller the problem is smaller all the better yes a little bit of trouble then go on to one of the other ones for a bit just to get in there got the force is 200 starts from rest at the bottom engine kinetic energy zero certainly it picks up speed but then it loses it again once it hits the spring all we care about is the start compared to the finish between point A and point D is all we care about any other parts that are zero outside force is being applied the collar doesn't move in the x direction if by x direction I assume you mean the direction of the force does not move in sideways no what work is being done by that force look at that force force doing some work to figure out how much force that work is doing the change in those energy terms that we're interested in change in height we've got M, we've got G what's delta H that's the only thing changing height technically the spring the center of mass goes up a little bit as the spring compresses but we're not going to mess with that we're not going to worry about that the collar will go up a certain height just how far the center of it will go up to 450 so it's going to finish right level with V on it because 75 millimeters of the collar is above that center point and that's how much spring squish work is going to allow so the collar is going to finish with its center right level so how much I did it actually game 450 plus 75 which is how much if you have something that speaks up all I can do is these little mumbles and then I have to keep asking for you to repeat thank you so now you're mumbling again oh boy isn't that big news hairline is going back hair is turning white I heard it all the time and I've got to hear from you when I'm getting home do the units work out is there a minus sign and so what's that come to be 6,000 millimeters a little bit of geometry work on that one but basically we got that the K we don't know that I know it's going to come into the equation for us to find it let's see del 1 is 0 and del 2 is the limit of 75 millimeters if K is in newtons per meter this will have units of newtons meters what's 0.057 squared 0028 1 and that happens in there as well centimeters if K is in newtons per meter itself we already worked out the units of K we're reducing this down to a very simple equation with a single unknown so that's the right hand side all set 0.1 so that integral will work out that way then all it depends upon is what's that delta s is it this distance that the collar moves trouble with that is if we do it by that the force on the collar changes it's first here then it's there its angle is changing all the time the dot product is going to change all the time the integral of that I think would be kind of nasty to do is there an easy way to figure out what work was done by the force the way to do it is to figure out how far that force actually moved itself in its own direction a force being applied in that direction if it moves in that direction that's got to be how much work it does isn't that this distance minus that distance is how much of the cable was actually pulled over the force would move that far in that direction just 200 newtons what is this distance 225 squared 450 squared actually 450 plus the little bit extra what is this distance 583 so in meters do you use the 5 to the when you're doing that do you use the 5 to the 450 well it depends on your right triangle is right there on whatever the right triangle is 225 across the top across the bottom it's 450 plus the 75 572 572 that's millimeters so we'll turn it to meters we want the units to work out here but it ends up with still the 225 over there that will give us units of newton meters what's that come out to be then how much path 69.35 meters close enough 69.3 the whole equation all the units checked all the minus signs checked now you've got a really simple little equation to solve for one single unknown and you already know the units are going to work out because they had to back here to give you newton meters in this 61 is 332 divided by 0.0281 right just because they had to here to give us newton meters in that equation that again is a pretty simple little equation to solve you put everything in there all at once you're going to miss unit signs units squares that's the other thing that goes very easily under pressure I think Wednesday will be ready for even harder problems