 Alright, if you remember, we're looking at now kinetics of particles. That's the deal where we've got some interest now in not just what the accelerations, velocities, and positions all these things are, but how to actually cause a particular acceleration or achieve a certain velocity. So the first of our kinetics solution methods was F equals ma. So we worked on that for a couple days, worked through it in a couple of the different coordinate systems because all of this depends still upon all the kinematics that go into it. In fact, it's necessary to bring kinematics into these often as the extra equations needed to find the solutions for these problems. This works real well for general type problems that directly involve the acceleration and or the forces. We're going to look at the second method today, very much in the same way we looked at in Physics 1 last year. A little bit different than the book approaches. The book kind of winds around a bit and then gets to this a little bit later, but generally I think it's, the pieces are all straightforward enough that if we get right to it and you read the page or two on each of the little parts of it, it goes pretty well that way. There's much more detail we can look at as we go through it. And then in a little bit we'll get to the impulse momentum equation. And then we'll do all three of these again once we look at rigid body motion. But right now we're doing just particle motion, treating everything as if it's a particle. So the work energy method, and you can go through the development of each little part of it in the book. But most of it is so straightforward, I think it's a little more useful just to get to the use of it once we've established a little bit of the parts that we're going. The work term, actually our book uses not a W but a U. Work is defined as the integral from some point to another, and that's very important to remember, that work as we're defining it is not being done if there isn't movement from one place to another of some type. There's usually a force causing that movement or involved in that movement somehow. Friction doesn't cause these movements, but it does certainly affect the movement. So that's clearly one of the forces we'd be using. Dotted with the differential displacement vector. Now there's a couple really important parts to all of this that we really need to look at. The first is that dot that we have in there. If you remember from what we looked at in Physics 1 some time ago, as we have some force acting on an object that's undergoing some movement, and I'll arbitrarily take a direction for the force and a direction for the motion. Bless you. This guarantees for us, even if F is not constant, this guarantees for us this dot product that we're only using the component of the force that's in the direction of the motion itself. That's what the dot product does. The component of the force in the direction of the motion gives us exactly this. So for this little example then, the work is simply the component in the x direction times however total much movement there might be overall in the s direction. Getting to that point is assuming that the force is constant so that it makes the interval very simple. That's real important that we keep an eye that we have to have the component of force that's in the direction of the motion. The other component of the force, F maybe we'll say v in the vertical direction, does not contribute to the work. It may affect the work that's being done because that component of the force might increase the normal force which might increase the friction force. If the friction force increases then the work done by the friction increases. So it's not that we can ignore that but it does not contribute your life directly to the work being done by F itself. The other part, well there's two other parts that are real important to us. First is just the integral. That means if this force varies with this position, we need the area under the force position diagram. And if it varies in direction or has components of direction, we can do that in either of the component directions. So this is also very important as a notion for us especially if we have rather well behaved forces. Forces that are constant for a certain section then change to a different constant or ones that change uniformly in a linear function or some other well known function which makes it much easier to do these integrals. Either calculate the area directly or actually do the integral analytically if you wish. The other part that's very important, this may be as important as any other part of this and is also the more subtle part of what's in here. It's that this is only going to give us anything if we take into account that we're going from one place to another where one and two or whatever locations are that we had on the integral there. In physics and engineering we call something like this where it's the integration from one point to another and in fact to highlight that I'll even put subscript indices on here that says the work done going from point one to point two, place one to place two. We call a function like this a path function because the work done depends upon the path taken. Of course you can see that if we have to have some force position diagram where there's some dependence upon the force whatever it may be as we go from point one to point two. If the work done is going from there from one point to another depends upon the area then it also depends upon how we get from one place to the other because the path determines what the area is. If we take two different paths from one to the other then we're going to get two different areas and we'll get two different amounts of work being done on whatever our object is by a force that varies in that way. This isn't something you don't know from your personal experience sometimes when there's something you need to do it's easier to go one way than the other and quite often you think about this a little bit especially if you're doing something simple like trying to move from some furniture from one place to the other. So it's very important that we keep these three ideas in mind. There's a deal of the dot product, the fact that the force can vary with position so we need the area under there. We have other ways we can deal with forces that vary with time. This is the easiest way to do it if it varies with position and also remember that it's very important not only that we're going from one place to the other but how we do that. All three of those affect the value we're going to get for the work being done upon some object as we look at the work energy method. Another place it's a bit subtle as we think about these types of things is that there might be friction in a problem. If we're moving left to right then the friction is going to be acting right to left and that is also part of the work being done. It's convenient for the work being done by the friction. It's convenient that it's always parallel to the direction of motion so that it comes to be minus because it opposes the direction we're moving. That would come out directly in the dot product anyway times the distance move. That again also is assuming that the friction doesn't change. It's very easy to come up with problems where the friction does change from place to place. You simply do that if you're pushing something across the floor and you go from the linoleum to the carpet and then to hardwood if you will. Any of those are going to change the friction. Easy enough to just separate that into three different sections. Calculate each of them independently and add them up. So the friction in our case is always does negative work. We're going to have to keep that in mind. It's easy to misplace that a little bit. For the units on any of these, of course we have units of Newtons times distance, force times distance is going to be Newton meters or foot pounds or pound inches or something like that and we use this opportunity if you remember to define the customary unit of work, the jewel. We'll look to one Newton doing work for one meter in that same direction. Same direction of the fourth, whatever it might be. So any component of a force will look to work. One Newton will look to do some jewel work. Alright, that should be a bit of a refresher from Physics 1, I hope. Make sure you read through the book to see some of the more details on it. But now let's look at the entire work energy equation as sorted out piece by piece in the book over several pages. Read through those just for a reminder of what each of the pieces are. We'll talk about the pieces as we go through them and use them now, especially since some of the letters used are a little bit odd on some of these as our author chose it. Plus I use a slightly different notation to help remember some of the other ideas. There's a couple things we can do with the work that we do on an object. We can change its speed. So we have a change in kinetic energy term here. Most of you remember that kinetic energy is 1 half mp squared, so this is the difference in that velocity or the kinetic energy contribution of the velocity between the two places, whatever two places in a problem we're talking about. Why our book chooses T for kinetic energy? I'm not real sure, but that's what they do, so it's easy enough for us to do that too. Yeah, right? Okay, that is helpful with me. The units on this, since it's equal to the units over here, which are Newton meters, I don't know that it's of great value to go to Joules and come back again, so often it's not saying Newton meters. Worth the units on the kinetic energy? I hope you instantly know, because if these two values are equal in any way, they better have the same units. But you can double check since mass is in kilograms, velocity is meters per second, so that will be squared. We'll have a kilogram meters per second squared, an extra meter, and we'll get Newton meters. That's real important. We've got to watch the units on every one of these pieces to make sure that they all go together right. We're going to have some much more complicated problems than we had in Physics 1, but they can all be broken down into very small problems that are actually rather easy to do. We can change the speed of an object with work we're doing on it. We can also raise or lower it through a gravitational field. So our book happens to use a V for potential energy. This is the change in the gravitational potential energy. And mostly remember that the gravitational potential energy is MGH. So this is MG delta H, where H is simply the height change in the direction of the gravitational field to the gravitational field vector. That sometimes tricks students up a little bit because it's certainly possible to look at some problem where we move a mass from one place to some other place that's above and over from the first one. The only concern with the gravitational potential energy is the change in height. There's no concern with that sideways motion because the sideways motion is not parallel to G. This is work being done against the weight vector. And the weight vector is always straight down, so we're working against that. We only have a concern with vertical changes in position. No horizontal component comes into it. And the last little bit of the work energy equation is if we have an elastic medium, which in our case will be a linear spring, that's the change in elastic potential energy. Now there's a lot of subtle pieces in here. We're gonna have to pay attention to all of them, but some of that will come out with experience. This is one I do a little bit differently than the way it's done in the book. Only in notation though, just because I think there's a chance for error to really creep into these problems using the notation they use in the book. So I use a slightly different one. The potential energy of a spring depends upon how much its length has changed from its rest length. Every spring has some length attributable to it when it's unused, fresh, and new out of the box. Call that length L0, will work as good as any. That's the rest length. Any change from that rest length will increase the potential energy of the spring. And that's what we're measuring here is the changes in the potential energy. If a spring is stretched longer or squished shorter, either way increases the potential energy of the spring. So we also then have to have some idea of the working length of the spring. In some problem it may increase. By some distance we can also squish springs up and decrease them. We don't need to use any different indicator for that because we're not going to use those lengths directly. What we're going to use is the change in length, which means change in length from the rest length. I like to use a little del. The book does this the same way. Only he may use an X in there, which I think is problematic because a lot of times X would be in the position in the horizontal direction in a problem as it is, which means there can be confusion with the different X's in the problem. We just define it as the working, the current working length in a problem minus whatever the rest length was. And then the change in kinetic energy is the difference between whatever that is at the second point squared minus whatever it was originally in the problem. Either one of those may be, the spring is at the rest length, but either one of those may also not be at the rest length. It depends on the problem. It's very easy to set up any problems and do any kind of those. What are the units for this term? Why do you laugh? Because you know, if these are all equal and add a bolt to each other, add a bayonet bolt, add dab of weight. What? If we can add these things together, they must all have the same units. So we know this must have units of Newton meters or joules. Let's double check, though. Remember what the unit is on the spring constant or the spring strength or the spring modulus? Newton's per meter. Newton's per meter. How many Newton's it takes to stretch that spring or compress it one meter? We're dealing with linear springs that have the same spring constant in either direction. It takes a Newton to squish it a meter. It takes a Newton to stretch it a meter. And of course these are units of length, but they're squared. So this as well will have units of Newton meters. It's very important we watch the units on each one of these as we go through these. We're going to look at some pretty complicated problems. And if you don't watch this stuff carefully, if you get one little minus sign wrong or forget one little term that needs to be squared or have one term in the wrong units, you're doing a different problem. Couple other things. On each of these, the potential energy, sorry, the kinetic energy term will always be positive if the speed increases. If we're going faster at two than we were at one, it'll be positive. That's important, too, because we're always going to need to know what the sign is on these. It's a good chance to check it. It's also true that if we have several things changing speeds in these problems, we just calculate this term for each one of them and add them up. In fact, it could be useful to us if we put a summation sign in front of each one of these terms to indicate we may have one or more things undergoing any of those changes at any time. Positive speed increases and, of course, negative if it doesn't, or zero if there's no change in the speed. So it's important to look at any of those possibilities as we go through these problems. So I won't write down that it will be negative for the decreases. There's zero if it doesn't change. I'll just say end, etc. I won't pay attention to these as we go. Notice there's a difference in the characteristic of this term than there was in the work term. Remember I called the work term on this side a path function, meaning that what path was taken between points one and two was of vital importance to the actual value of this work term. This depends only upon the speed at point two and the speed at point one. It doesn't have any dependence whatsoever on the speed in between there. It can go up to the speed of light, come back down to zero, go to the negative speed of light, do whatever it wants. All we care about is what was it's speed when it started? What was it's speed when it stopped? We don't care about any other point. That then is known as what we call a point function. A path function depends upon what's done in between the start and the finish. A point function does not actually makes things an awful lot easier. It's a lot easier to just look at the two points, see what the velocities are there, and not have to worry about much else. Not sure what the work term, but it is true of that term. The gravitational potential energy term is always positive if the height of our object or objects, there may be several different things changing heights, but if the height increases then that term will be positive. These are very useful things. They help us anticipate what's going to happen. Also is most important, helps us double check all these things as we go along. And of course, I can add and etc, meaning the same kind of thing as then over here. Positive if height increases, negative if it decreases, zero if it doesn't change for any one of the objects in there. If we have more than one object, we just do this calculation for each one of the objects and add them up. It's very much like an accounting problem. You just have to keep track of the different items, add them all up at the end and see what's left over. Is this a path or a point function? Point function only depends upon the height when we started and the height when we finished. Nothing else in between matters. Makes it much easier to use. And that part, that's due to the fact that the gravity vector, gravitational field vectors only in a single direction. This one's a little bit more difficult than the other one is because there's a lot of possibilities of whether or not it's going to be positive or not. So I can't really make this kind of statement here because there are situations where the length will increase, but the potential energy of the spring will decrease. Can you think of one that would be the case where the length of the spring would increase, but the potential energy would decrease? If it starts heavily compressed and decompresses a little bit, its length will increase, but its potential energy will decrease because there was more potential energy in it when it was very compressed, not nearly as much potential energy when it's less compressed. So this one, you have to pay a lot more attention to it. I don't even know what I can say here. I guess I'll just say an et cetera right from the start because I can't explain before that. We have to pay attention to that problem by problem as we go through it. So kind of vague, kind of wimpy. That's due to the nature of springs that they increase in potential energy on either side of the rest length. Path or point function? At one vote, no votes. Point function. It only depends upon the length before and the length afterwards. Another thing of great value and interest to us in this is it only depends upon the length of the spring. It does not depend at all on the angle of the spring, the direction in which the spring is pulling. It depends only on the length of the spring. It depends only upon del at the different positions. One other thing, and let's see I have, I think I have 11 people in this class, one of them they're all here. 11 people my experience says that at least two of you will fall for this simple trap. Notice that we have a v squared minus v squared term there. We have a del squared minus del squared term there. So I give everybody this warning, two of you, two out of 11's about my experience, two of you will get caught by this at some time during the next week or two. I'll just use a general variable for the v and the del because it's an algebraic truth that x2 squared minus x1 squared is not equal x2 minus x1 squared. Two of you, yeah don't give me that face. Two of you will get caught by that sometime during the term and the rest of us thank those two for taking the bullet. Would anybody like to volunteer right off? Thanks Phil. Says Alan. If I remember a physics one partner. You never did it. I made my photocopy everything you ever turned into me. I'll go back and check it. Don't get caught by that. It seems incredibly simple a thing to be caught by but those are sometimes the things most easily doing the catch because it concentrating on the harder stuff the easier stuff slips through a lot of times. All right I think that's all the general setup we need so it's time to start doing some problems that are more complex right from the start that we would have done in physics one. Leave that up here we might need to call it Alan's law. You couldn't say that they mean for something a little more complicated. Well since we're taping I didn't want to use your last name. All right so let's let's start setting up some more difficult problems. All right here's one just to get us started jet plane you can tell it's a jet plane not my car as far as wheels and not wings unless I push the button and the wheels come out and the wings go in. A jet plane launching from an aircraft carrier is going to have two two sources two forces exerted on it to get it up to some speed. First the engines will be on and they'll be on full so let's say that that's 140 kilonewtons and we'll take that to be a constant the thrust is constant which it pretty much is they they open the engines full before they do anything to get started. Let's give a mass in the plane there you go crest like that that's what you saw in the book that means 10 to the 6th oh did you did you think this may be met 10 to the 6th kilograms because this is 10 to the 6th grams just like a kg is 10 to the 3 grams. That's a mega gram. This is a mega gram. It's like a box of 24 chocolates on Valentine's Day. That's a mega gram as well. There's also if you've seen aircraft carrier launches they attach a catapult hook to the bottom of the plane to help get it up to greater speed in a much shorter distance than the thrust alone would do and that catapult force varies with position along the along the deck. It actually starts out low comes very quickly to a peak and then pretty much drops off linearly. This first part happens very quickly so we'll ignore that for this problem and just assume a linear variation for the whole thing but that varies between about 65 kilonewtons up to a maximum of about a hundred and it does that over about a 90 meter length of ship deck. Let me double check. I think that's all the pieces. Again we'll assume a constant that initial part of the oh sorry this is a hundred this is a 1,100. 1,100 kilonewtons here is the final force. All right so we want to find the final velocity the launch velocity once the catapult's done its part then what is the velocity. Now we could do this by finding out what the acceleration is except the force varies with position which makes finding the acceleration that is not constant a little bit tougher problem but if we use the work energy equation it becomes pretty easy. Work energy equation is very easy for things that are position dependent especially if the force is position dependent. Start with the work energy equation. Are any parts of this zero if so we're already doing a smaller problem. Is there any work being done by any forces like oh I don't know thrusts and catapults. These catapults are generally not an elastic type thing they're usually steam driven. So this is certainly not zero because we do have forces acting over a certain distance. Is there any change in speed. Of course that's a whole point we're trying to get it up to launch velocity in the short distance of the deck. Any change in gravitational potential energy. Look at nothing more than did it change height did anything in the problem change height. Nothing did so we're already a smaller problem and this catapult is not a spring activated mechanism so there's no elastic medium in it either. So we're already working on a very much smaller problem. So a work being done by two components we'll just calculate them separately and add them up. One is being done by the force the thrust and one is being done by the catapult. We can just add them up calculate them separately and add them up that's one of the things that makes this work energy problem methods so easy to use it's just you can break it into as many small pieces as you want and then just add up all the pieces at the end. All you have to do is keep track of all the minus signs all the squares. Since the force of thrust is constant and it acts over the same 90 meters of the launch it's and in the same direction it's just simply the constant thrust times the distance the thrust acts. If the thrust shut off early or starts late or something we just change that distance. Calculate for only the distance it actually works. Now we've got catapult force that changes with position linearly so we need to integrate this but if we're assuming it's linear then we don't need to do anything more than calculate the area under the graph. We'll let you do that see if we get the same thing. The first one's easy we just have to watch the units since the thrust is in kilonewtons and so does the catapult force might we just leave that and we'll have kilodule rather than kilonewtons. That one's easy it actually is the area under the thrust graph it's just a constant doesn't change so it's very easy to calculate that area. Alright so I'll let you calculate the force sorry the work done by the thrust it's positive because it's in the same direction as the motion is during other things like friction doing the wheels in the deck and air resistance of the plane because I imagine they pick up pretty good speed and they have to practice these launches and landing so my brother-in-law came to our wedding he needed to practice some landing so it went to the naval base borrowed one of their planes and flew out to the wedding in a US government jet plane and practiced the fake landing in Syracuse and on the tarmac they have an aircraft carrier painted you practiced grabbing the landing hook then right in the car came to the wedding show off I still sold today you figure out the work done by the catapult and then I'll do it we'll see if we get the same thing we're still just working on this left hand side now watch your positive signs watch your units watch your squared terms if any there's a temptation just to dump everything into the equation as a whole especially when it's small like this it's not too busy deal but when these get bigger when all four terms are here these problems get very very complicated and it's tremendously easy to lose a minus sign screw up a unit squared term it shouldn't be your vice versa ignore ignore this very very sharp part of the front here just assume it's linear from the 1100 newtons all the way to the end of the launch we're finding let's go we still have 65 newtons of force on it but here at the end of the catapult end of the ramp by then end of the deck all you need to do is integrate this function got something Travis same as Chris you have the work term so why you're okay on this you're already over there all right let's see what we get for the work term what's this function we figure that out we just integrate it however might be a little bit easier to just calculate the area under the graph that's all the integration can give us anyway so the area of a triangle is one half the base times the height that's one half the base is 90 meters the height is 1100 minus 65 that's what 103 5 is that right 103 5 1100 my 65 and what's that equal to everybody have it 46 575 that's killing newton meters who got that for the work done by the catapult nobody building a lot of people did not get this got something else got about twice that Chris was you get you get this we're not doing that yet you didn't even have to do this integral well fine now we can't check the individual parts we need to check this part Travis you've got this number do you Anthony yeah Samantha this number is not right you got something else what you do did you do something else besides this this this is remember this is the thrust this is the catapult and this is just the area under the triangular part about this area that area is also under the graph also needs to be included so that's 65 times 90 65 kilo newtons for 90 meters the area under the graph doesn't just stop where there's some kind of change in the function it's all the way down to the axis and if this was a negative force we'd have negative area and still work out this fun what's this value anybody have that the 65 times 90 bill five thousand eight meters so that's over a 10% error neglecting that part down there all right now we've got all the pieces the entire area under the any of the force diagrams that includes this one from the 140 all the way down to zero all of that adds up to be 65 again 60 yeah about 65 65 mega joules 65,000 kilojoules or kilo Newton meters all right so that's the first term here then done now we have the delta t part one half m which we have v2 squared minus v1 squared starting from rest so that part zero we're okay with that that means delta t is going to be positive is that right yeah we only have to look at one thing is it picking up speed yes it is and so we can figure out that term watching the units 24 actually we leave it in mega grams then we're gonna have kill a new to all way and the v2s what we're looking at is that going to give us the right units kilo Newtons meters if we leave that in mega grams we leave this in mega grams will we get kilo Newton meters from this assuming that the velocity is in meters per second let's see 24 10 to the third kilograms that right is that a mega gram times whatever the velocity is in meters per second squared so the kilogram meters second second squared that's a Newton meter so this is a kilonewton meter so if we leave this like that then we're already okay because we just have to change this back to get kilonewtons that's a Newton meter that's going to be a kilonewton meter so okay and now we have a very simple problem it just says 65 mega joules actually we want that to kill with joules because everything else was 24 v2 squared we already know that's going to be killed so that's a pretty simple problem to solve should get about how many meters per second yeah by 74 meters per second check your numbers y'all need to do the square and square root what did you make Tom cruise with miles all right simple problem I know some of you just wanted to put the one app in these Delta Delta v squared in there solve for it you came if you want but if you mess up a minus sign or a squared or a unit conversion somewhere it'll come one in debate we're going to do harder problems now if you put them all together everything goes into the work energy equation you're going to make a mistake sometimes because you're doing one really big problem instead of four really little problems all right so here's here's the next thing we'll look at okay over against solid wall we have a spring attached to a collar that rides on a run a bit so far now attached to that collar is also a cable it goes up to a pulley where it's pulled by a particular constant force right so some of the other things will need spring constant 80 newtons per meter the mass of the collar 50 kilograms the force up here 300 newtons the rest length of the spring is a meter three dimensions three distances so from here to about there is 1.3 sorry 1.233 meters the collar goes from there to right below the pulley obviously not to scale that's a distance of 1.2 meters growing not quite to scale but as long as we have all these values we're okay and the pulley is at a height of 0.9 meters over the rail itself you're gonna believe it starts from that position okay so it goes from point A to point B along the frictionless rod find the velocity at point B if it starts from rest very depend this problem depends a lot on the position of everything we've got springs we've got velocities and different distances so start with the work energy equation are any parts of it zero let's make the problem smaller right now it's a four-part problem let's see if we can make it into a two or a three-part problem any parts of it zero is there any work being done by things other than the gravitational force or some elastic force which would take care of somewhere else any other force but one of those is this force doing some work on it of course it is pulling on it and it moves some particular distance so we know that terms not zero is there any friction in the problem no this is this is we've got it from your house it's all greasy is there any change in speed yeah that's the whole problem it's starting from zero we need to find the final velocity that's how we're going to get it is there is anything changing height at least anything that has a significant mass yeah some of the cable is changing height because it gets pulled up over the pulley but it's an insignificant mass compared to the 50 kilograms so nothing is changing height problems a little bit smaller and do we have an elastic medium in the problem whose length changes from start to finish and yeah we do so that term is not zero so let's do one at a time that way we can watch their signs in their units and pick out any other particular problems in these this one's a little bit tricky because the force that force P is in that direction for a little while and it ends up like that and we have to integrate the that vector that force vector dotted with the position the distance as it moves along in the horizontal direction sound easy to do who wants to do that integration for a set it up well use somebody set it up and somebody else will do it split the effort there probably just gonna wait for somebody to do it there's a much easier way to do this anybody see what it is David you see these kind of things sometimes instead of doing this integration where the vectors the same in magnitude but it changes in direction as we move in that distance so remember at every place and I'm only drawing a few of we need to know this continuously every place we need to know the force in the direction of motion the vertical part doesn't contribute anything to the work because there's no movement in the vertical direction by the man but there's a much easier way to do this problem anybody see it huh area of what that triangle I'm gonna trust you guys don't take this off Chris you see an easier way to do that animal not a pleasant I don't know the setup I'm pleasant you're gonna do David the function of force seems to be the derivative of a circle circular equation I don't know it could be that doesn't sound easier though okay that sounds harder by easier I mean really no that's constantly changing though that's the trouble it continuously changes from whatever that is down to finally zero right there and that's the trouble that the component in the direction of motion is continuously changing now we could figure out how we can figure out that you know that's that maybe the cosine of the angle and the derivative of that but there's a much easier way to do it now you see this kind of stuff all the time the easier ways to do sometimes you're wrong but it is easier grab to be around this is the force doing the work that force is gonna move from right here is some position right there could we figure out that distance because that would be that force moving in that direction there'd be force times distance can we figure out that distance delta s yes it's just the difference in length between the cable there and the cable there it's very easy to figure out so this is going to be p delta s that's the we don't have to worry about the fact that the force on the collar changes because this is the same force up there doing the work we just need to know what delta s is so that's 300 newtons delta s let's see it it was let's see 1.2 meters squared minus 0.9 meters squared square is that right that's that's this length originally is that right just just this right triangle you know we're ignoring the thickness of the remember this is particle motion we're looking at minus 0.9 meters minus 0.9 meters oh yeah I don't want to subtract it want to add it I was getting set to subtract it there sorry that's that's the hypotenuse here this original distance maybe we'll call that B is that right now and the difference between those two let's see you know the set of brackets in here so that's the force times the distance that's how much of the cable gets pulled up over the pulley and moves along in the direction delta s for those of you that want to check go ahead and integrate this you see it's the same I think I'm not going to do it I did my head too I did this way all right to the units work we get newton meters is it going to be positive should it be positive sure why sure that's where you're pulling on it that's where you want it to go because the force and the movement are in the same direction so yeah that's positive so what's that come out to be 180 newton meters take one half m v2 squared or vb squared if you wish starts from rest so that's easy we're looking for v2 we've got everything else all we have to do is check the units 25 kilograms times meters squared per second squared will be newton meters so that's all that that term disappears to be we have the spring constant some of these problems we're going to look for are either of those two terms zero it's got a little bit of stretch in it came out of the box of one meter we've got it attached to start the problem at two point one point two two three and then we stretch it all the way out even farther so that is not going to be either of those are going to be zero so what's del two what is the working the change in length from rest length when the color is all the way out of B our last point it's these two added one point four three three minus the rest length so that's one point four three three square that's L2 minus L0 and then that's square and that'll give us newton meters we have news per meter times meter squared and the original del is one point two three three minus the rest length don't forget this is not the length of the spring in any point it's the change the difference between it rest length the total distance here is two point four three three but minus the rest length this is L1 minus L0 it sends a break back to 80 what it sends a break back to 80 what do you mean sends it right back to 80 79 oh that's what this all equals yeah more coincidence than anything I believe is that newton meters yeah we have newton's meters now we have a very easy problem to solve should that be positive should this term be positive oh we didn't should this one be positive too when you check both of them are we going further if we made a mistake on the sign you might as well fix it before we go to solve it this should be positive picks up speed that's all we need to look at this is stretched to start with stretched even more to finish with so that should also be positive and that works out so now our problem becomes very very simple newton meters from the work term equals 25 v2 squared it's also newton meters as long as the velocities and meters per second plus a newton meters that's a pretty easy problem to solve a very small problem now if you wanted to and I know some of you do because there's at least half the class that never never follows me on this you work energy problems loves to put everything together right from the start and work it out from there I don't understand how you're you're elastic there right there in the middle there that's the equation with everything put in the original work energy term all the way across and you got to make you got to solve this for vb or what I call v2 up there without making a single negative error a single square error or a single unit error anywhere it could be done but we've got more complex problems than this one coming up too and if you want to put them all together and like this and then guarantee you won't make a single small error like here we checked small problems didn't make any mistakes just a lot less hazard solving this problem is not making any minus sign errors and it was would be solving this one in its entirety it's doable I don't like doing it that way I don't like making minus errors Allen you still didn't see why this is positive well up here when you had your work total it just seems like this with the springs gonna be pulling against your yet 80 Newton's Newton meters pulling against your 300 at the top so I don't see how it was positive this this is not this is the potential energy of the spring this is not to work the springs doing on the peaks that's the opposite of this and you'll see that developed in the book yeah yeah yeah potential energy of the spring right I was seeing it as a potential negative energy the potential to pull it in the opposite direction we're trying to go so it seems like that well if you want to redefine this yeah you can this is the potential energy stored in the spring which is different than but it comes from the work done by the spring on the peaks right yeah just I was just trying to figure out how the potential energy was not this potential energy stored in the spring look at it that way okay that's the end there