 Hi, I'm Zor. Welcome to Unisor Education. Today's lecture is about work, mechanical work, and I will be talking about definition of this concept. I will try to justify it as much as possible, because usually if I open some textbook, not all of them, but many, they're just giving the definition as a formula, and basically that's it without much explanation. I would like to provide certain basis for this definition. Now this lecture is part of the course called Physics for Teens, presented on Unisor.com. It's a free course, free for all, and also on the same website you can find the prerequisite course for this Physics for Teens. It's Math for Teens. Mathematics is very important and for Physics and definitely something like vectors and calculus are used all over the Physics. So I do suggest you to get comfortable with Math, maybe using the Math for Teens course on the same website or anywhere else, and that would actually make your life much easier as far as the Physics started. Alright, so about the definition of work. I will present four different problems and in all these four problems, the same formula, if you wish, for work actually will be presented as basically being the basis of the problem. And that would, in my view, justify the definition. Now, here's my first problem. Let's consider a straight line movement of some kind of object of the mass m and you have the force, constant force f, acting on this. No friction, no any kind of a complication. Now, my purpose is to accelerate this particular object from speed equals to zero to some kind of a speed equals to b. Now, the perfect example is whenever you start the car, you basically accelerate during certain amount of time. You do it during certain distance and then basically it goes with this distance more or less permanently. Well, obviously we are talking about no friction, no traffic lights or anything else, so all we are doing we are accelerating the car and then it goes by inertia all the way. So, what's the purpose of our acceleration? What's the purpose of applying force? What's the purpose of accelerating force to the car? Well, to reach certain speed. Now, more powerful engines are capable of increasing the speed from, let's say, from zero to 60 miles an hour in, I don't know, six seconds. Something like Tesla can do it, let's say, in four seconds. So, we are measuring the performance of the engine, so to speak. Certain, let's say, time which is needed for this particular engine to reach particular speed of a particular car. Now, instead of time, we can use the distance. For instance, one car needs, let's say, 100 meters to reach the speed 100 kilometers an hour. Now, a little bit weaker engine in the car it will be necessary, let's say, 200 meters to reach the same speed. So, basically what I would like to do is to have some kind of a dependency between the force which is applied to a particular object, final speed which I would like to achieve and the distance I have to really apply this force to achieve this particular speed. So, that's my purpose. I would like to have some kind of a dependency between F, S and V, well, and M, obviously. The heavier object, the more difficult it is to accelerate it, right? So, what can I do to establish this dependency? Well, let's just apply the regular second law of Newton and the laws of kinematics and we will do the following. First of all, the first is this formula. Now, obviously this is the second Newton's law. From this, we can define acceleration. Now, knowing the acceleration, I can find the time which is necessary to cover this distance. So, with this acceleration, now, this is a regular kinematics formula. I hope you remember it for the movement with a constant acceleration, right? So, it's equal to F t-square divided by 2M, right? From this, we can find, let's say, t, which is equal to 2M S divided by F-square root. Am I right? So, A is F over M t-square and t from here is equal to 2M S divided by F-square root. Now, to find the dependency, we obviously have to connect V and A. If you have an acceleration, now, final speed is V. The beginning speed is 0. So, that's basically the kinematics of the final speed, which is equal to F over M and t is equal to this square root of 2M S divided by F, right? Am I right? Okay, now, let's square it. V-square is equal to F-square divided by M-square F-2M S. Now, M is canceling. This is canceling. So, it's 2F S divided by M. So, this is my formula, basically, which connects the V, final speed, which we would like to achieve basically the result of applying the force F. Now, what is the final result? What do I apply it for? Well, I'm applying to achieve this particular speed. So, and this is the formula which relates my final speed with the force and the distance during which I have to apply it. Now, what I would like you to do to pay attention to is this product. Let me just put this product as a letter W. So, that would be equal to 2W divided by M, or we can resolve it for W, which means W is equal to MV-square divided by 2. So, it can be interpreted in both ways. If we would like to achieve certain speed, then we would like to spend our resources, which I denoted as a letter W. It can be one particular force and one particular distance S. Now, we can double the force and shorten the distance by half, exactly the same result, right? Because it's the product of force times distance, which is important in this case. So, that's all I wanted basically to do. So, if you have certain amount of resources, then you can find certain amount of speed, which is, well, certain speed, which will be the result of applying these resources. Now, if you have given certain speed which you would like to achieve, that's how you can find out how much resources you need to achieve this speed. And resource is the product of force and distance. So, either you apply one force and one particular distance, or you can double the force and shorten the distance by half, or you can half the force and double the distance, whatever it is, or one tenth of the force and ten times the distance, whatever you decide to do, so you can do it with a weaker engine on a longer distance or a stronger engine on a shorter distance. But anyway, it's the product of these force and the distance, which is very important to achieve certain speed in this case. Now, let me just instead of achieve certain speed, I will rephrase it to achieve certain result. So, if you would like to achieve certain result, as basically result of the application of the force on certain distance, that's basically the formula which gives you this type of thing. So, it doesn't really matter how you achieve it. Again, with a weaker engine on a longer distance or a stronger engine on a shorter distance, result will be the same. It's the product of force times distance which defines the result, or defined result, if given a result, then you can basically find out what is your product of the force time distance. And basically, now I can tell you that this thing is called work. Now, this is a work which is performed by this force during or applied on the distance of s. So, if the force is applied on the distance s, their product is called work, and it's from the work your final result depends. So, if you want to achieve certain result, then you can find out what's the work to be applied. Or, if you have the work, then that will be your result. Okay, this is my first problem which basically introduces the concept of work. And what I'm going to do is to give you a couple of more examples which basically either make a little bit more universal this formula or just confirm it in some other cases. Okay, first of all, to make it just a little bit more universal, what if you have the force applied at an angle? Fine, if this is the force. Okay, let's say you have a toy train, for instance, which is going on the rails. Okay, this is your toy train. And the child is pulling the train at an angle. So, what happens? Well, if he doesn't really put it too strongly to completely derail it, the force actually should be divided into two. It should be represented as a sum of two. One component will be along the track, another would be perpendicular to the track, but if the track actually is made of something which resists this thing, obviously there is a resistance. So, it does not derail. So, this force is balanced with the reaction of the rails. So, you can completely ignore this. And only this one will act in the direction. And this force is equal to, if this is phi, it would be f times cosine phi. So, basically it's exactly the same problem as before, except we have to consider, instead of force f, as in the previous case, we have to consider f times cosine phi. So, a little bit more universal formula would be this. Cosine phi. That's my second problem. Again, that's not to emphasize, basically, the correctness or validity of this definition. I'm just trying to make it a little bit more general to cover this case. Okay, so that's my second example. Now, my third example is completely different. So, what I'm going to do, I'm going to use the incline to pull up with a certain force onto the height h, some kind of an object. And I will try to do basically exactly the same. What's my purpose? My purpose is to lift the object onto the height h. Now, I would like to find out what kind of resources, basically what kind of work I have to perform to do it. Okay, how can I do it? Well, obviously, there is a weight, and the weight must be, obviously, represented as some of two vectors. One is perpendicular to the incline, and another is against the movement up, right? Now, obviously, this force would be balanced, this force would be balanced to a reaction of the plane, and both are balancing each other. Now, to move it up, I have to have the force f to be equal to this one, to r, right? If I would like to move it with a constant speed, no acceleration, just make it simple. So let me shorten this vector, so it's equal to this one by magnitude. So as soon as my force is equal, we start moving, no friction, right? So what is the value of this force which is resisting the movement? Well, if this is an angle phi, then this is an angle phi. So my r magnitude is equal to weight times sine phi, right? Okay, so I have my force. By magnitude, it's equal to this. Now, what's the length which I'm basically using to, what's the distance I have to cover to reach the height h? Well, that's s which is equal to h divided by sine of phi, right? This is the height and this is the length of hypotenuse. And again, f times s which is equal to w, which is work by definition, the force acting along the distance s and it's equal to sine is cancelling out, which will be p times h. Now, that's very important that angle is completely irrelevant here, which means you remember how it was with the car, slower car, longer distance, stronger car, more powerful engine, shorter distance, but whatever it is, the result which is achieving certain speed depends only on the work. Well, NMS, obviously. In this case, we also have very, very similar, philosophically similar result. I can use a smaller angle, smaller slope, more horizontal, so to speak, which would actually require less of a force, right? Because if phi is smaller, sine phi is smaller, so the f will be smaller. But the distance would be so much longer. And the product is constant. It's the weight, which is mg, if you wish. Again, it depends on the mass, obviously. And the height. Height is always the same. So the purpose of applying the force and the work are basically related to each other without any additional parameter. It doesn't matter how I'm lifting my weight, my object. I can lift it up, and it will be exactly the same thing. It will be, force would be equal to weight, and the height would be equal to h, so it would be again p times h. Or I can use a slope, a more steep slope or a less steep slope. It doesn't really matter. The product will be exactly the same, which means amount of work. I have to spend, the amount of my resources I have to spend to achieve my goal, and the goal is to reach the certain height. It will actually be exactly the same. So again, the relationship between the purpose, the result, the goal of acting with the force, and the work, which is supposed to be spent like resource to do it. Very important relationship. Again, result and work. The purpose of applying the force and the work, which this force actually performs during certain distance, applying on a certain distance. That's my third example. And the force one, again, different. It's related to rotation, and I will show you that in that rotation it's also basically the same thing. Everything depends on this product of the force and the distance. Okay, let's imagine you have a carousel, and you have tangential force F. Let's say, well, children are on the carousel, and it's a manual carousel. It doesn't have any kind of engine behind it. And one of the parents actually holds the carousel at the rim and moves with the carousel trying to speed it up. So it's kind of an equivalent to the first problem when my car actually was speeding to achieve certain linear velocity. Now, in this case, I would like this very good parent to speed up the carousel to reach certain angular speed and let it go for a while so the kids will enjoy it. So we are talking about a goal of achieving certain angular speed. Now, okay, let's recall our rotational dynamics. Now, the angular speed is equal to acceleration, angular acceleration times time in exactly the same way as linear speed is equal to linear acceleration times time. Okay, now my linear speed would be omega times r, right? My linear acceleration would be alpha times r. Sorry, yes, yes, linear acceleration of the rim, of the point on the rim, right? So this is the linear speed, but this is variable because omega is variable. This is constant because alpha, the angular acceleration is the same time radius, all right? So this is radius, obviously. What else? Now, we are talking about dynamics. Remember the equivalent, the rotational equivalent of the second Newton's law. It looks like tau, which is torque, which is equal to f times r. It's equal to moment of inertia times acceleration. So this plays the role of mass in this moment of inertia and this plays the role of linear acceleration for second Newton's law. But this is basically the rotational equivalent, all right? Now, I will do exactly the same thing as I did for linear acceleration of the car. I will try to find the relationship between the fours, the final angular speed which I would like to achieve. Now, instead of mass, I will use moment of inertia and obviously I have to have the distance. This particular parent will follow the carousel on its rim to speed it up to the angular speed omega. So I have to find out this. Well, again, the dependence is very much like in the linear case. In the linear case, if you remember, it's a times t squared over 2 where a is linear acceleration. In this case, my s would be equal to r alpha t squared divided by 2 because r alpha r is linear acceleration, right? So I know this. So let's say how I do it. I'm taking alpha from the first one. So alpha is equal to f times r divided by i, okay? So I'm substituting to this and it's equal to r times alpha which is f r divided by i times t squared divided by 2. Am I right? Now, instead of t, I can put this. t is equal to omega over alpha which is omega over f times r times i, right? From here, alpha. And I can substitute it here and let's see what happens. rf r, again, t squared which is omega squared i squared divided by 2i and square f squared r squared. Am I right? So r, r, rf, f, i, i. So s is equal to i omega squared divided by 2f. And from here, again, times f both sides. f times s which is my w is equal to i omega squared over 2. By the way, very much like in the linear case where I had mass times v squared divided by 2. So again, my final result, the purpose of why I'm doing this which is omega and work are very much related in the same way as we were basically deriving it before. So it doesn't matter actually what concrete force and concrete distance, what does matter is their product which is work. So again, you can have a weak parent who exhorts a weaker force but it will take him longer distance to travel to speed up the carousel to the same angular speed omega. So my final result and my work are related. So f and s separately are not as important in this case as their product which is important. So they can change in some way while retaining the same value for their product and that's okay. The result will be exactly the same. So that's why a concept of work is so important. And that's why I think it's perfectly justified to basically define a new physical entity which we call work and we can define it as product of force and the distance this force is acting. So we are just introducing new concept by definition basically. All I was talking about before was to kind of justify that this does make sense because it's from the work actually the result of the action actually depends more than on individual force. It depends on the work. Okay, now this is in case the force is acting along the trajectory. Now if it's not, if it's under angle you have to add the angle phi. So basically that's where many definitions in many textbooks end. Well the problem is it's good for a straight line movement. Now when we are talking about a curved trajectory well first of all the cosine obviously is changing all the time and so it's not exactly as... So what is correct thing in this particular case is to take an infinitesimal piece of the trajectory and whatever force acting at that time and multiply them and the cosine of the angle. And what's even better is to consider this piece of trajectory as a vector in which case we don't need the cosine because this is what is called scalar product of two vectors. So if you have a vector F and you have a vector differential basically infinitesimal piece of the distance of the trajectory at which point this force F is acting and you have their scalar product that would be a definition of the differential of the work. So on each infinitesimal piece of the trajectory we can define separately amount of work we are basically trying to perform and then if you want the entire trajectory well we have to do integral from 0 to some kind of S maximum whatever S maximum is. So again back to mass for jeans you really have to understand basic concepts of vectors and the calculus if you would like to address things in the most rigorous general form. I mean this is good too but obviously you understand this is for a very limited number of cases. Whenever your trajectory is not really a straight line you have to really do something else and the proper definition of the work is using the differentials. So the F is a function of each point at this point it goes this way and this point it goes this way it doesn't really matter it's some kind of a function and differential is another concept which you have to know this is a piece infinitesimal piece of trajectory as a vector and then their scalar product is differential of the work which you have to integrate if you want to know the entire work spent. Okay basically my purpose was to not just to give you this formula which is kind of obvious but to give you some justification and generalization in a more rigorous case. That's it for today I do suggest you to read the notes for this lecture on this Unisor.com website so the site is free obviously. Well that's it for today. Thank you very much and good luck.