 When we think about our dynamic universe, we're often drawn to the concept of energy. But what is energy? Perhaps we picture in our minds the roiling plasma of the sun, bathing us in the warmth that we enjoy on our third planet here in the solar system. Or maybe we think about the roar of a crowd and the crack of a bat as a baseball player who drives a ball hard into the outfield going for that home run. Or maybe we think about the sense of wonder and awe at the rumble of the rocket engines that blast fuel and metal off the planet aiming for orbit with the ambition of one day driving our species from this world to others. It is energy that connects all of these things and inspires in us a sense of wonder about the universe. But what is energy? Let's begin to explore that concept a little bit in this portion of the course. We're going to shift perspectives now in the course. We're not going to abandon the concepts that we have been exploring so far, concepts about space and time, forces, for instance Newton's laws. We are going to expand into a new frontier, introduce some new ideas, energy, and the related concept of work. And then we will tie those concepts back to the foundations that we have already laid for ourselves intellectually in this course. The key ideas that we're going to explore in this section of the class are as follows. We will recognize that motion has an associated quantity, a kind of energy which can be quantified even if it's conceptually difficult to define. We will come to understand that forces acting on objects in part changes in energy and that objects that change energy can exert forces on other things. We will learn to describe some kinds of energy and especially to relate the force and energy concepts. As I said, we are not walking away from the foundations that we have established already. Rather, we're going to look at the cosmos far larger in many ways than we've been looking at it before, but from the perspective of a new quantity in nature. And finally we will begin to describe not only variable forces and their effect on energy, but time-varying energy changes themselves. The first thing we have to grapple with a little bit is to try to define energy in the first place. If I say the word energy to you, you might think about an electric switch, a light switch being flipped on and a light bulb turning on, or a lightning strike, or maybe some kind of glowing ball of gas, some picture will form in your mind when I say the word energy. It's really easy to visualize energy in a kind of almost dramatic way, but often when we do this we are picking a very specific phenomenon to use to represent energy and it leads us down an incorrect path. It's extremely difficult to describe in an accurate and concise way what energy actually is. And oddly enough, I'm going to begin with a more advanced and more precise definition of energy, because ultimately if you keep going forward in the study of energy, matter, space, and time, if you continue to think more deeply about the universe from the perspective that physics offers, you are going to encounter a fundamental concept and that is the state of matter. So for instance, the state that an atom finds itself in while bound up in a lattice of other atoms that forms a material, or the state that an electron finds itself in, orbiting a parent atom with other electrons and a nucleus of neutrons and protons deep down inside the atom. State is an advanced concept and it's really often beyond the scope of an introductory course like this one. But nonetheless, I want you to have at least heard the term and understood the implications of state and energy together before you move forward in this course. So what do I mean by a state? What I mean is a physical description of an object that really summarizes all of its distinct measurable properties. And when you get down to the atomic and the subatomic realm, there are only a handful of numbers that at any moment, when written down, can fully define the state of a piece of matter. Energy therefore becomes a quantity that's associated not only with a state you can write down, but the changes in state, they are the things that are related to changes in what we call energy. So a change in the energy of a material object is really a change in its fundamental state description. So for instance, we have some notion already from exploring matter from the course so far that an object that's in motion probably has more energy than taking that same object and placing it at rest, say with respect to the surface of the earth or a table. If the object is moving, we immediately jump to the conclusion that it has more of something than that same object sitting at rest. In fact, changes in the state of motion, which is a concept that we have encountered before, it means changing the velocity of an object, that is then associated with a change in the energy of the body. If velocity labels the state of a material object, then changing the velocity changes the state and therefore the energy changes in response. Now let me back off to a more hand-waving definition of energy. Energy you could maybe more broadly and less deeply think of as a property associated with a material object that can change, but can neither be created without cost nor destroyed without consequence. Now I didn't really touch on that when I was talking about the state concept, but that still applies even to states. You can't change a state of a material object without paying a penalty somewhere else. You don't get something for nothing. And that is really one of the other deep realities that's tied to the energy concept. You can't create energy, you can't remove energy without consequence. So for instance, an accelerating object, it could be speeding up, its velocity could be increasing, it could be slowing down, something might be working against its direction of velocity and reducing its velocity. Accelerating an object cannot be done for free. Something else has to necessarily give up or store energy for this to happen. And we will see examples in this part of the course where one body gives up stored energy to another and say increases its speed, but also one body can give up stored energy to another and has other effects on it as well. So it's entirely possible to cause both kinds of things to occur by storing and releasing energy. Let's begin with a look at the energy associated with motion, which is also known more formally as kinetic energy. The reason we begin with this is tied to the notion I introduced a moment ago. We have some preconceived feeling that an object in motion has more of something than an object at rest. And so it's quite natural to begin to think about the energy concept from the perspective of moving objects and different states of motion, different degrees of motion, different speeds. But how did we ever figure out how to relate this energy concept, which is perhaps much more abstract than we are used to grappling with in the real world, to more concrete things like changes in space with respect to time, speeds, velocities? Well, to understand this, to build some familiarity with the ways in which an experimental scientist can conceive of an idea and then attempt to, even if one cannot directly measure the concept that goes with that idea, infer a measurement of the concept from other things one can see in the world, it's important, I think, to look a little bit at the history of physics, mathematics, and philosophy, which for some time were not distinct subjects. I think it's useful to reproduce an experiment that was itself reproduced, but then more importantly popularized in an extremely important way by the mathematician, physicist, and philosopher, Emily du Chatelet. Now, the fact that Chatelet in her time, she lived from 1706 to 1749, this is pre-revolutionary France that she is engaged in society in. It's already impressive for a woman living in her time in her society to have had any notoriety as a mathematician, physicist, or philosopher. Certainly back then, societally, such avenues of investigation, such intellectual pursuits were often closed off to people who were not wealthy, for instance, or of a certain race or gender or socioeconomic status. And so as a result of this, it's remarkable already that she is known at all for the things that she was able to understand about the natural world. Among her accomplishments, and perhaps the one she is most recorded for, is that she translated Isaac Newton's Principia, his fundamental work in mechanics and gravity, into French. But she didn't just translate it, she added her own commentary to the volume. And in her commentary, she added her own discovery. And that was that the energy associated with motion, moving objects, is not proportional to linear speed. That is, whatever that energy is, it doesn't go as V, which is something, for instance, that Isaac Newton had surmised. But instead, it went as the square of the speed, V squared. That was a remarkable insight into the natural world, one, that we are still essentially reeling from the revolution of. Now, what's cool about this is we can reproduce her discovery quite easily. It's hard to come up with ideas the first time. But once you've seen the elegance and the beauty of them, it's actually quite simple to go and try it yourself. And that is one of the most wonderful things about science, is it's not for one person to do. It's for everybody to do. All you have to do is have the idea and have the tools or see the idea and have the tools. And you, too, can engage in investigation of the natural world. In doing this, we will see how she resolved a question that alluded some of the greatest minds, like Newton, that came before her. What is the internal property of a body associated with motion? But more importantly, how does that property scale in size with the degree of motion of an object? So here is the essence of Dushatale's experiment. She dropped metal spheres into a floor made of soft clay. So what's going on here? You hold the ball above the soft clay floor, which I've illustrated over on the left with this gray box. And in the picture here, we have the moment where, for instance, Dushatale is suspending the object over the floor some height h1 above the top of the clay. So the ball is not moving right now, and the clay is unblemished. Now, we can get that ball moving by simply letting go of it. We can let gravity act on the ball. And we know that since gravity wants to pull things down toward the center of the Earth, and since the ball has weight due to gravity's influence, the ball will fall, and it will accelerate at g, 9.81 meters per second squared. So the fact that motion at the time that Dushatale would have been investigating it was associated with some change and some internal property of matter was a fairly well-established, at least philosophical, concept. And even mathematically, there was some sense of a concept that we're going to get to later in the course, and that is momentum, some property that does, in fact, scale linearly with the speed of the object. But Dushatale's experiment revealed the exact relationship between this internal property, which we now call energy, and the state of motion, that is the velocity of the object. So again, here's the idea. Drop a ball from a height h1 into a block of soft clay. The ball will accelerate until the bottom of the ball reaches the top of the clay. We can determine exactly, using Newton's laws and equations of motion, what the speed of the ball is just before it encounters the top surface of the clay. Now, once it encounters the clay, it will then dimple the clay. The weight of the ball will exert a force on the clay. The clay cannot resist this force perfectly, and the clay will compress. And the degree of compression of the clay is somehow a measure of the energy that the body had before it encountered the clay and now has deposited it into the clay by deforming it. So whatever was present in that body due to its motion just before encountering the clay is now lost to the clay and compressing it. And this experiment should reveal to us something about the character of that internal property related to the motion of the ball. We know the maximum speed that the ball will achieve just before striking the clay. You can work this out from a consideration of gravitational acceleration and equations of motion. And you will find that the velocity that that ball has just before it strikes the clay, the maximum velocity we'll ever have is the square root of 2 times the acceleration due to gravity times the height from which it was dropped above the clay. It will then penetrate into the clay some depth in this particular experiment, which I'm labeling D with a subscript 1. So drop from a height h1. It achieves a maximum speed, v1, and penetrates into the clay a depth, d1. I can now repeat the experiment, but increase the height of the ball above the clay. I can drop it from a much higher altitude above the surface of the clay. This is what Duchâtelet did. So now we have a new height h2. This achieves a new maximum speed, v2. And this penetrates into the clay some different depth, d2. So this picture now shows us the second experiment. We're going to gather data from this. And I'll show you what data we're going to get in a moment. But basically, we can analyze from motion considerations how the height affects the motion of the ball. Height determines maximum speed. The depth of penetration tells us something about that internal energy of the ball before it's depleted by compressing the clay. And we can try to see what the relationship is between heights and penetration depths. And from that, because heights determine speeds, figure out something about how that energy given up to the clay relates to the speed of the ball just before it strikes the clay. So let's run through the math of this. Let's take a look at Duchâtelet's experiment using the foundations that we have established already in this course. We know that the initial speed of the ball in both experiments is 0. It's dropped from a height h1 or h2, but it starts at rest. We know that the ball, in each case, is displaced by a height down. It starts a height h1, and it finishes at 0. So final minus initial is 0 minus h1. So the delta y in the first experiment is negative h1. The delta y in the second experiment is negative h2. And I'm measuring this from the underside of the ball, the bottom of the ball. Now we can use the equations of motion to find the speed of the metal ball at its last moment of freedom before making contact with the clay. And that is given simply by this equation of motion. The final speed of the ball after falling squared is equal to the initial speed of the ball before falling squared minus twice the acceleration due to gravity times the vertical displacement. And if we plug in the vertical displacement and the initial speed of 0, we find that v1 is the square root of 2gh1, as was indicated on the previous graphics. This is where it comes from. You can repeat this for the second experiment. The second speed squared just before it strikes the clay is the initial speed squared minus 2g times the second displacement. We rearrange this and solve. We find that v2 will be equal to the square root of 2gh2. And so we learn something fairly profound. If we can measure the ratio of the heights from which we drop the ball, which we can absolutely do very easily, then we learn something about the ratio of the velocities of the ball just before striking the clay. This is all determined by gravity, which Galileo studied in great detail, and Newton came to understand in even greater detail. All of this was well-established before De Châtelet's day. We are reproducing it in a modern sense here, but this is nothing that wouldn't have been known to Emily De Châtelet. And so in fact, we find that the relationship between the speeds of the balls just before they strike the clay and the heights is a square root. In other words, v2 over v1 is equal to the square root of h2 over h1. Or v2 squared over v1 squared is simply equal to h2 over h1. I can write this equation in any of a couple of ways. They're all equally valid. They say the same thing, but basically they tell us something about the relationship between height and speed just before striking the clay. Now to compress the clay, the ball must be exerting a force. That force is something that we can make an inference about that somehow the compression of the clay scales with the amount of internal stuff that the ball has as a result of its speed in each case. So let's run with this idea. If we think of the internal property of the body, which we now call energy, as being some numerical quantity, something we can eventually figure out a number for, we might label that quantity e1 in the first experiment and e2 in the second experiment. And in each case, this quantity is given up to the clay in compressing it. And in each case, that compression depth will be a measure of that internal energy. In the first experiment, the compression depth will be d1. In the second experiment, it will be d2. And we can make the argument that the ratio of e2 and e1 is proportional to the ratio of d2 and d1. We will use these penetration depths as a measure of this internal energy of the ball that has been given up to the clay in compressing it. And the question we want to answer is the following. In nature, in the real world, in the universe, what function of h2 over h1 will yield the observed ratio of d2 over d1? I'm not presupposing the relationship between these. I'm going to measure h2. I'm going to measure d2. I'm going to measure h1. I'm going to measure d1. I'm going to then take the ratio of h2 over h1, and I'm going to compare it to the ratio of d2 over d1. Now, Emily Duchât-Delet did a far more careful version of this experiment than I will do. We will get the sense of her results from my experiment. But we are interrogating the natural world itself to find out what the correct answer is. We are not guessing it. We are not presupposing it. And in fact, presupposing it didn't serve Duchât-Delet's predecessors very well because they got it wrong. We are going to measure it. And in measuring it, we are going to learn what the correct answer actually is in nature. And because the speeds are dictated by the heights thanks to equations of motion and gravity, whatever we find out is the relationship between d2 over d1 and h2 over h1, we can map that onto the relationship to v2 over v1. So let's do the experiment. Let's see how this works. Let's look at our results and let's find out what we have learned from nature itself. To try to reproduce in miniature, Duchât-Delet's much more extensive experiments dropping objects into a flat surface made out of soft clay. What I've done is I've simply gone to a craft store and I've bought some modeling clay, some cheap, off the shelf, clearance sale modeling clay. I use the blue stuff because it's SMU. I've patted the modeling clay into a little pancake and I flattened the pancake out so that it lays relatively flat and smooth on a flat surface, a table. Now, I have rusted a two meter stick on the top of the clay, clamped in place so it's not pushing down into the clay and similarly, I have a copper tube that I'll come back to in a moment, also clamped so that its bottom rests just on the top surface of the clay. Now, I tried dropping little steel balls but they don't have a lot of mass and I wasn't getting very good results. It was hard to measure the dimples they were leaving in the clay but it doesn't really matter what I drop. So instead of dropping little steel balls, I found these lovely metal cylinders. Here's one of them. It fits perfectly inside the copper tube I mentioned a moment ago and it can be dropped from a height and the height can be measured through slots in the copper tube using the two meter stick that I mentioned also a moment ago. I can take these metal cylinders, so one of them in particular, load it into the copper tube, let it fall down to a certain height and stop it, measure that height and then let it go so that it falls down and penetrates the clay. Then I take the metal cylinder, back up to the top, drop it again from a different height and then again, let it penetrate the clay. And in both of those cases, I just used a paper clip dipped gently into the dimple in the clay and then compared to a meter stick to figure out what the penetration depth was. So the two meter stick gives me the height from which the metal slug was dropped and then I carefully work the metal slug out of the clay. I take just a paper clip that's been straightened out, use it to estimate the depth of the clay, rest that against a meter stick and then measure the depth of penetration of the clay. Let's look at this experiment again, unfolded here in real time. So here is one of the metal slugs, I'll just turn it around to show you and what I'm gonna do is I'm gonna step up here and I'm gonna load the metal slug into the top of the copper tube which has the meter stick next to it. Now I'm gonna carefully lower it down so we can see where the bottom of the slug is compared to the meter stick, about 124 centimeters above the top of the clay, drop it, that happens fast and then go down and then we can lift the tube up and we can take a look at the penetration of the slug into the clay. So here is the slug, you can see it's sitting nicely in the clay and I'm just trying to gently work it out of the clay so I don't press it further into the clay and don't ruin the hole. There we go. And there is the dimple in the clay which we can see, let me take the clay out here very carefully. And now I'm just gonna rest the clay on the surface, I'm gonna get my trusty paperclip and I'm gonna use the paperclip as kind of a depth probe into the dimple. So I just gently put the paperclip down, rest the edge of my finger on the top of the clay and then press them together and now I can see where the bottom of my finger and the tip of the paperclip rest against a meter stick and we see that this experiment here for 124 centimeter drop is about an eight or nine millimeter penetration depth into the clay. In fact, when I did the experiment carefully, I found it was about eight millimeters. Now finally, I went ahead and crunched the numbers. I did a drop from a height of 66 and a half centimeters and found a penetration depth of four millimeters. And then the drop I just showed you was from about 124 centimeters and I had a penetration depth of eight millimeters. The ratio of the heights of the drops is about 1.9. I haven't put any uncertainties on these numbers so there's probably some error on that but I'm not quite sure what it is. And then I calculated the ratio of the depths of penetration and that came out to be about two. So the ratio of the heights is about two and the ratio of the penetration depths is about two. The ratio of the heights and the ratio of the penetration depths both seem to be about the same. When I increase the height by a factor of two, I increase the penetration depth also by a factor of two. What we'll learn from this experiment is that the penetration depth of the metal object into the clay is linearly proportional to the heights. In other words, if I double the height from which I drop an object, I double the penetration depth. So this tells us that H2 over H1 is equal to D2 over D1. But remember how the speeds went with the heights. The speed ratio goes as the square root of the heights. But since H2 over H1 is equal to D2 over D1, it tells me that D2 over D1, that penetration depth, the energy deposited into the clay to compress it, that also goes as the square of the speeds. So if I double the speed of an object, I don't double that internal property. I quadruple it. The dependence of energy of motion, kinetic energy, is quadratic with speed. Double the speed, increase the amount of kinetic energy by a factor of four. It's thanks to this kind of careful work by Emily Deschatelais and also others who not only were working before her, but inspired by her work continued this investigation, that led to the firm establishment that the kinetic energy of a body, which we will denote as a capital K, is given exactly by the following equation. The kinetic energy of an object moving at a speed V is given by one half times the mass of the object times its speed V squared. This is incredible. Because it tells us that this internal quantity that a body possesses due to its motion, for instance, is not a vector. It goes as the square of a vector. And as we know, the square of a vector is its length squared. So unlike velocity, which appears in kinetic energy, kinetic energy itself contains no directional information. It is a scalar quantity. It's a pure number. So it is unlike velocity, or force, or acceleration, or displacement. It's a number. And this is part of what makes energy concepts so great. Rather than pushing vectors around on a piece of paper, which can often feel clumsy, you're pushing simple numbers around on a piece of paper. Now, as you'll see, these numbers can be positive, they can be negative. So you have to be careful with the signs, but they contain no directional information. They merely tell you about whether or not the energy of a situation is increasing for an object or decreasing for an object. It doesn't tell you which way the object per se is moving, which way, for instance, it's accelerating. It contains no directional information, but it summarizes a tremendous amount of information and is an extremely powerful concept. Kinetic energy is one half MV squared. That's the first concept we need to walk away from in this part of the course. Now, this also tells us the units for energy. From the equation that we just wrote down, we can glimpse the standard international unit that represents energy. It is mass times velocity squared, or in other words, energy is equal to a kilogram meter squared per second squared. Now, this is a relatively awkward quantity. It doesn't exactly roll off the tongue like poetry. So it has its own name, its own unit associated with it, which because it includes the kilogram meter in second is itself an SI unit. And so we assign it the traditional name that's used for energy, and that is the joule. And by definition, one joule is one kilogram times one meter squared divided by one second squared. One kilogram meter squared per second squared is one joule. So whenever you see kilogram times meter squared divided by second squared, you can replace it with the letter J, capital J. Or whenever you see capital J and you wanna work out units, maybe by figuring out if units cancel out, expand it into its kilogram meter squared per second squared glory, and use that in its place to do some algebra with units. But the joule is the standard international unit of energy. Now let me make a final comment on Emily de Châtelet. Because it offers us a glimpse into a far greater future in terms of human knowledge about the foundations of the natural world. Now de Châtelet herself and the groundbreaking work that she did, as is always the case, was built on the foundations of others. Nothing that I do as a scientist is always and ever entirely my own. How could I make progress if the people before me had not made fundamental discoveries in the universe? I'd have to reinvent those discoveries just to get started. A good scientist builds on the discoveries of others, but in doing so comes up with ideas and makes discoveries of their own. And certainly she was inspired directly by people like Isaac Newton, Leonard Euler, who may be known to you from the mathematical world as a very famous mathematician, and Gottfried Leibniz, whom we've encountered in this course before as the co-inventor of calculus. These are of course just a few of the people whose works that she had clearly read and clearly pointed to is inspiring her own thinking. But it was her own work that would become the foundations for a new generation of giants. In formulating motion in terms of energy and finding out the relationship between the degree of motion and the quantity of energy, de Châtelet inspired the later work of scientists like Joseph Louis Lagrange and Leonard Euler himself, who inspired her, who went on to reformulate Newton's mechanics not in terms of the force concept that was laid down by Newton and that was built upon the work of Galileo and others, but rather in terms of the energy concept. And in fact, any more advanced mechanics course, say 2,000, 3,000, 4,000 level or above at a place like SMU, that mechanics course in a department of physics, including our own, starts from the energy concept or gets to it fairly quickly and then basically shows you that it's possible to absolutely re-derive everything we have committed to memory so far, equations of motion and so forth from the first principles of energy and energy conservation. And in fact, because the energy concept fundamentally boils down to how numbers, scalars are changing, recasting mechanics in terms of energy and changes in energy, affords you the opportunity to handle far more complicated situations than the kind that we're encountering in introductory physics relatively easily once you get the hang of the tools. So let me close this part of the lecture simply by noting that I've cherry-picked one person from the history of science, Emily de Châtelet. I've shown you the experiments that she conducted more carefully and more methodically than I could do it for this lecture video that revealed the V squared, the speed squared character of kinetic energy and led eventually to the formulation of kinetic energy as one half times the mass of the object times its speed squared. And this quantifying energy, quantifying this internal property of a body that itself cannot per se directly be measured led to a complete revolution in the formulation of mechanics, a revolution we still live on the foundations of today. Modern computational techniques for doing mechanical calculations are built not per se on Newton's way of formulating mechanics, but rather on the Lagrange approach, the Euler approach, or what's also known as the Hamilton approach to mechanics which all involve energy concepts at their foundations. ["Energy Concepts at Their Foundations"] Now that we've explored kinetic energy, the energy associated with motion of a body, let's go a little bit deeper and look more generally at energy associated with a force acting over a distance in space. Forces acting over distances have already played a role in the consideration of Duchâtelet's experiments with the falling ball and the clay that can be compressed by the impact of the ball. We've already encountered the second half of the series of concepts that are to be introduced in this part of the course, and that is the concept of work and specifically work done by a force. So for instance, when the falling ball, which has achieved some speed V just before it strikes the clay, then encounters the clay, at that moment it possessed a degree of kinetic energy given by one half MV squared. And then just a moment later, it has now made contact with the clay, it begins to compress it, and then eventually it becomes motionless. It has given up all of its kinetic energy to compressing the clay. There is no more kinetic energy and the ball stops. During that time in between, when it's compressing the clay, it is exerting a force on the clay, and the exertion of that force depletes, if you want to put it this way, its reserve of kinetic energy. That allows it to do work on the material of the clay, but at the cost of its own motion. So work is just another kind of energy, and it's specifically the energy that's associated with the action of a force over a distance. Now when a body loses energy, we can say that it does positive work. When a body gains energy, we can say one of two things, that it does negative work, there's nothing wrong with that, energy can be positive or negative, it's changes in energy that matter, and we'll reinforce that concept with other energy related concepts in the next section of the course. Another way of putting this is that when a body gains energy, we can say that it had work done on it by another force. So often when we are talking about work and forces and kinetic energy, there may be one force acting, there may be multiple forces acting. The work associated with one force could be positive, in which case it means that that force is doing work, it's giving up energy, perhaps from another form to do that. But if something gains energy, it has negative work done on it, then some other force has acted upon it, depleting the other force of some reserve of energy, giving energy to the body that has negative work done on it. These signs will reinforce through problem solving in the course, but this is just to get you, at least familiar with the possibility that energy can be positive or negative, depending on the circumstances in which the changes of energy are occurring, and the sign has meaning, it has physical meaning. Let's take a look at a formal definition of work, a formal mathematical definition. Work is just another kind of energy. Kinetic energy was converted into work done on the clay by the ball when the ball encountered the clay in Duchâtelet's experiments. Therefore, because it is another kind of energy, it has the exact same units, jewels. Now, if work can cause a change in kinetic energy, then we can ask specifically what quantity of energy causes the following to happen. I begin with some object with a mass m, with some initial speed v subscript zero, or v zero. The kinetic energy of that initial state is one half mv naught squared. Later on, I observe the object again, and I notice that its speed has changed. It has now some final speed that's perhaps different from its initial speed. That means it has a new kinetic energy associated with it, one half mv final squared. The mass hasn't changed in this case, but the speed has, so the kinetic energy has changed. What is it that has caused the energy between the initial condition, v zero, and the final condition, vf, to occur? What caused that change? Well, we can glean an answer mathematically to that question from one of our equations of motion. If we treat this whole question as all occurring in one dimension, the object is only moving along one coordinate axis. We could choose that to be x or y or z. It doesn't matter, but all the motion, all the acceleration, all the excitement, happens along only one dimension. Then we can just start from one of our one-dimensional equations of motion. And the one I'm going to start from is the circumstance where there's some kind of acceleration that's causing a change of speed. I mean, after all, if speed is changing, there must necessarily have been an acceleration. And we know from Newton's laws of motion if there was an acceleration, there was a force. So let's explore that a little bit. We have an equation of motion that relates the final speed squared to the initial speed squared plus the activity of the acceleration over the displacement. So between our initial measurement of the speed and our final measurement of the speed, the x-coordinate could have changed, time might have marched onward, some things changed. And in this particular equation, it's space that we can think about, displacements. So we have our final speed, our initial speed, our acceleration a, and our displacement. If I multiply this entire equation by one half times m, I'm dangerously close to recovering kinetic energy in an equation. In fact, I have exactly the final kinetic energy on the left-hand side. I have the initial kinetic energy on the right-hand side, but I also have this other term, one half m times the quantity two a times the displacement, delta x. Well, I notice that one half times two is one. So that's nice. They kind of negate each other, and I just get a one multiplying this term. And then I notice I have m times a. Let's look at that. So I have kinetic energy in the final state, kinetic energy of the initial state, and I have mass times acceleration times displacement. Well, what is mass times acceleration? We appeal to Newton's second law, its force. So whatever it was that caused the kinetic energy to change from the initial state to the final state, it's associated with the action of some force, m times a over some displacement, delta x. So I'm recouping that equation here, using what I just said to substitute for ma with f. And finally we see what it is in this thought experiment that must be doing the acting to cause the kinetic energy to change. And that we label as work. It is another form of energy. We see that f times delta x, which we define as work, has the same units as kinetic energy, joules. This is all in one dimension, so this is the work of a force acting along the x direction over a displacement of delta x on that dimension. Well, we can take that thought experiment just a little bit further. So let's imagine a situation now, not where we make one observation at a later time of the speed of an object of mass m, but two observations. So we have our initial observation, the initial speed, a follow on observation where we observe that the object speed is now v one, and a final observation where we observe that the speed is now v two. Let us simplify this thought experiment by saying that whatever these intervals are, and let's do them in space, that they are equal in size. In other words, let's say that I start my whole experiment at zero meters along the x direction, at which time I observe, for instance, that the initial velocity is zero or some other number. And then one meter later, I observe the object, and I notice that it has a different speed, v one. And then one meter after that, at two meters in the x coordinate system, I make my last observation, and I notice that it has a speed v two, which is different from v one and different from v zero. And maybe I notice, for instance, that the speeds keep increasing. I notice that v one is bigger than v zero, and v two is bigger than v one. I can conclude, based on everything I just thought about a moment ago with the equation of motion, kinetic energy, forces, work, f equals ma, that there must be an external force that's causing this acceleration. Something is imparting kinetic energy through its action on this object. How can I find out the work that's being done by the force on the object? Well, we could break the problem into pieces. We could first compute the work that's being done in that very first interval, maybe that first meter of travel. So I just go ahead and write it down. The work done in that interval must be the difference between the final and the initial kinetic energies. So that's one half m v sub one squared minus one half m v sub zero squared. And this must necessarily be equal to whatever that force is that's causing this change in kinetic energy acting over the displacement, which for now, let's just say is one meter. I'm gonna leave it as delta x on the slides, but if you like concrete numbers, again, we're making our observations from zero to one meter, and from one meter to two meters, our delta x's are one meter each time. Great, that gives me the work done by the force in the first interval. Now I look at the second interval where we go from speed v one to speed v two. I again compute the work now by the difference in the final and initial kinetic energies. That's this equation here. And I find out that this, of course, must be related to whatever force is causing this change in kinetic energy acting over this distance. How do I get the total work done by the force? Work is just a number, just like all the other kinds of energy. Kinetic energy is just a number. Work is just a number. So I can just add them together. There's nothing that prevents me from doing that. I could have just as equally have said that I take the difference between one half mv two squared and one half mv zero squared to get the total work, but it's equivalent to having added the works together from measurement number one and measurement number two, w one and w two. And so I find that the total work done by this force over this total displacement of two meters of delta x plus delta x is just the sum of the work in the two intervals. So the total work is w one plus w two, which must be equal to f one delta x plus f two delta x. And again, I've kept my intervals the same. So these really are the same delta x's. I can write this a bit more compactly in mathematical summation notation. It's a short sum. It's the sum from i equals one to two of f with a subscript i, which labels each of the segments of the measurement times delta x. And finally, this must be equal to the total change in kinetic energy that the object experiences delta k. Another thing that we have observed here is that there is a relationship between the work being done on an object or the work done by an object and it's change in kinetic energy. And this relationship is known as the work kinetic energy theorem. And it's something that we'll return to later to use in our future investigations of work done by various forces. What if I wanted to make more than two measurements over some total displacement of this object? So imagine that I revisit my experiment from a moment ago. Let's imagine that I have an object that goes from zero meters to 10 meters. And I make two measurements just like I did a moment ago. One at five meters and one at 10 meters. I measure the speed at five meters. I measure the speed at 10 meters. I calculate the kinetic energy. I calculate the changes in kinetic energy. I get the work in each of those intervals. I add them up, I get the total work. And this is represented graphically over here. Here I find the speed at five meters and I use that to compute the force that must have been acting over the distance of five meters in order to get the change in speed from zero to five meters. That's represented, the force is represented by the height of this blue rectangle. And the width of the blue rectangle represents the displacement over which I am measuring the distance. The initial measurement is on the left side of the displacement. The first measurement after the initial measurement is at the right side. And then again, I do another measurement at 10 meters and I use this to compute the force. And here's the force. It's the height of this different blue rectangle. Same interval, five meters in this exercise. And I can see that indeed it looks like the force per unit distance is increasing from here to here. From the first unit of measurement to the second unit of measurement. This is a very discrete measurement. What if I wanted to increase the precision with which I am making these measurements? So for instance, what's imagine I up my game and I then do five samples. So instead of measuring at zero and five and 10 meters, I measure at zero and two and four and six and eight and 10 meters. And for each of these five displacements, I calculate the force that must have been acting over that displacement in order to give me the change in kinetic energy I observe. One of the things I'll call your attention to is the height of these rectangles is the force, the width is the displacement. The product of the height and the width is the area of each of these blue rectangles. The area is the work. Work is force times displacement. Force is the height, displacement is the width. So height times width for this particular exercise is exactly equal to the work. So if I wanna know the work done in the first displacement, I need to know the area of the blue rectangle. If I wanna know the work done in the second displacement, I need to know the area of the second blue rectangle and so forth for the third, fourth and fifth blue rectangle. Now I haven't said anything about this yet. I'll come back to this red line in a few moments. But what we are seeing here is now where the trend before was very hard to spot. It was definitely true that the amount of work done in the first interval was smaller than the amount of work done in the second interval on the previous slide. On this one we can see a clear trend of increasing amounts of work done by this force, changing the kinetic energy of this object more and more and more. So this is what it looks like to do five samples in this exercise instead of just two. Note also that what I mean by delta x has changed. Delta x used to be five meters when I did two samples, but for five samples it's two meters. I'm still only looking at this object in between zero meters and 10 meters. I've just increased the frequency with which I'm making my measurements. So I'm in principle getting a more precise picture of exactly how the kinetic energy and thus the work is changing as a function of its displacement. Let's double it again. Let's do 10 samples now. This is what it looks like to sample this process over 10 samples. Now each of them is one meter wide. I do 10 of those here. One, two, three, four, five, six, seven, eight, nine and 10. Again, the work in each of these is the area of the corresponding blue rectangle. So to get the total work, I just keep adding up the areas like I did for my two sample experience. Let me call your attention to something else. If I go back to the two sample demonstration, notice that the area of these two rectangles added together, which gives me the total work, gives me a number, 187.5 joules. Now every time I increase the sampling, I'm still looking at the same object moving across the same 10 meter distance. I'm just repeating this experiment with more precision. I'm sampling more finely. And so when we go from two samples to five samples, we see that the work has changed. And that's because we're sampling more finely. And so there's no guarantee that when we do this finer sampling that we'll get the exact same number we got with two. With two samples, we were doing a very coarse sampling of the physics of what's going on here, the change in the speed and thus the kinetic energy of the object. With five samples, we're getting a more precise sampling at moment by moment what's going on. So we get a different number. But watch what happens as we go from two 187.5 joules to five 170 joules to 10. We double the samples again. Now we've only moved a little. We've gone from 170 to 167.5 joules with 10 samples of the velocity of this object in one meter intervals. Let's do 10 more. 10 times more intervals. So now let's do 100 samples. And now we've gone from 167.5 to 166.675 joules. Notice also that as I add more samples, indeed my precision is growing. I'm able to quote more decimal places for what I think the total work done by this force over this displacement actually is. Notice something also that's been happening quietly as I've added more samples. Here it was very hard to say with any certainty that the blue measurements had anything to do directly with the red. For all I know, the red line doesn't have anything to do with the blue histograms, the blue boxes that are on here. But as I sample more finely, we notice the blue boxes really do trace out the red line more and more clearly as I sample more and more finely. Until we get to 100 samples, you can barely see the daylight in these samples. There are places where white peeks through. The gaps between the red and the blue. But for all intents and purposes, by the time we get to 100 samples, this is looking like a pretty good representation of whatever this red line is. And finally, if I go to 1,000 samples, I take my computed work from 166.675 to 166.6675 joules. And we notice that we're trending toward some kind of number here. We started off at 187.5, and then we moved about 17 joules to 170. Then we moved about two and a half joules to 167.5. And then we moved a fraction of a joule down just slightly less than one joule to 166.675. And then we moved hundreds or thousands of a joule to this number where we have 1,000 samples and we're adding up the area of each of the blue rectangles to get the total work. What's going on here? What's going on is what is known as integral calculus. We approximate a function shown in red, which could be a very complex function by sampling the function coarsely over wide bins. And then we imagine increasing the sampling of the function. Why are we doing this? Well, the work is the area of these blue boxes. And correspondingly, the true work being done by the true applied force is the area under that red curve. I want the area under that red curve. But to get it, I need to have a sort of attack vector. I need to have an approach to get the area under the curve. How could I do that? I've started by slicing the curve into two coarse regions and just calculated the average height of the red curve in those regions. And then I've taken the areas of the rectangles and added them. And I got a number, 187.5. And that turns out not to be a very accurate number. As we see by adding more samples, if I now do this five times, I get a number that moves quite a bit from the original number. And after you do 10 times, I get a number that only moves slightly from the previous number. If I increase my sampling by a factor of 10, rather than a factor of two, then I see that I've greatly improved the accuracy of my number, or at least the precision of my number. Accuracy would have to be determined by comparing it to the known true value of the area under the curve. If I go to 1,000, I get to this number, 166.66675. It turns out that the exact area under the red curve is this number, 166.6666666. It just keeps going. It just repeats itself out to infinity. Look how close these numbers are. I did 1,000 samples of this, calculating the average height of the red curve in each of those little bins, and then taking the area, the rectangles area, and adding up all the areas of all the rectangles. And I went from 187, basically, to 166.66675 by doing 1,000 samples instead of two. And now these numbers match each other up to the 1,000th place, and then they diverge at the 1,000th place. I could multiply the number of samples again by 10, or 100, or 1,000, and I could continue to improve not only my precision, but it turns out my accuracy moving toward the true value. This is what integral calculus does. Now I did this numerically in a computer. Integral calculus defines a procedure to do this so-called analytically by actually using functions directly to get the answer that you need, rather than approximating them in this methodology. Work, which, when you're doing discrete measurements, is just the product of the force times the displacement. For the case where a force, for instance, could have a complicated dependence on position, it could be changing continuously throughout the displacement, that you need to use integral calculus to solve for. And in general, work is the so-called integral of force over distance. Let's take a look at an example of that. The methodology of integral calculus is just like the methodology of derivative calculus or differential calculus. In differential calculus, what you're trying to do is to understand the behavior of a function at an exact moment, a location in space and time along the function. In the case of integral calculus, you're trying to do something slightly different. You're trying to figure out what the area is under a curve. So instead of trying to find out information about only one part of the function, you're trying to figure out gross information about the entirety of the function. That's, in a sense, the distinction between differential calculus, the calculus that tells you about what's going on at one exact moment, an integral calculus, the calculus that tells you an aggregate about what's going on in a function. The methodology of determining the area under a curve by summing up tiny slices, like I just showed you, is, in fact, exactly the basis of integral calculus. So in the limit that the number of slices tends to infinity, so the number of times I slice up the function goes to infinity, and the thickness of each slice, delta x, tends to zero. So I make the slices thinner and thinner and thinner, even while I multiply their number. It's in the limit of those things occurring, the slice size going to zero, and the number of slices going to infinity, that you then sum up the product of the force and the displacement in each of those intervals. And this becomes what is known as the integral. A sum is a discrete addition of little units. The integral represents a continuous sum of infinitesimal pieces. So the dx represents the infinitesimal displacements that we're now considering. They take the place of delta x, because they represent teeny tiny infinitesimal little things. dx is not the product of two numbers, d and x. Rather, it represents an infinitesimal chunk of the x-axis. dx is a single thing. This represents the sum from some minimum location on the x-axis to some maximum location on the x-axis of a function of x over those little intervals. And in integral calculus, what you learn is that this thing is equal to a relatively simple object. It's the difference between a number, which I'll just call f with a tilde over it for now. I'll explain what that is in a moment, but it's just a number f tilde evaluated at x max minus f tilde evaluated at x min. I'll come back to that in a moment, but let's define what this f tilde thing is. f tilde is known as the antiderivative of f. It's the function for which the following statement is true. The derivative of f tilde, whatever it is, with respect to x, its dependent variable, yields f, the thing that appeared inside the integral, or so-called integrand. This thing here, f of x, is the integrand. f tilde is the antiderivative. If you can find the antiderivative of your function, f, you can solve your integral very fast, and that is what makes integral calculus so amazing. It takes something which would otherwise require a ton of labor, if not a computer, to do, which is slicing up a curve into tiny pieces and adding up all the little areas of all the little slices, and it converts it into an exercise in simply knowing how to use derivatives to then invert and get the antiderivative. I'll give you an example of this in a moment. Again, this thing on the right-hand side over here, f tilde with the vertical line x max and x min, this is a symbol that represents the following operation. You evaluate f tilde at x max, you evaluate f tilde at x min, and you simply take the difference between those two things. That thing right there is the area under your curve. In our case, it's the work. The work done by the function, which could be a complex function, changing with displacement over the full distance that the function acts, which in the example I showed you a moment ago was 10 meters. Using the integral therefore lets us handle forces that have extremely complicated behaviors. They continuously vary as a function of x, for example. If we know the force is constant over large swaths of x, then you can just use the discrete sum. That's easier. But if the force varies continuously over x, if f has a different value at every value of x that you can plug into it, then trust me, the integral is a much easier approach to solving the problem, and in fact, it's the necessary approach. So if you're dealing with constant forces over large parts of your displacement space, just use a discrete sum. Sum a couple of things together to get the work. However, if your function of force changes the value of force at every value of x, you have no choice but to use integral calculus to solve that problem. So maybe that will help you in thinking about what's an appropriate way to set up and solve these problems. Here's an example of a continuously varying force. So imagine we have a force acting on an object. It's a function of distance f of x, so here's the force as a function of x, and here's the exact functional form. Mass times a constant b times x squared. So it's a quadratic dependence. If I double x, the force goes up by a factor of four. Here, m is the mass. I'm gonna set that to one kilogram. Nice convenient number. b is a constant that makes the units work out. So it multiplies x squared to yield an acceleration. After all, f should equal ma, so it must be true that b times x squared equals an acceleration. And here that means that b has to have units of per meter per second squared. That's the only way that this is gonna work out. So I've chosen to set b to be a number, 0.5 per meter per second squared. So I picked some numbers to plug in here, but that's not the point. The point is to do integral calculus to solve for the work independent of what m or b are. We wanna be able to find the work done by this force on the object of mass m. And as I said, we wanna go from some initial position, x min, to some final position, x max in space. I've chosen x min to be zero meters and x max to be 10 meters. Very similar to the exercise I showed you a few minutes ago, slicing and dicing a function from zero to 10 meters to figure out what the work, what the area under the curve is in that case. So let's begin by writing down the formal calculus-based definition of work. Work is the integral from the minimum value of x, zero, to the maximum value of x, 10, of the function of x, which is the force in our case, dx. Now I can plug in for the function. This must be equal to the integral from zero to 10 of mbx squared dx. So I'm about two steps away from completing the problem at this point. So here's our integral again, just to remind ourselves. To solve this integral, we need to answer the following question. What is the function f tilde of x such that the first derivative of that function gives us mbx squared? Well, if you stare at this for a little bit and you think about, okay, I had a function originally, f tilde. When I took the derivative of it, I got mbx squared. Let's see, the derivative of x to the n always gives me n times x to the n minus one. So if I reverse that thought process, it must have been that since the function I wind up with is just mbx squared, the original function must have been something like x cubed because after all, the exponent will decrease by one unit when you take the derivative of f tilde. So the original function must have been x cubed, but I can't just get x squared by taking the derivative of x cubed. I should get three x squared from the derivative of x cubed. So I gotta get rid of that three and the only way to do that is to divide by one third. And so if you play around with this, if you reverse the derivative, if you think backward through the process, you'll find the following function absolutely satisfies these requirements. F tilde is equal to one third mbx cubed plus a constant and that constant is known as the constant of integration. It's a number, we don't know what it is, but it absolutely doesn't depend on x so that when we take the derivative of this, the derivative of c with respect to x is zero. So go ahead, take the derivative of this thing and show that you get back mbx squared. Then you know that you have successfully found f tilde, the antiderivative of f. So we have our antiderivative. We know that this integral will be equal to something involving the antiderivative. So let's go ahead and get going. So now we use our antiderivative to evaluate the integral. The work is the integral from zero to 10 of mbx squared dx and that's gonna be equal to the antiderivative evaluated at 10 minus the antiderivative evaluated at zero, the limits of our integral. So plugging in our function for the antiderivative of f tilde, we find out that at x equals 10, we have one third mb10 cubed plus c and we subtract from that one third mb0 cubed plus c. Well notice that we have plus c here and then a minus c here so these constants of integration cancel out so they didn't matter anyway. We could have dropped them but to be thorough I put in the constant of integration because sometimes constants of integration actually play a useful role in other aspects of physics. We're left with the answer. If I plug numbers into this, m equals one kilogram, b equals 0.5 per meter per second squared, I find out that the work that is done by this force over the displacement of 10 meters is 167.6666 dot dot dot dot dot joules. If you flip back in the video a little bit, you'll see that this is exactly the work I told you that we were trying to approximate by slicing and dicing that red curve into little rectangles and adding up the areas of those rectangles and that procedure was getting closer and closer and closer to this number. Integral calculus yields exactly the answer to all decimal places. We see that the red line was in fact this quadratic force that I was describing and have written the function for here. So this nice little exercise illustrates how continuously varying force like mbx squared whose value is different at every value of x is best handled using integral calculus. The only mental leap that you have to make here is to work backward. Instead of taking the derivative of a function and getting a function from it, you are given the resulting function and you have to find the original one whose derivative yields the function you already have. So it is a little tricky because you have to think in reverse in calculus in derivative calculus, differential calculus, but it's good exercise and going backward and forward through the derivative. It will give you good practice at getting good at this kind of mathematics. So we've seen what happens if we have a complex force that changes with every point in the displacement. We need an integral calculus to solve that. We've also seen what happens if we have a constant force over a large displacement. That's much easier to handle. We can just take the force f and multiply by the large displacement delta x. But what if we have forces acting in two where three dimensions and displacements in two where three dimensions, now things seem pretty complicated. And in fact, a more general version of the work definition can be employed. And this is true both in the integral form of work as a definition and in the discrete form of work as a definition. This vector product works just fine in both cases. In general, in more than one dimension, the work is the dot product of the vector f with the vector delta r. Delta r is the displacement in two or more dimensions and f is the force in two or more dimensions. You'll have to go back and look at vectors part two if you want to review the dot product but basically you take the product of each of the components of f and delta r. So you take the x components, multiply them, the y components, multiply them, the z components, multiply them and sum those products together. That's the dot product. It's not really algebraically complicated. With rehearsal and practice, you'll get good at doing that. So the good news is, is once we embrace this general definition of work, we have all kinds of options available to us. If we have forces acting in two or three dimensions and displacements, therefore, in two or three dimensions. We can, for instance, have a constant force with x and y and z components acting over a large displacement delta r vector in which case the dot product just becomes the sum of the x, y and z components multiplied together, f i, delta r i, where i is labeled from one to three. One is the x component, two is the y component, three is the z component. On the other hand, there's a more compact definition of the dot product. If you know the angle between your force vector and your displacement vector, theta, then you can very quickly calculate this sum that I showed you here by just taking the magnitude of f times the magnitude of delta r times the cosine of the angle between f and delta r. That can be often much easier if you already know the angle between them. Of course, if you have a force that isn't constant but rather varies as a function of the displacement, you must use integral calculus. And then it just becomes a sum of three integrals, the integral from x min to x max of f x dx plus the integral from y min to y max of f y dy plus, surprise, surprise, the integral from z min to z max of f c dz. It looks just like this sum, but now it's the sum of three integrals of these products. And again, for each of these integrals, you would just do the exercise in integral calculus. What's the antiderivative of f x? What's the antiderivative of f y? What's the antiderivative of f c? Plug in the limits, take the difference, you've got your answer for each component. You'll get a single number out at the end and that is the work. Let's take a look at the work done by some specific forces. One, we've encountered many times before, that's gravity. So this will be a friendly way to visit this question about what does work look like done by a specifically defined force in nature. But we'll also use this as an opportunity to introduce a new force we haven't looked at before. And that is the spring force, the force associated with the compression or stretching of an object that wants to restore itself to its original size. So let's think about gravity first. And specifically, let's think about work that's done by or against the gravitational force. So consider the picture shown here on the right. We have a block illustrated in gray. It is free to slide up or down a frictionless inclined plane. So there are surfaces in contact in this picture, but there is no friction between them. So there will be no friction force in this discussion. Therefore, the only force that can act on this block is gravity. There is nothing to oppose it along the surface. Of course, we have the normal force, but again, that's a response to the gravitational force. So the normal force is the surface keeping the block from passing through it, essentially. If we're thinking about sliding, if we're thinking about moving up the incline or down the incline with some displacement, say delta R, we only need to concern ourselves with the component of gravity that the weight that lies along the surface of the inclined plane. So keeping in mind the fact that gravity is gonna be pulling straight down to the center of the earth, that's the weight force. Here, however, is the normal line, the perpendicular line to the inclined plane. The plane is inclined in an angle theta. So using similar triangles, you'll find that this angle here between the perpendicular line to the surface and the weight is also theta. We can quickly write down the mathematical components of the weight, the one that lies along the inclined plane's surface is the one we care about. But what is the work done by gravity on the block? All right, well, there's no friction, so there's nothing to prevent the block from sliding. So the block will begin to slide down the inclined plane. It's accelerating. Therefore, there must be a change in kinetic energy. The speed is changing, so kinetic energy must be changing. Something is causing that change to occur and that something is weight, gravity in this case. We know from considering changes in kinetic energy where the final and the initial kinetic energy are not the same, that means that there's a nonzero work being done. And if there's a nonzero work, that means that there's some force acting over some displacement. And I've drawn here the most general definition of f vector dotted into delta r vector. The force that's doing all the work in this case is weight. And specifically, it's the component of weight that lies along the incline, which is given by w sine theta, or mass times the acceleration due to gravity times sine of theta, where theta is again defined over in this picture. So if we treat the surface of the incline as the x-axis of our problem and just ignore everything else for now, this really just becomes a one-dimensional problem, which is nice. The change in kinetic energy along the x-direction is given by kf minus ki, 1 half mv final squared minus 1 half mv initial squared. And that's going to be equal to mg sine theta delta x. This is the force of gravity, the weight along the surface of the plane. And this is the displacement delta x along the plane in the x-direction. Now, since the block speeds up, the final kinetic energy is greater than the initial kinetic energy. So this difference is positive. It's a number greater than zero. And so it must be that this number is also greater than zero. And we can say that the work done on the, work must be done on the block because its kinetic energy is increasing. And we can also say that work, therefore must be being done by gravity on the block in order to make this happen. So let's revisit this question one more time, but let's consider a slightly different situation. It's again one involving gravity and weight, but a different thought experiment. So let's say you're holding in your hand a solid ball, and that's depicted over here in the graphic on the right. Your hand and the ball are initially at rest. And they start down here at the lower part of the picture. You then raise the ball a height into the air. So you displace it vertically up like this. And at the end of this exercise, you stop your hand and the ball stops with it. And so the hand and the ball again are at rest. Thus, if we were to write down the difference in the final and initial kinetic energies, well, everything started at rest and everything ended at rest. So the final kinetic energy is zero and the initial kinetic energy is zero. And so it must be that the change in kinetic energy is also zero. But that also means that the work done by any forces in this problem must also sum to zero. Well, there are two forces acting in this problem. There's your hand pushing up against the ball and lifting it up higher into the air. But at all times, of course, there's also gravity, the weight of the ball, the weight of your hand that tends to want to pull things down to the center of the earth. So we know that there's work done by the hand and we know that there's work done by gravity. And we know that whatever they are, their sum must be zero. Because after all, we start at rest and we end at rest. And so what we learn by doing this simple little thought experiment is that whatever the work is done by the hand, it must be the negative of the work done by gravity. That's the only way this situation could be true as we've described it. Whatever the work was that was done by your hand is the negative of the work done by gravity. So if you find out that the work done by gravity is negative, then the negative of that will give you a positive number and it must mean that the work done by your hand was positive. So if this relationship is true, that in this example, w hand equals the negative of w gravity, we can see again that the work done by the hand against gravity is the negative of the work done by gravity. And the work done by gravity, we can find. It's the force of gravity times the displacement. And since the force here is the entire weight of, say, your hand plus the ball, that must be negative mg. Points down in the negative y direction. But the displacement was positive. You went from zero to a height h above where you started. H is a positive number, representing the height above the starting location. So mass is positive. G is a positive number. The minus sign that represents the fact that gravity accelerates down is out in front here. And h is positive. So this whole number here is a negative number. And so it must be that your hand did positive work in order to make this happen. In other words, your hand lost energy in this process while gravity gained energy in this process. And we'll revisit where that energy is going what that energy is later on. But it's enough to know that based on this thought experiment this statement must be true. So since it's true that w hand equals negative w gravity we've learned that w hand must be equal to positive mg h. So the hand does positive work. In other words, the breakfast that you had this morning was made it possible for your muscles to contract in such a way that it could lift the ball against what gravity wanted to do which is drag it down toward the center of the earth. You made it go higher up into the air. That cost you. Gravity gains something in the process. Gravity has done negative work and that means work has been done against it. Energy has been stored somehow in gravity. Now where that's been stored we'll learn more about that later but it's enough to know that the bank accounts are balanced. You had to give something up but gravity gains something in the process. We can reverse the situation. We can start with the ball now at the height h and we can drop it back to a height of zero. So in this case the situation is exactly the same. The initial and final kinetic energies are still zero. So it's still true that w hand is equal to negative w gravity but what's w gravity this time? Well, this time the displacement was negative h. The weight still points down so we have the product of two minus signs negative mg times negative h. The work done by gravity now is positive. Gravity gave something up to make this situation happen. On the other hand, because the work done by the hand is the negative of the work done by gravity we see that the work done by the hand is a negative number. Work has been done on the hand. The hand has gained energy in the process. Gravity has now done positive work while the hand has done negative work. In other words, work is done on the hand by gravity. In principle, you've just stored energy in the tendons in your arm gained by sapping it from gravity somehow. And we'll revisit that concept a little bit later in the course. So let's consider some of the lessons that we've learned by thinking about the work done by or against gravity. We've learned to assess a problem involving the gravitational force but now from the perspective of work and kinetic energy. And I hinted this earlier but as I said this relationship between the work done by a force and the change in kinetic energy that we've explored so far, the so-called work kinetic energy theorem can be applied here. And if there's a change in speed associated with gravity acting on a body like an object sliding down an inclined plane and accelerating on the process, then it must be true that the change in kinetic energy is equal to the work done by gravity. We can relate now weight, displacement, kinetic energy, speed. In fact, without doing too much work now, just knowing the weight and the angle of the incline for instance for the sliding block, knowing the mass of the block, you can very quickly solve for the speed of the block at any displacement along the incline. Energy lets you get to certain things that otherwise take a lot of effort with Newton's laws and equations of motion much more quickly. Now in the case where a force acts in addition to gravity, consider the problem of your hand raising a ball up or lowering the ball back down. In the case where a force acts in addition to gravity, think about the situation we just looked at where your hand is holding a ball. It raises it up in a gravitational field or lowers it down in a gravitational field. It's still going to be true that the change in kinetic energy is equal to the total work done by all the forces involved. But now we've got a couple of forces. We've got the hand, we've got gravity and so there's a summing of the work done by each of them that has to be considered. Okay, one of the works here is the work that is done by gravity. Now in the special case that the change in kinetic energy is zero in the case we explored with raising the ball up or lowering the ball down, we're able to relate the work done by gravity to the work done by other forces and we learn that the work done by gravity is the negative of the work done by the other forces in the problem. Now there is absolutely nothing at all special about gravity and in general these lessons that we have learned from thinking about gravity and its role as a force in cases of changes in kinetic energy or work done with other forces, they apply to any force. So let's use this as an opportunity to dig into a new force, one we have not yet encountered in the course, the so-called spring force. Before we dive into this, we have to look a little bit at the fundamentals of the spring force. The law that relates the force exerted by a compressible object, a spring and its degree of compression, that is delta X, how much it's displaced from the place where it's relaxed, it's neither stretching nor squashing. That was determined by Robert Hook. He lived from 1635 to 1703. Now he determined that up to a mechanical limit of a compressible material, it is possible to pull on a material or compress a material too much. But up to that limit, it will be true that the force exerted by the spring in trying to restore itself to its equilibrium size, the shape that it takes where there is no outward expansion, no inward compression that's net. He found that it's true that the force exerted by the spring is proportional linearly to the displacement from equilibrium. That is that the spring force is equal to negative K delta X and negative because the spring attempts to oppose changes in expansion or compression. If you stretch a spring, you have a positive delta X. You've moved it outward away from its equilibrium position. The spring force will attempt to pull it back. If you compress a spring, it wants to restore itself back to its larger size. So the force now points outward, trying to stretch the spring back out against your inward compressing action. The constant of proportionality between the force and the displacement, the expansion or compression of the spring from equilibrium is known as the spring constant. It has units of newtons per meter so that the product of K and delta X gives us newtons. And it's determined ultimately by the material that you're stretching or squashing. And that means it's determined by the way its atoms are arranged, the way they bond, et cetera. So it's chemistry, it's physics, it's quantum mechanics, it's material science, it's all of this stuff acting in concert that ultimately details us how springy or not a material actually is. Ultimately, that spring force derives from the inter-atomic forces. Atoms are bonded to one another. They are comfortable at a certain distance from each other. If you force them to get closer, the electrons and the orbits around the atoms will probably act to repel one another and push the atoms further apart. That's an attempt to restore from a compressed state back to the comfortable equilibrium position. If you pull on the atoms, because they are in fact chemically bonded together, they're sharing electrons, they have covalent bonds or ionic bonds, something like that, they want to restore themselves back to the comfortable equilibrium place where they were before. It is atoms, atoms, atoms, and the complexities of their electromagnetic interactions that ultimately is the origin of the spring force. But for us, it is sufficient to note that the spring force on a large-scale macroscopic material sense has this relationship to the displacement from the equilibrium position delta X. It is linear and proportional. So let's consider the work that is done by stretching or compressing a spring. I have depicted a spring, a coil of metal wires over here on the right. One end of the coil is affixed to an unmoving object. In this case, I've depicted a wall. I want you to throw gravity out. Do not think about gravity. We're doing this on the International Space Station. There is no gravity. It's a microgravity or no gravity situation. All we have is the material the spring is made from. We attempt to compress it. It pushes back outward. We attempt to stretch it. It pulls back inward. It wants to get back to that comfortable equilibrium position where all its atoms are kind of in their happy place. So that's what a spring is going to try to do. So right now, I have it resting in this picture at its equilibrium position. It is neither compressing or stretching beyond the place where it is in a happy equilibrium. And so you can say that its delta X is 0. If I pull on the spring to the right with a force and attempt to stretch it out, if I define its equilibrium location as X equals 0, then pulling on it to the right is a positive delta X. It's a positive displacement. I'm stretching. I'm expanding the spring outward from its equilibrium. That force that is exerted back on my hand by the spring is negative K delta X. If I'm pulling to the right and positively displacing this thing to the right, the force of the spring is pulling to the left, attempting to restore it back to its equilibrium point. F equals negative K delta X. What do you notice about the spring force? It is not constant as a function of displacement. In principle, the harder I pull on the spring, if I double the displacement, I double the force that this thing is acting with on my hand. I've tripled the displacement. I've tripled the spring force. It's pulling back three times harder than it was at the beginning. The force depends continuously on displacement. It depends on X coordinate. We're going to have to employ calculus to analyze this problem. So this is an excellent moment to dust off our calculus, integral calculus, one more time and try to apply it, see if we can work our way through this problem. So let's do that. Let's analyze this problem using the work kinetic energy theorem. If the spring's right end is originally not in motion, and then I stretch it and I hold it there, so it's not in motion again, then it is not in motion at the beginning of the problem, and it's not in motion at the end of the problem. So it has no kinetic energy at the end, no kinetic energy at the beginning. The total kinetic energy is 0. Now that must mean that 0 is equal to the sum of all the work that is being done by forces in this problem. Well, one of the works is going to be contributed by the spring, and one of them is going to be contributed by me. Well, I can write the work done by the spring as an integral from x min to x max of the spring force, dx. I'm just going to leave the work that I'm doing, the external force, that's me, as just w with a subscript external. We're going to try to solve for the work done by me in terms of the work done by the spring. OK, let's keep going. If the equilibrium length corresponds to the case where the right end of the spring is at x equals 0, I'm going to define that location, equilibrium, as x equals 0, then I start at x equals 0. That's my minimum x location, and I stretch out to some displacement, x, away from 0. That's my x max. That's my maximum displacement. So my integral has two limits, the minimum, 0, the maximum, x. I don't know what x is. x could be 5 meters, it could be 10 meters. It's a variable. We're just going to leave it that way. My spring force is negative kx. So if I plug that in, I now have an integral from 0 to x of negative kx dx. Now, it's antiderivative time. The derivative of what function gives me negative kx? The answer, if you think this through, is that it must have been the function negative 1 half kx squared. Think about that a moment. If I take the derivative of negative 1 half kx squared, I take the derivative of that with respect to x. It's like taking the derivative of x squared. The 2 comes down. The exponent, 2, gets reduced by 1 to 1. So the derivative of x squared with respect to x is 2x. And then I wind up with 2x times k times negative 1 half. And I wind up with negative kx. So this works. Now remember, there's this constant of integration. The derivative of this constant with respect to x is 0. So it has every right to be here. And again, we'll find it will cancel out in the following action. And so indeed, we finally wind up with the following relationship between the external work and the work done by the spring, which is negative 1 half kx squared. And so we take this one step further. And we rearrange this equation. And we solve for the work done by me, the external force. And it's equal to 1 half kx squared. Now that's still equal to the negative of the work done by the spring. In the previous equation, we had the work done by the spring, which happened to be negative 1 half kx squared, added to the work done by the external force. That's me. It's still true that work external is equal to negative work spring. It's just that the form it takes now for the external force is a positive number, 1 half kx squared. So not a big surprise. I grabbed the spring. I pulled it to the right. The spring doesn't want to go to the right. It wants to stay where it was at equilibrium. And so, of course, I had to put energy in in order to get the spring to stretch. My work is positive. The work done by the spring will be a negative number. x squared, even if it's a negative displacement, will be a positive number. k is a positive number, and 1 half is a positive number. The product of these three things is always going to be a positive number. So the work done by an external force, either stretching or compressing a spring, is positive. You have to do work in either case. You want to compress a spring, it costs you. You want to stretch a spring, it costs you, because it really doesn't want to do either of those things. It wants to stay at equilibrium. So it's still true that the work done by the external force is the negative of the work done by the spring. It's just that whether you want to compress it away from equilibrium or stretch it away from equilibrium, you have to do work. Now, of course, if it's already away from equilibrium, say it's compressed, and then you let it expand back to equilibrium, you're going to get that energy back. But if you take it any direction away from equilibrium, it costs you. Equilibrium is the lowest energy state of a spring. Anything else after that, you're storing energy in the spring, you have to give up energy to do that. Something has to do work in order to make that spring compress, because it does not want to compress or stretch if it doesn't have to. So there's some deep lessons from these examples, and they all have to do with the changing form of energy. At work in all of what we have done so far has been an assumption, which it turns out is actually an observed feature of nature. There's a deep reason why this feature is true, and it would eventually be the work of a brilliant mathematician named Emily Nother who would figure this out. But the fact that we observe that energy changes from one form to another, you can go from work to kinetic energy, but it's neither created nor destroyed. This is the principle of the conservation of energy, and the reason for the conservation of energy has at its roots a deep connection to the mathematical symmetry or mathematical simplicity of nature, and its relationship to time. Let's consider the gravitational force. It acts to change the speed of an object by accelerating it. Think about the block sliding down the inclined plane. The block is going from low or no speed to some higher non-zero speed. The gravitational force does work, giving up energy from somewhere to do that. The object increases in speed, it's gaining energy in the form of kinetic energy. What one has lost, gravity, the other has gained, the block. Now this can go the other way of course, but it's a zero sum game. Something has to give up energy for something else to gain it, and the energy that you give up may be converted into another form of energy in the other object, but it's a zero sum game. From considering springs, we see that we can store energy in the compression or stretching by acting on the spring with an external force, whether I stretch it away from equilibrium or compress it from equilibrium. I'm storing energy in the spring. I have to do work, and that means the spring gains energy. So the external force does work, that costs it, the stretching or compressing stores it. If we let go of the spring, that energy should change form again. After all, that spring wants to get back to equilibrium. And what we'll observe, it's a very simple experiment to do, is that the spring will attempt to move back toward equilibrium. It will overshoot it. It will compress if it started stretched, and we let go of the spring. It will overshoot equilibrium. It will compress. It will come to a stop. It will re-expand overshooting equilibrium again, and then it will stretch, and then it will stop, and then it will overshoot equilibrium again, and compress and stop. And this cycle will simply repeat over and over and over again. Now, if there wasn't collisions with air molecules, if the spring wasn't twisting a little bit where it's connected to say the ceiling or the stand from which it's hanging, then we wouldn't be losing energy to friction or drag. But of course in reality, drag is acting on the spring and the weight attached to it. Friction is acting at the place where it connects. This wears down its motion over time. No system can continue forever. It's dissipating its energy, its kinetic energy, in other forms. It's losing it to heating up the air or heating up the stand where the spring rubs the hook where it's connected. Nonetheless, for a long time, the system continues to bounce back and forth overshooting equilibrium each time. We've changed stored energy in the spring into the kinetic energy of the spring. Energy is a fantastic concept. It already gives us access to far more complicated situations than we've been able to touch before with just equations of motion and Newton's laws. And what's great about it is it's a scalar quantity. I can't emphasize that enough. Sure, it can be positive or negative, but it's a number and those numbers can simply be added or subtracted as appropriate. It's derived from vectors, force is a vector, displacement is a vector, but work is a force product with displacement which yields a number. It's either just a positive or a negative number that you just have to add up and or equate. So it's really quite an elegant concept and while it's struggled to gain quantitative traction in its origins with breakthroughs and the understanding of this internal property of matter with, for instance, the kinetic energy discovery of Emily de Châtelet, we have the foundations here of a fantastic way of viewing the universe. And in fact, modern mechanics, a more advanced mechanics course would do basically everything from energy conservation and solve problems using that concept to great, great effect. The last concept that we're going to look at in this section of the course is power. Power relates to work done over an interval of time and specifically it attempts to address the question, what is the rate at which an applied force does work? The answer to this is the concept of power. So for instance, we can define the average power as the ratio of two things, the work done in some unit of time delta t and the unit of time delta t. The ratio of these two things, work per unit time is the definition of average power. Now, this being the average work done per unit time and work having units of joules because it's energy and time having units of seconds, we can automatically define another standard international unit for this concept. And the standard unit for this, one joule per second is known as the watt and it's named after James Watt, who lived from 1736 to 1819. He was an inventor, a mechanical engineer and a chemist and over his vast quantity of work, he improved on the steam engine incredibly in terms of efficiency and performance and his work fundamentally resulted in a deeper understanding of energy. And in fact, it was the development of better and better steam engines, engines that can convert chemical energy into heat energy and heat energy into mechanical energy that led to the explosion in our understanding of heat as a kind of energy, so thermodynamics, and fundamentally led to a convergence of ideas in mechanics and thermodynamics that revolutionized our understanding of the universe in the late 1800s. Now, similar to the concepts we introduced earlier, such as average velocity versus instantaneous velocity or average acceleration versus instantaneous acceleration, it's also possible to define the instantaneous power. If for instance, you can write the work done by a force is a function of time, then at any given moment in time, it's possible to find the power that is the rate at which the applied force does work. And that is simply given by the first derivative with respect to time of the work W. So instantaneous power or simply power is DWDT, the change in the work done with regards to the change in time. Now finally, combining all of these concepts together, vectors, instantaneous quantities like power, we can actually connect the force to its effects on an object, given the orientation or relationship between the force acting on the object and the velocity of an object. So we can actually connect together in a very sort of brief mathematical exercise, the concepts of power and force and velocity by starting from the instantaneous power definition. So for instance, let us imagine that we have a force whose magnitude and direction are constant with time. But the displacement is changing with time. Perhaps the place over which the force is acting is moving with time. These are two vectors, force is a vector, displacement is a vector. The force doesn't have to point only along the entirety of the displacement. The force may act in an angle to the displacement. For instance, if this object has some velocity already and then a force acts on it, the displacement and the force may not point in the same direction. So they may have between them an angle theta. So let's start from the definition of instantaneous power, the time derivative of work with respect to time. Now the work done by the force can be written out as the force times the displacement times the cosine of theta. After all, work is the dot product, the scalar product of the force vector and the displacement vector. So vector F and vector R. But the dot product of those two things can be written as the magnitude of F, which is constant. The magnitude of R, which is not constant, and the cosine of theta, which we are going to take to be constant in time for this exercise. The force is not changing direction, the displacement is not changing direction, only its magnitude is changing. Well, since force and the cosine of theta are constants with respect to time, we can pull them out to the left of the derivative and move the derivative operator over so that it's only acting on the thing that does depend on time and that is displacement R. So we wind up with an equation F cosine theta times the first derivative of R with respect to time. Well, if you dig way back, the change in position with respect to time is, of course, velocity. And so we have an equation, F the force magnitude times V the velocity magnitude times the cosine of the angle theta between the displacement and the force. But, of course, displacement and velocity, by definition, always point in the same direction. This is something we've seen earlier. It's no different now. It's no less true now. And so this actually just represents the scalar dot product between force and the velocity of the object. The force acting on the object and the velocity acting on the object. So the power, the rate at which work is done by a force, can also be written as the product of the force and the velocity. In fact, the scalar product of these two vectors. We can learn a bunch of things from this right away. For instance, we learn that if a force and a velocity vector point in opposing directions, the force points backward, the velocity points forward, so that the angle between them is 180 degrees. It tells us that the object has energy sapped from it. Why? Because this dot product winds up being a negative number. The cosine of 180 degrees is negative one. So if you have a force and a velocity pointing in the extreme case in exact opposite directions, you wind up with a negative power. That is, there's a negative rate of which work is being done. And as a result of that, that tells us that this object is having energy sapped from it. Well, this makes sense. If I'm driving a car forward and suddenly I encounter a huge headwind that pushes back against the direction of my car, that's going to sap energy from my car. I'm going to have to step on the gas harder to get it to go faster. I'm going to have to put more energy in to compensate. So a power being less than zero is not a bad thing. It's not a weird thing. It tells us that energy is being drained from a system over time. On the other hand, if those two things, force and velocity point in the same direction in the extreme case, cosine of theta is one, the angle is zero, then it means that energy is being put into the object by the external force. The force is pushing on the object in the direction it's already moving, increasing its kinetic energy. Power is greater than zero in that case. And again, I've given you an extreme case. There are many angles for which power will be greater than zero. And finally, there's the specific case of a force that's acting perpendicular to the velocity. In that case, theta is 90 degrees and the cosine of 90 degrees is zero. In that case, no energy is transferred to or from the object by the force, at least in that direction of motion. So we can see that then there is in fact no power in that case due to the external force acting on the object. Let's review the key ideas that we have explored in this section of the course. We've recognized that motion has an associated quantity, energy, which is difficult to define, but which can be defined. And even if it's difficult to define, it can be quantified. For instance, reconsider the experiments in dropping objects into soft clay that Emily, the Marquis du Châtelet, conducted during the second half of her life that revealed the V squared character of the kinetic energy, the velocity squared dependence of the kinetic energy. We've understood that forces acting on objects in part changes in energy. We've learned to describe some kinds of energy and especially to begin to relate the force and energy concepts. That happens specifically through the concept of work. Work is a form of energy. It is a product of force over the distance that the force acts. And finally, we began to describe not only variable forces. We thought about what it would mean for a force to vary with distance and how to handle that in our calculation of the work. But also we've looked at time-varying energy changes themselves, the rate at which work is done by a force on a system and explored a little bit some of the basic first principle ideas of that concept. That is the concept of power.