 Welcome back to our lecture series, Math 1220, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. Lecture 7 in our series, we're going to start talking about the concept of work. Now, you might be thinking, well, haven't we already been doing a lot of work in this class? Wouldn't you say? By the meaning of work in this context, we're referring to its interpretation in physics. That is, work is a force applied over a distance. As this section is going to involve a lot of physical terminology, it's worthwhile mention a few things for those who might not be up to date on their basic physical mechanics here. Imagine that we have an object whose location, position is given by the function s of t, which we see right here. We've used this before to refer to a position function. Given a position function, recall that the force of an object, which we typically denote with a capital F, the force of an object follows the rules of Newton's second law of motion, which tells us that the force to accelerate the object is equal to the mass of the object times the acceleration that it begins to have. Oftentimes, Newton's law of motion is thought of more in the following way, that if we take the force and divide it by area, that'll tell us how quickly the object will begin to accelerate and things like that. But of course, if we rewrite it in the following right here, if we have an object which we know it's mass and we want to accelerate it as certain acceleration, then this tells us exactly how much force is needed to do that. Now, the relationship between, of course, acceleration, a of t, to position is that, recall that acceleration is the second derivative of position, because after all velocity, which is the derivative of position, it's derivative is acceleration. So for our purposes, force is just mass times acceleration, and acceleration is the second derivative of position. Now to the idea of work. In the physical sense, work is a force applied over a distance. And so if we denote work typically with a capital W, work is going to be the product of force times distance. So the idea is if you have some horizontal distance and you have an object that you want to move it along that distance, well, you start exerting a force on that object, right? That force is going to be based upon the object and how hard you're pushing it, yada, yada, yada. That's the second law in action here. But as we exert this force, the force will transverse a distance, right? And the product of those two things gives us the work done to move that thing. Now work, unlike some of the other physical quantities we talked about, is a directionless object. This is what we refer to as a scalar quantity. There is no direction when it comes to work. The idea is if you were to push something up versus pushing it down versus, and when I say up and down, I mean like if you push it north or push it south, push it east, push it west, doesn't really matter. The work is going to be the same irrelevant of the direction. This is a difference between like position function, which if we have our x and y axes, zero. Position, like if you go three to the right, that's different than going three to the left, which we put a sign attached to it. So position, S of t here, is often what we refer to as a vector quantity, meaning that the direction does matter. We have a positive direction, we have a negative direction. This is also true for its derivative velocity, and it's also true for its derivative acceleration. And as likewise mass, mass is likewise a scalar quantity. There's no positive mass or negative mass. Going mass to the left is the same as going mass to the left, right? Right and left makes no difference there. As mass is a scalar quantity as well, force will be a vector quantity. Force does have a direction. So like with our little force diagram right here, a force going to the right is very different than a force going up or a force going down or something like that. And these forces add together by the usual laws of vector addition. We will only worry about one dimensional vector, so it's not going to make much of a difference. But for our purposes, force has a magnitude and direction, but work just has a magnitude. If you were to push your car all the way around campus in one giant circle, that would not feel like you did nothing, that the net effort is not zero because the direction doesn't sort of cancel each other out there. But with this formula in mind, there's really not a lot to some of these basic linear work type problems. So for one example, imagine we want to compute how much work is in lifting a 1.2 kilogram book off of the floor to put it on a desk that's exactly 0.7 meters high. So to calculate this, we're going to first calculate the weight of the book, which is its force due to gravity. So in that regard, by Newton's law, mass times acceleration, we have the mass of the book, which is 1.2 kilograms. And then the acceleration due to gravity is going to be 9.8 meters per second squared. That's just a constant here on planet Earth. That will give us the weight, the force of the book, which if you multiply those together, you end up with 11.76. And the units here would be kilograms by meter per second squared. That's kind of a mouthful. So instead, we refer to the standard units of force as a Newton named after Sir Isaac Newton, of course, for his work in mechanics here. So be aware that one Newton is a single kilogram by meter per second squared. So you can think of it as one force if a force is equal to one Newton, that's the force necessary to accelerate a one kilogram object by one meter per second squared. This does not give us the work. This gives us the force. This is just how heavy the book is. The work is then going to be the force, in this case, the weight multiplied by the distance. So we get 11.76 Newtons. And then we times that by the height, which was 1.7 meters. And so when you multiply those together, you end up with 8.232. The unit here would be a Newton meter. But again, to kind of simplify things, the unit that's often used here is a joule, capital J there. And there we go. If this was a scientific problem, we might need to worry about significant digits. But this is the math class, so I don't care. The work to lift your book is just going to be the weight times the distance it has to travel. And if you're working with scientific units, do make sure you use kilograms, meters, seconds, Newtons, joules as appropriate. And remember that one joule, remember, is just a single Newton meter. Now if we were to do a similar type of exercise about using the English units, like pounds and feet and things, how much work is it done in lifting a 20-pound weight six feet off the ground? Now one thing to remember when you're using these American-British units that pounds is not actually a measurement of weight. Pounds is actually, sorry, I said that wrong. Pounds is not a measurement of mass. Pounds is actually a measurement of weight. Pounds is a force. It's not a mass. This right here is a force. As such, you don't have to multiply 20 pounds by acceleration due to gravity, because the 20 pounds has it already built into it. What I'm saying is the force due to weight of this object is already 20 pounds, like so. And so if we want to do the work, which is just force times distance, then we would take the 20 pounds, which is the weight of our object here, and then we times it by the six feet that we want to lift it. And that ends up with 120. And here the unit actually is called a foot pound. And so we just put the two things together. It takes 120 foot pounds of work to lift the 20 pound weight up by six feet. And so for very simple work problems, this is all that one has to do. You calculate the force. You multiply that by the distance. And as long as force is constant along the whole journey, that's all one has to do with these work type exercises.