So it's really useful to stop, and now we're entering the third segment of the course. The courses you'll discover is really in three parts.

That's why we have three exams. We did kinematics first, and you still want to keep in mind as you study, and I encourage you, so for this exam, it will just focus on the last three weeks. You'll probably not have to do any kinematics on it. You'll look at the old exam from last year, and you'll see there really wasn't any kinematics,

but by the time you get to the final, you'll be expected to combine force problems with kinematics problems. So now is the time to start. As you study, look at the problems, ask yourself where might they go together, but keep in mind that kinematics was basically our description of motion, and it was mostly done with projectile motion problems,

and the common question is where and how fast, and sort of did I make it or did I just make it, right?

Those were the type of questions. You see those sort of words. Now you'll see there's going to be a little bit of overlap. The where and how fast can occasionally show up in what we're about to do, which is energy. So you'll want to look at, okay,

when did I use kinematics? When did I use energy? And what we'll find is that if no direction is involved, you'll almost always want to use energy. It'll be easier. It's a scalar, but if direction is involved, if it's not just how fast, but it's also where am I going, you're probably going to have to use kinematics because that's the vector version of describing motion.

Now what we just finished were forces, often called dynamics, and what you'll want to be looking on in this exam, right, this was always about the causes of motion or lack of motion, and we were almost always asking questions about

just about to blank, blank, blank. Notice it's a little different. The word just is there. We had just make it in kinematics. Now we have just about to slide, just about to stop, just about to fall over, just about to, you know, do various things. We also had the simple ones where we were literally just asking

what is the acceleration or force? You know, that was a more straightforward one, and we, like I said, the limit of this of things not doing anything was what we did last week, which was static equilibrium. It was still about causes of motion,

in this case, no net force, no net torque, so we don't accelerate or start to move, right? Those were the two pieces we've had so far, and you want to be able to recognize a force problem versus a kinematic problem because they're going to come together in the future. You want to be able to separate those out.

Now what we're doing, I'll categorize as energy and collisions. This is the final piece. We're going to do chapters seven, eight, and nine in the next three weeks, and a key thing for these type of problems, what we're going to be looking at

is a change in state of the system. We're going to talk about words like the system and the state of the system and how that changes, and one of the key features of that is almost always in the problem there's going to be a clear initial

versus final state or what we might think of as a before and an after. So whereas before in projectile motion we said it launched with maybe 30 meters at an angle of 45 degrees, where does it land?

We might now talk about you compress a launcher on a spring, and then you release the spring. How fast is it going when it's launched? There we have the before, when you had it compressed on the spring, and the after, the launch speed. We didn't worry about how it got from one to the other. We just wanted those two points in time.

With projectile motion, even though we may not have calculated it, we were thinking about it as flying from here to there. There was kind of the whole where was it going, and it's flying. So that's one of the hints that you're going to be working with energy and collisions when the problems start getting put together is that you're always going to be identifying

this initial final before or after. So as we get into that, you really want to look for that when you solve problems. Questions, answers. So how does this break down?

This week we'll basically do Chapter 7, and the focus will be introducing the concept of work, which is the cause of a change in state. And we're going to have to figure out what we mean,

start figuring out what we mean by a system. And then we're going to go into Chapter 8, and the focus of Chapter 8 is energy. And we're going to learn all different types of energy, which basically describes the state of the system. And the great thing about energy is it is a scalar, so it does not have a direction.

And you should be excited by that because usually what you get wrong is the direction or you forget to use the direction, and not having to deal with direction is a great thing. And then finally, in Chapter 9, we'll really think about collisions again. We've kind of talked about collisions occasionally,

and we've talked about them more in the abstract case of collisions as just interactions between things. Now we'll do real like things colliding and either sticking together or bouncing off and what happens. You really, the number one thing going into these three weeks

that tends to cause most students the trouble is really being careful in making sure at the beginning of each problem you pick your system. So you really want to watch for that in the lectures, watch for that in the examples, look for that in the problems. What do I mean by my system? How do I pick it? Fundamentally, your system is the objects you are interested in.

By the time we get to the end of Chapter 8, we'll find that most of the time, that's what we're going to do. We're going to put everything we can inside our system, and we won't worry about anything outside our system, and we'll just be able to use energy conservation.

When we start out, we're going to have stuff outside our system, and that's going to do work. And sometimes you need that, sometimes you don't. So keep that in mind as we go through the lectures. The other piece that we're going to talk a little bit about today

is we get to do our first integral and a dot product. So this is the definition of work. We're going to come into it in more detail, but notice we need to figure out why is there an integral, what does that mean, and what does this dot mean?

And those are the two pieces we're going to want to look at, because if you notice, force is a vector, this is a little vector, I put the arrow over it, there's no arrow here. Work is a scalar. So we're going to see what that means as we go through it. And then we'll look at the special case of constant force,

where work becomes a little more simple. And that's the equation you get. We are going to introduce one type of energy today, the kinetic energy.

And we'll talk a little bit about why it's one-half mv squared. But really, this is one of those things that's more of an experimental fact. We found that this is a useful way to characterize the motion of a system. And so we just want to accept this as a definition that turns out to be useful.

It's not really a concept to understand. It's a definition to use. So that's something to keep in mind. And as I promised, we're going to introduce one more force to our list of forces we have to keep track of.

And again, this is a definition of a spring force. So I kind of wanted to give you that big overview of the pieces that we're going to be getting this week and in today's lecture. I'm going to review most of these concepts in today's lecture,

give you kind of that overview. And then on Wednesday, we'll do some example problems with them. And then on Friday, you'll take your test on last week. Questions so far? Remember, there is a quiz due Wednesday.

All of this is leading to our next really big equation. This is as big as F equals ma, but not quite as famous. And this is work. And as with F equals ma, this equal means cause.

Work causes. And our delta means change. Changes in energy. Now, for chapter seven, our system will always be a single particle.

What that means is that there's only one energy it can possibly have. If we have a state of a system, there are two types of energy. The first type is motion.

That's our kinetic energy. That's our one-half mv squared. A single particle can move. That's allowed. And this is an important part. If you have more than one particle in your system, more than one object,

then they can have a relative position. And then they can have potential energy. And so this is a warning as we go forward to chapter eight. We're not going to talk about potential energy in chapter seven,

but it is one of the more confusing points as we switch from single particle systems that just have kinetic energy and we talk about work causing changes in kinetic energy to the more complicated collection of particles where we have more than one object and we have potential energies. And one reason this is important

is it helps you remember the few formulas we do have to have. Kinetic energy has velocity in it. All of these, because they have to do with relative positions, will have some sort of position in the formula when we get to that.

So now that I've put you to sleep with this nice introduction, we get to ask the question, if work is going to cause changes in energy, so work will equal change in energy, and for our single particle,

the only energy is kinetic energy, one-half mv squared, then work, remember, change is final minus initial. That's another thing that's going to be important. You have to keep that order straight. And I can factor out my one-half m.

So we see work will cause a change in speed. So let's ask ourselves what things can cause a change in speed.

Now, what I want you to notice in this case, I want you to think about it. A is a force in any direction. B is a force in a direction of motion only. C is energy. D is work. E is more than one is correct. If you think A is correct, then obviously B is correct.

But if you think B is correct, A does not necessarily have to be correct, because that's any arbitrary force. So as you think about your answer, keep that in mind. Now, that's a safe guess in this case, right? So unfortunately this is a good question,

except it gives me very little information. So somebody tell me, more than one is correct. What are the more than one that are correct? So let's redo this. Raise your hand, and we're going to risk, because nobody ever raises their hand, but raise your hand if you think A is correct.

Raise your hand if you think B is correct. Raise your hand if you think C is correct. I like how people are looking around to see if anyone else raises their hand. Raise your hand if you think D is correct. Interesting. So raise your hand if you think the two that were correct are A and B.

Interesting. A lot of people, okay. How about A and C? B and C? C and D? B and D?

That is really odd, because that doesn't add up to the like 80% of you that actually answered more than one is correct. Interesting. So you were just guessing on that more than one. So let's look at it. Think about making something move. If I want to change its speed, what do I need?

Suppose I'm moving in the X direction, and only in the X direction, so my velocity is VX I hat. How do I change my object's speed? What do I need?

See, we have a test on this on Friday. I need a force. And what direction should that force be in? Let me ask you this. Suppose the force is in the Y direction. What's going to happen to my motion?

I will change my what? Direction. Right, that aspect of my velocity will change. What happens in uniform circular motion? Does my speed ever change?

No. Is there a force on me? Yes. Which direction is it? Friday. Friday, you've got to know this by Friday. Which direction is my force? Radially inward. Practice those words. Same as you go to sleep.

The force is everywhere radially inward, which is always what compared to my velocity? Perpendicular. So a perpendicular force cannot speed you up. It cannot change your speed. So a generic force will not work. You need a force in the direction of motion to change speed.

And that's the way forces work. And remember, we can always break everything into components. We can always arbitrarily pick the direction of motion, say to be the X direction, to make life easy. And any arbitrary force we can write with two components,

one of which will change the speed, and the other of which will change the direction. And if the force is only in my direction of motion, do I turn? If the force is only in the direction I'm moving, do I turn at all? No. I only speed up.

And if it's only perpendicular, do I speed up at all? No. So you want to really start thinking about that whenever you're describing a motion or seeing what's going on. It is the force in the direction of motion that causes my change in speed. Energy does not do it because energy describes my state.

My energy is one-half mv squared. And if my speed changes, yes, my energy will change. But that's just because that's how I'm using things to describe my state. My energy can't cause a change in my speed.

Now, work, however, we just said, does cause changes in energy, so it will cause changes in speed. And that's the heart of what work does.

Any questions on that? Now, what does that mean for work? First of all, we're going to have to figure out how to find the force that's parallel to my motion so that I can figure out what part of the force changes my speed

versus my force perpendicular to my velocity, which just changes my direction. And the mathematical way we do that is something we call the dot product.

And if you think about it, how do I find out the component of the force parallel to the velocity? What do I do?

The math I told you that was the most important math for this class. Trig. And so which component of my force is parallel to the velocity? Let's say it loudly with confidence. Which cosine or sine do I use?

Cosine. That's the one that's parallel. The magnitude of f cosine of theta is the part parallel, and so we introduce this concept called the dot product, which tells me if I have two vectors, the dot product is the magnitude of one times the magnitude of the other times cosine of theta.

The other way I can do it, if I only know the components, not the magnitude, I just multiply the components and add them together. Notice this makes perfect sense because what it's saying is

if there's any f in the same direction as this part of v, that's parallel to part of v, and if there's any f in the same direction as this part of v, that's parallel, and then we add those pieces together, and that's all that there could be. Notice if I line my x-axis up with my velocity,

then the velocity only has an x component, and this reduces to the component of f in that direction times v, so it gives me the right answer. These are the two ways to do it. If you're given the components, you do that. If you're given just the vectors with the angle between them, you do this here.

It's a lot easier than the cross product because we don't need to get a direction. We don't need to use our right-hand rule. We just use our formula to get a single number. Now, let me ask you this. Can it ever be negative?

Can the dot product ever be negative? So I saw a lot of pointing at the screen. Can the dot product ever be negative? Yes or no? Let's vote. A is yes.

B is no. So we have a close split. You didn't do very well discussing, did you? So someone who said no, why not? What do you see up here that convinces you it can't be negative?

Now you're getting suspicious because I'm asking the no's first. Be honest because I need to know why so we can correct the problem. Now you're unwilling to admit it, right, having said that.

Anyone who said it could not be negative, why do you think it could not be negative? No one willing to volunteer? Yes? You could change the coordinate system. Notice, however, the angle between them will never change,

no matter how much they change the coordinate system. It's a fixed physical number, right? If these are my two vectors, that's them. And they're just sitting there in space, force, velocity. And no matter how much I change my coordinate system, there is some angle between them. And the angle between them might be this, right?

Because my velocity could be that way, my force could be that way. If that happens, what happens to the cosine of theta? It becomes negative because it's bigger than what? It's bigger than 90, right? It's on the other side. So, from the top formula, cosine can be negative, so clearly it can be negative.

From the bottom one it's harder to see, because yes, I can change my coordinate systems, and it may go from negative to positive, and positive to negative. So depending on my coordinate system, it might look like it can change, where in fact it can't. If you do it very carefully, you will find that no matter how you change the coordinate system,

the number and sine of f dot v can't change, and you know that from that formula. So fx and vx, fy and vy will always change in exactly the right way, so you get the same answer no matter what coordinate system you pick. It's not worth it in this class to go into that math,

but you can see it from the other formula, because this one has nothing that depends on the coordinate system in it. What does it mean physically for it to be negative? Work causes a change in energy. If work is negative, what happened? If work is negative, the change in energy is negative, because it equals that.

And if the change in energy is negative, what happened to the energy? It went down, because the final minus the initial, the final was less than the initial. That's how I get a negative number.

So negative work decreases your energy. That's all that's going on physically. Notice it's not a vector. The negative work doesn't tell you the direction of work. There's no direction here. It just tells you if work, which is change in energy, is less than zero, your energy decreased.

That's what went on. Any questions on that? Now, if you recall, when we make small displacements, our velocity is always in the direction we're moving, right?

So delta x is in the direction of velocity. We want to get our forces in the direction that we're moving, and also we want to think about it. If the force acts for a longer distance, then it will obviously cause a bigger change in energy. And we can recall one of our kinematics formulas.

This just came from the definition of describing motion. But, now using F equals ma, if I multiply by m, I get a piece here that's my force.

If I subtract this and divide by my two, I get my one-half m v final squared minus one-half m v naught squared equals F dot delta x.

So to be consistent with what we know about describing motion, we now see why we want to define work by the combination of the force, the displacement, and the dot product. So the dot product gets us the right direction.

It gets us the force in the direction that we moved, our displacement. And the amount of the change in energy is caused by how big the force is in that direction and how big of a displacement we make. Now, this is not necessarily a derivation or a proof, but it shows you the deep connection,

because we're going to go beyond this to work causing all sorts of changes in energy, not just the change in kinetic energy. But it kind of helps you remember the formulas and helps you know why we put a dot product in there if you go back and look at that kinematics formula. So now the fun question.

Why do I have an integral? I told you at the beginning, work is really the integral, and I'll write from a to b of f dot dl. And what does that mean? It means if I'm moving along from some position a to some position b,

dl is all my little steps. So these are all my little dls. They're little pieces of what I'm walking along, and at each of these, I have some force.

Aren't we excited? We're doing integrals. We're breaking lines up into little pieces. This really is a little bit more for cultural edification at the moment, but it's helpful to understand this, because you're going to have to do something like this in 3b.

Again, there's lots of things I stick into here that you may not do this quarter, but you have to do something like it in 3b. So this warms you up for that. So why, if from up here, we seem perfectly happy with not doing an integral, why do I say this is the real definition?

Anyone remember what an integral does for us? Okay, yes, that's one of the main things you do in calculus. You do it over and over, the area under the curve. Am I taking any areas here? No. So that didn't help us any.

But a good answer, I like it. What else does an integral do? And the hint is, does anyone know what letter that is? It's an S, yes, and it does a sum. An integral tells you to add up all the little pieces

that are underneath the integral. So it says, I'm going to take a bunch of f dot dls, take a bunch of little f dot dls, so that's what I'm showing here. At each little place, I take an f dot dl, and then I add them all up. And the integral is a mathematical way

of doing that adding up. And that actually, in physics we do use integrals to find areas under curves. And we didn't really do it much back in kinematics, we could have there. That's a place for doing integrals. However, its more common use in physics is to add up little pieces of things.

So in 3B in the next quarter, you're going to have to add up little pieces of charge to get electric fields. You're going to have to add up little pieces of current to get magnetic fields. So this idea of, and I like the force one, because just think about it. Each little place you're at, you get a little force that increases your speed

a little bit or not, and you add up all those little pieces and you get the total work. Are we okay with that sort of? So let's just make another picture of this. Let's suppose you're moving like that

and the only force on you is gravity. So everywhere gravity is down, so think about it. At every little piece, the F dot dl will just equal the force of gravity times whatever little change in y you have.

Because F is only F sub y, which is mg down. That's our force, is gravity. So if I move from here to here,

all I care about is adding up all my little f and little dy's. These are little dy sub i's. That should look to you like the formula for the integral. When I add up that sum, this is a constant. So it comes outside,

and I just sum up all my little delta y's. Well if I add up all the little delta y's, I get my one big change in height. And so I will get an mg delta h. And that will start to look like my formula for when I have a constant force.

F dot delta x. So this is one of the easier cases to see what this integral is actually doing because I don't actually have to do the integral really. But if I sum up a bunch of little pieces of delta y, I get the big delta y.

If F was more complicated, if it was at some strange angle, I'd have to be summing up something else a little more carefully. And I'd have to know how to do the integral a little bit differently. And we'll do one other example of one of these integrals that's relatively easy to do. That's a spring force. I want you to see this. You won't have to do any of these integrals in this class,

but I did want you to see it once before you go into 3b where you'll have to do it. Question. And then, but... No. Nope. The w just at that point

takes whatever force you have at that point and includes that part of it. So if you have a big force causing a big acceleration, you get a big jump in kinetic energy at that point. Well, no, right. Yeah, the work is always about telling you whatever you did during all this time,

I'm going to tell you the difference between your kinetic energy at the end and the beginning. And average isn't quite the right word. It kind of adds it all up. If we divided by something out front, that would be averaging. This is really just adding it all up. And that's why I said, notice, work and energy are all about

the before and the after. That's why we do an integral from a to b. It takes us from some initial point to some final point. We good? That's your aside into integration.

We're going to do one more integral later, but first, we now have to wake up again and ask a clicker question. So now that we've completely reviewed all the definitions of work, there I am, holding my book. Oh, wait, did I turn it on?

Probably not. So we've talked about it being positive, about it being negative. There we go. When I hold my book here, am I doing positive work, negative work, or zero work?

So I'm holding the book up. And I'm tired, I'm done, so I'm going to go put it down over here, and you can keep answering the question. So, 54% of you caught the trick. It is zero.

Why is it zero, anyone? There's no change in position. Work equals force dot delta x. If this is zero, zero times anything is zero.

This might show up on a test. This is the acceleration at the top of your throat question for work, right? Work is zero if there's no displacement. Any questions on that?

We good? We awake now? Excellent. So my next question, not a clicker question, just a general question. How many of you know what a spring is? I know that's a surprise question. Like six of you, okay.

It's a small amount of water running through a ravine. Usually it's kind of clean, we get to drink it, right? No, here's a spring, right? Your basic spring. I brought one here just in case many of you had not seen one.

Now, what is the basic behavior of a spring? If I stretch it, it does what? Well, besides stretching. When I let go of this thing, what's going to happen? It'll go up, right? Very exciting. The force from the spring is always opposite the direction you move it.

It's a little hard to do, but if I compress this spring, what's going to happen? It'll go back down, okay? So this is a spring. We all good? That was the demo for today, I'm sorry. Not much exciting, I know. But it's important because this is our next big force law.

They're often drawn this way. Sometimes they're drawn vertically. And sometimes they'll be at an angle. The key feature of a spring, it always has an equilibrium position

defined by the place where the force from the spring is zero. And then, as we move the spring, let's make this the positive x direction. We move it in the positive x direction, the force ends up being a negative force involving the magnitude that you move it.

If you move it in the negative direction, the force ends up being a positive force. Obviously, that's the opposite of the direction you've moved it, so if I make x equilibrium zero, this would tell me that the force is minus k times x.

A more common way to write it often is we don't worry about x equilibrium being zero. It can be whatever it is, and we write it relative to the equilibrium position. Now, this is a case where we can ask how much work does a spring do

when we move from, say, the equilibrium zero to some point x sub b. We take the force, and now that we're moving only in the x direction, dl becomes dx, right, because all of my motion is in the x direction.

And this is an integral you should be able to do. What is the integral of minus kx? Not quite kx squared. We need our one-half, right, so that when we take the derivative, we get our answer back, and we're going from zero to x sub b.

So we get the work done by the spring is minus one-half kx sub b squared. And if you think about it, the minus sign makes sense here because if I am moving with some speed,

so when my mass is here, here's my equilibrium position, right? If I move through my equilibrium position with some speed, what happens to my speed on the other side? What do I do? Am I slowing down or speeding up? Slowing down until I stop,

and then I'll start to speed up as I come in. So notice, if x is getting bigger, I'm losing my energy, I'm slowing down, so there must be a negative sign in my work. And if x was getting smaller, I'm getting less negative, which means I'm getting big again.

So the minus sign tells you about the fact that springs tend to slow things down when you move away from x equals zero

and speed it up if you move toward x equals zero. Because if I do this integral again,

going from point xb to zero, moving towards the equilibrium, right? I get the same integral here, but now I'm going from xb to zero. What does this end up being?

When I do that definite integral, how does that turn out? Remember how to do our definite integrals? I do minus kx squared over two, where x equals zero,

minus my minus kx squared over two, where x equals xb. This will be zero, and my two minuses will cancel out, and I'll get kx squared sub b over two. So when I moved from xb to the origin, I sped back up.

So we're going to use springs in some of our problems, and this is the one case where you won't be required to do the integral because it does have a nice formula associated with it, but if you're comfortable with the integrals, it helps keep you get the minus sign correct.

So that's just something to keep in mind. Make sense? Now, we have one final clicker question associated with one final problem. Now we're going to go

and lift a book five meters high, and I'm going to lift with 50 newtons of force, and the book has a mass of five kilograms. So if I draw my free body diagram, the weight, mg, is 50 newtons,

and I'm pulling up with a force of 50 newtons. That's my situation, and what I want you to answer is which of those will be doing negative work?

And so we're going to ask that in the more general way, negative work occurs when the work is in the negative direction, when the applied force is opposite the displacement, when the energy of the object decreases, both A and C, both B and C.

A lot of people left for their next class. There we go. Most of you went with E. Think about it for Wednesday, but you'll also need it for the quiz that's due Wednesday. You've got to figure this one out.