So let's talk about energy. Okay, so first of all, what is energy? I'm going to give you a nice, compact description of what energy is. In the back. Can we keep the chattering down? Thank you.

In the back. Raise your hand to change your mind. Okay. Ability to do work. Ability to do work. So then what's work? Transfer of energy. Force over distance. Force over distance. Okay. Is that the only definition of energy, though? Kinetic energy well described by force over distance?

No, it's not. Do you have another comment? No. Okay. Okay, so energy is sort of an abstract concept, right? It's not something easy to nail down physically. And that's fine, but we have various kinds of energy, right?

We observe in the world. There's kinetic energy, which is energy of motion, right? And we all know kinetic energy is one half mv squared.

And then there's potential energy, which is like stored energy. And for example, potential energy due to gravity is mgh.

Of course you know what m and gr. H is the distance, the height we're talking about. So potential energy due to gravity is relative to two positions, right? Where h is the distance between those two positions. So these are just two examples, right, of energy.

Energy is a very useful concept because we know that it's conserved. Meaning that energy of the universe as a function of time is some number.

So when we say energy is conserved, strictly speaking we mean energy in the complete universe, right? So if you draw the universe, this is my best description of the universe,

and you say, okay, well what's the total energy in the universe that's some number, right? This energy is conserved. Now it's very unusual to be solving a physics problem that involves the entire universe, right? Typically we're dealing with a small fraction of the universe, okay?

So this is Irvine, a relatively large fraction of the universe, from our perspective. And of course there can be energy blow into Irvine and out of Irvine, right? And what can we say about the total energy of Irvine?

Well not much, right? We can say that it's part of the universe, we can say there's energy flow, the energy can be varying, right? So it's not terribly useful. However sometimes you work in even smaller systems, right? And in a smaller system which is decoupled from the rest of the universe, there's some sort of barrier or separation or something of this smaller system from the rest of the universe, right?

If there's no connection between this system and the rest of the universe, then you can say that the energy is conserved in this smaller system as well, right? Because there's no energy transfer from the two subsystems, this little system here and everything else, okay?

So why is that useful? Well, you know, imagine an example problem. Don't want to get read. Say you have a bookshelf in your room, right? And on top of it you have a book, okay? And you want to think about the kinetic and potential energy of this book as it falls off the shelf and hits the floor, okay?

So let's think about three time points. So it's on the shelf, it's halfway down, or it's just before the floor.

I'll say on the floor because I should write just before. It shouldn't be lazy to be precise. Just above the floor. And then let's think about what is the kinetic energy and potential energy in each case, okay?

So kinetic energy, what's the kinetic energy when it's still sitting on the shelf? Zero, right? Because it's not moving, so there's zero kinetic energy. And the potential energy when it's sitting on the shelf, this is height h, is just mgh, right?

mgh. And so let's say that this book has this mass one kilogram and this height is one meter, right? Just to make the calculations easy. So then this is one, this is one, so this is 9.8, right?

Units of energy, so joules. Okay, now when it's halfway down, what's its potential energy? Well, it goes down by half, right? So it's linear. This is very convenient, the potential energy is linear with height.

Makes it very easy to calculate how it changes. Right, this is 4.9 joules and here of course it's going to be zero, right? Potential energy just above the floor, so epsilon perhaps. Now the kinetic energy, halfway down, what is the kinetic energy?

Well, without knowing the potential energy, this would be a little bit tricky to solve, but knowing that the total energy is conserved, right? So the initial energy was zero plus 9.8 and that half of it is now being used by the potential energy, which tells us immediately that this is 9.8 minus 4.9, right?

The rest of it has to be in kinetic. And here, all of the potential energy has been turned into kinetic energy. So we don't have to know anything about kinematics in order to understand the kinetic energy of this object. And this is very convenient because we can then calculate the velocity.

We know the total kinetic energy of this object. We know its mass and so we can derive its velocity, right? So we can use energy to solve kinematics problems. And in some cases, it makes the problem a lot easier to solve. Okay, let's do another one, another one with exercise.

Imagine you have an elevator, okay? And then you have an engine for the elevator. Into the elevator steps a very fat man.

This very fat man is a student at what? USC. At USC, alright. We have a very fat USC alum. Steps into the elevator. Now, the elevator starts out at zero height, right?

So it has no potential energy, okay? But there is potential energy stored in this system in terms of the fuel inside the engine, right? So there's potential energy stored in here. This one currently has no potential energy and no kinetic energy, right?

Now, after some time and some amount of struggle by this engine, it manages to push Mr. Fat Booster up a certain amount of distance, some height, and now he's gained potential energy.

Where does that potential energy come from? It's come from the fuel stored in this engine, right? So you can transfer potential energy into potential energy or you can transfer potential energy into other forms of potential energy. Okay, now how much energy has been transferred from here to here? Well, this is what we call the work, okay?

Now, this is very confusing if you use your brain when you try to think about it because work is also a word we know well in English, right, which means something. It means like I have a job or boy, that was a lot of work, right?

It means something in English. Now, forget that it means something in English. We're going to give it a new definition of a word that has a very similar idea, okay? So like I said, very confusing. So think of this as like the technical definition of work, not the colloquial or English definition of work, okay? Maybe we should like spell it differently. Here we go.

W-E-R-K. W-E-R-K, there you go, okay? Work, work, work, work. The German, right? Okay, so work is defined as force times distance, okay? You apply a force over a certain distance, you've done work, okay? This also happens to be the amount of energy you put into the system.

So in the case of the elevator, the work here is M-G, where M is the mass of Mr. Fatty, G is what we know, and H is the distance over which the elevator is pushed up our heavy gentleman, right?

This is the work that's done. It's also the amount of energy transferred. Notice, what is the potential energy of our heavy gentleman after he's been pushed up? Well, the potential energy is M-G-H, right? Well, notice that's exactly how much work it takes to push, to lift somebody up a distance H.

Okay, so you put some potential energy into this person, you've done that amount of work, transferred energy from here to here. So work is really just energy transfer, okay? So here, I write this in one dimension, all right?

But often we deal with multidimensional problems, and so we need to generalize work to a two-dimensional system. In order to do that, we have to have one mathematical concept to make sure we're all clear on. You may or may not already know this, but I just want to make sure that we're all thinking about it the same way.

Okay, so what is a dot product? Sometimes also called an inner product. Well, if you have two vectors, B and A, for example,

then my intuitive notion of a dot product is that it tells you how much of one vector lies along the direction of the other. Essentially, how much do these two have in common? So what you do is you take the projection of B. Now B is a vector, you can break it up into any pair of orthogonal axes, right?

And in this case, you can choose the axes along A and perpendicular to A, okay? All right, so B is the sum of these two blue vectors. This one here, the length of this is the dot product, okay?

This is the magnitude of A dot B, is the length of this vector, is the length, is the distance from here to here. Okay, the fraction of B, which lies in the A direction, okay?

And there are ways to calculate this. I mean, if you just think about the length of this vector, this is B cosine theta, this is B sine theta. So the magnitude of this is AB cosine theta, okay?

Okay, so this is how we do the dot product. Now you can also do it in components, right? You can say A dot B is AXBX plus AYBY plus AZBZ, right?

And if you think about the dot product of some basic objects, just to make sure we're thinking about it clearly,

the dot product of any vector with itself, sorry, with the unit vectors isn't normal vectors, right?

So if you have a unit vector, I, such that the magnitude of I is one, then the dot product of I dotted with itself is equal to one. If you have, so this is I and this is J, if you have an orthogonal unit vector, I dot J is equal to zero.

The two vectors who point in perpendicular directions have a zero dot product, right? Imagine if A and B were orthogonal, then the component of B in the direction of A would be zero, right? Similarly, if you have two vectors which point in exactly the same direction, then the dot product also is simple.

In the case that they're magnitudes are both one, the dot product is just one. Okay, before we do dot products, let's just test our intuitive concept of energy, okay? So if you had to answer this question, which of these candies contains more energy, what would you guess?

Stationary. Stationary. Remember, your answer is not graded.

Okay, everybody ready? Let's see what your guesses are. Same, huh? Somebody presses B. All right, well let's see. So this is not the kind of experiment you could actually do in the classroom,

but fortunately somebody has already done it.

That's a gummy bear. Gummy bear has a lot of sugar in it.

So now do the M&Ms. Take a little longer, get lit.

So here they're turning the energy stored in the fuel, right? Sugar and fat, whatever, in the M&Ms,

the gummy bear, into light and heat. Anyway, as you can see, both of these have a tremendous amount of energy. It looks to me like they're about the same. So why don't we just power the cars with gummy bears, right? How much energy does it take to create a gummy bear?

If we had enormous, vast natural resources of gummy bears, and we dug them up under the ground, I think that would be an excellent idea. Okay, so let's do another one. Okay, so here are two vectors expressed in terms of their unit components, right?

So 5i plus 2j is really just a vector that means 5 comma 2. 1i plus 5j is a vector that means 1 comma 5. So what is the dot product?

Oh, you pressed the wrong button. Thank you. Wow.

Okay, everybody seems ready. So let me first ask you a question. A dot product gives you, as an answer, a vector or a scalar?

Scalar. Scalar, why is that? Yeah, it's a number, right? It reflects the amount of overlap of the two vectors.

It doesn't have a direction. The dot product does not point in any specific direction. It's a scalar, okay? So what does that tell you that the answer is d and e? They have to be wrong, right? Because they're vectors, okay? So try that again, those of you who voted d or e.

Okay, very good. Somebody's insisting on d.

Okay, so this is straightforward, right? You just multiply the components, so a times a plus b times b. Sorry, so the x components multiplied by each other plus the y components multiplied by each other gives you 15. Okay, questions about that?

Okay, good. So we understand a little bit about work. We know how to do a dot product. So now we can combine those two and talk about work in two dimensions. Okay, because I wrote over there somewhere work is force times distance, right?

Now what happens if the force and the displacement are not in the same direction? So say for example you have a block on a surface and you pull this block with a force slightly up.

And the block slides some distance along the surface without leaving the surface. The force vector delta r vector, okay? Now first of all, how can this happen? How can it happen that you can pull on the block giving it an x direct, x accelerate, x force and a y force without it leaving the surface?

Well, like there's energy downward and the y component of the force that you're using to pull the force upward will not work on energy. It would just lessen the friction. Okay, but I'm confused because before I pull on it, it's sitting on the table, right?

The forces are balanced. All I've done is add another force. Why isn't that lifted off the table with any tiny amount of force? Because the force that you're sufficing with to take it upwards is taking away from the normal force. That's right. The normal force is reactive. It changes, right? The normal force balances all the forces on the surface.

If I'm not pulling on the rope at all, the only force on the surface is mg down. If I start to pull on it, there's some component here, Fy and Fx. If this one is weakening the impact of gravity, then the normal force decreases, right?

So if this force completely balances mg, then the normal force goes to zero. Could normal force be overcome? Sure. I mean, you ever crushed something, right?

You step on a beach, right? You imprint your foot into the sand. In our models, usually we assume the surface can provide any amount of force to avoid reform, unless you're explicitly told. In reality, of course, normal forces have limitations.

Otherwise, we would all be impervious to bullets, right? Which would be pretty cool. Okay, so in this case, it's just good to ask yourself these kind of questions. Like, do I understand physically how this makes sense? But in this case, we're talking about the work, right? So the object is dragged in this direction, the force was in this direction, so how do I calculate the work?

Well, the work generalized in n dimensions is force dotted with the displacement vector, okay? So force is a vector, certainly. You can have x, y, z components.

Displacement, also remember, is a vector. Here I write delta r to indicate that it started here and it went there. In this case, the force is fx, fy, dotted with delta r, which is, we'll call it delta r comma zero, which is like a magnitude.

So you see that only the x component matters here, because it only moved in x. You can also write, see how this is magnitude of f delta r times cosine of theta, right?

Which is just the x component of the force times the magnitude of delta r. Delta r is change of position.

Well, sure, r is our general way of describing position in vector coordinates, so delta r is just change of that. I'm not trying to create a new notation. Any other questions about that? So the work doesn't directly correlate to the actual amount of energy used to exert that force, right?

Because you could pull up, but that wouldn't be included in the work. It's just the amount of force used to take a certain distance. Well, if you pull up, you're having to apply a force, which feels like work w-o-r-k, right? But if you're not moving the object in the direction you're applying force, you're not actually giving it any energy, right?

Like, what is the kinetic energy in x and in y for this object? It's still zero, right? You haven't changed its potential energy or its kinetic energy in any sense, so you haven't done any w-o-r-k work on time. In y, only in x.

Remember, work is a scalar quantity, right? It's a number, it's a amount, it doesn't have a direction. It has contributions from motion x and from motion y and from motion z. But work itself is a number. Are you? w-o-r-k or w-e-r-k?

If there's no motion, there's no work. So where does that energy go? When you're pushing against the wall, how does it dissipate? What energy? The energy you're using to exert your force on the wall without moving it, where does it go? How does it dissipate? Because I know it's like heat and sound, light, but none of those are here.

Well, let's think about a different case. Let's say somebody gives you something really heavy to hold, right? And they tell you to stand there and hold it. And they say, well, this takes no work if you keep it at the same height, right? But you're holding like a thousand kilograms above your head. Obviously, you're going to run out of energy to do this, right?

So in that case, where is the energy going? It's going in heat. You are expending energy out of your body, right, to apply a force. Your muscles are burning energy and you're going to get sweaty. You're not transferring any energy from your system to the block.

Just like if I'm pushing on the wall, I don't even get sweaty and hot, but I'm not moving the wall. So I'm not transferring any energy to the wall unless I heat up the wall. So it's like you're supplying the normal force just to keep it up, but a table could do the same thing and not exert any energy. Yeah, it doesn't use any energy.

It's got its bonds, right, which act like a spring. We could replace you with a spring, which would last a lot longer, right? You're an efficient spring, I guess we can do. Okay. That's good. It's good to think about these situations and try to understand the energy to put it up. Okay, so let's do an example problem from the book, okay?

Chapter seven, problem one. Okay, it says a mass of 2.5 kilograms is pushed 2.2 meters by a 16-newton force,

16 newtons directed 25 degrees below the horizontal. Okay, so I think we need a diagram here. So obviously we have a surface. We have an object on this surface. There's a force applied below the horizontal, so that means that it's basically being pushed like this, right?

This is our force, and it's 25 degrees below the horizontal, right? So if you have an object on a surface, you're not just pushing it sideways. You're also slightly pushing it down. That's the configuration of this problem. Part A says find work done on the block by the applied force, okay?

So let's do a free body diagram, right? We have Mg down.

We have the normal force. Then we have this force here, right? So we have the pushing force here, which has components in x, f cosine theta,

and a component in y, sine theta. Okay, everybody agree those are the three forces in the block, which have been broken down into four forces because this one is in two different directions?

Everybody happy with that? Okay, good. So now it's asking us what is the work done on the block? So how do we do that? Yeah, exactly. So we can either do it the full direct way and say, well, it's f dotted with delta r, right?

And this is f cosine theta, f sine theta dotted with 2.2 meters comma zero, right?

f delta r cosine theta, surprise, surprise, right? Or you could say, well, look, I know the motion is only in one direction. I don't have to use the 2D value. All I need to do is take the one-dimensional, the projection of the force in the direction of motion and use that and say f cosine theta delta r. All right, so you can either use the full dimensional solution

or you can recognize that it's really a one-dimensional problem and go straight to that. This way always works, though, and I find it useful for thinking clearly about everything that's involved. OK, next part of the problem is it says find the work done on the block by the normal force from the table.

OK, find work on the block from the normal force.

So how do we do this one? It's zero, y. The normal force only points in y and the motion was only in x, right?

So we could write it out and say it's the normal force dotted with delta r, but the normal force is zero in x and mg minus f sine theta, no, plus,

plus f sine theta in y dotted with delta r zero. So this is zero because there's no components in common, right? So the normal force does no work, even though, if you imagine it was full of a bunch of tiny little weightlifters,

they would be getting sweaty, right? They're doing no work because it doesn't move in their direction. And then the next part of the problem says... Would you say they're orthogonal? Yeah, they're orthogonal. They have no... orthogonal just means perpendicular or not. Orthogonal also means that the dot product is zero, the definition of orthogonality. Okay, find the work done on the block by gravity.

Find work by gravity. Okay, same thing here. So the work is mg dotted with delta r, but g only has the y component, right?

The only thing that does any work is something that's pushing in the direction of motion, which in this case is x. Okay, and then the last part says d, find the net work.

What does that mean, the net work? Yeah, find the net work. I can't get in the internet. Just add up all the work, right? A plus b plus c.

So the net work is the work of all the forces, the pushing force plus the normal force plus gravity. Alright, but the only one of these, this is non-zero, is the pushing force. Okay, any questions about that?

Okay, let's do another one. As usual in physics, the number of concepts are pretty simple, but a lot of the wrinkles come out when you try to apply them to all the actual problems.

Okay, so I have a problem from the book. A raindrop of mass 3.35 times 10 to the negative 5 kilograms

falls vertically at constant speed, that's going to be important, under the influence of gravity and air resistance. If it falls 100 meters,

what is the work done on the raindrop by gravity and by air resistance? Okay, so let's draw our free body diagram. Forces on the drop are gravity, air resistance, which is going to be pointing up.

We know air resistance is minus bv, etc. But we're not told the velocity here, we're just told it's at constant speed. Alright, so what does that tell us? Constant speed means change in velocity is zero.

It's just another way of saying change in velocity is zero. Change in velocity is acceleration, so acceleration is zero, which means net force is zero, which means it's in equilibrium, right? Which means mg equals r, right? Which means magnitudes, I guess.

Okay, so now it's asking us what's the work done on the raindrop by gravity? Okay, so work by gravity. So what do we need to know to compute the work done by gravity? The force, right? So it's the force times the distance.

The force done by gravity is just mg, right? The distance is delta h, right? So this is just mg delta h, which you could have done either by saying work is force times distance,

or you could say, well, it's just a change in potential energy. And the change in potential energy is mg times the change in height. Two ways to get to the same answer. Okay, and if you plug this all in, you get 3.3 times 10 to the negative 2 joules.

Okay, and the next part of the problem says, what is the work done by air assistance? So what's the work done by air assistance?

Same thing? Opposite, why is it opposite? We know the network has to be zero. The network has to be zero, why is that? Because the air is pushing upwards, and the thing is moving backwards, so the distance traveled is effectively negative. Okay, or you could say, well, I know the work is delta r, right?

And these two guys point in opposite directions. Air resistance is pushing up while the drop is moving down. And this is going to give a negative answer.

What is the force? So I was sloppy a little bit here defining my axes. Let's be careful about that. I defined gravity to have, let's say, yeah, so it looks like I'm saying this is plus and this is minus, right?

So delta h is positive, mg is positive. So force due to air resistance is then going to be negative, right? So it's going to be negative mg. I don't really need two components here, but whatever.

And this is going to be delta h, right? So the work done by gravity and the work done by air resistance are opposite to each other. Why is that?

Right, they have opposite directions and they're equal, right? And of course the motion is the same. Is it always the case that air resistance and gravity will do opposite work? Or about when the drop is just formed?

It's still accelerating, right? In this case the drop is not accelerating. It's not gaining any kinetic energy, right? These two forces are balancing each other out. When the drop is first formed in the cloud, it's accelerated towards the ground and work is done on it by gravity.

It reaches terminal velocity when these two things balance. That's right, at terminal velocity the forces are equal and opposite, so the amount of work done is also equal and opposite. Okay, so up until now we've only considered constant forces.

But sometimes the amount of force you apply can vary. So say for example, you're pushing something and you're getting really into it and you get tired and you don't put as much force into it.

Sort of like your chemistry homework. And then, how do you calculate the force down here, if this is force, over this period? Well, what you can do is say, well, I'm going to approximate in some tiny sliver

of space that the amount of, that the force doesn't vary very much over this little sliver. And in this little sliver, I can calculate the work. Right, it's Fx delta x, right, force times distance.

And then the total work for the whole thing is just the sum over all these x's, over all these x's of Fx delta x. Right? So I can just add all these little slivers up.

Now, of course, sometimes the force is varying pretty quickly, so you'd have to take very small slivers to make this approximation valid. And so, of course, in the limit, delta x goes to zero, this just becomes the integral.

So in the case of a constant force, you can just say the work done is the force times the distance, because the force is constant. If the force is varying, then you take the force and you integrate it over the distance, from the initial position to the final position.

Now, if the force is constant, you can also start with this equation. What happens if the force is constant? You can just pull it in front of the integral, right? And then the integral dx just becomes x. So it reverses the same thing in the case of a constant force, right?

You'll notice this in physics a lot of times. You have a few versions of the same equation with different assumptions, right? Like we have the equation of motion for constant acceleration, we have the equation of motion for zero acceleration. There's often the most general case, and then useful simplifications that add assumptions. This is a more general case of the work expression.

Usually, work equals force times distance is accurate, because it assumes it's constant force. So we showed you that one also, so you don't have to re-derive it every single time from this one. But this is the more general case. It doesn't replace that equation, it sort of subsumes it. This equation turns into that equation and has the right assumptions.

So let's do an example.

This is from the book, Chapter 7, Problem 14. We have a force 4x in the i direction, plus 3y in the j direction, this many newtons.

And you notice this is a vector force, right? It's an x component and a y component. The x component depends on the x position.

The y component depends on the y position. So the larger your x and y values, the larger the force. So it's definitely not a constant force. x is on an object as it moves in the x direction from 0 to 5. So the initial position is 0, 0.

The final position is as it moves in the x direction, so it's 5, 0. Find the work done on the object. Now I've slipped into one more concept, right?

Not only are we doing an integral problem with varying force, but it's in two dimensions. So this is the one-dimensional integral equation. This is the multidimensional one.

See, force dotted with delta r. It's an integral where the infinitesimal is a vector. So that's going to look confusing, like what does that mean, how do you integrate d or r? Well remember, this is just a shorthand. This is just a way of writing several things at once. So it's the integral dx fx plus the integral dy fy.

Because this dot product takes the f component and dots it with dx. dr here is just a vector of infinitesimal.

Just a shorter way of writing it. So let's use that equation here. The work is an integral of f dot dr.

So let me plug in the values I know. This is 4x comma 3y dotted with dr, which is dx dy. This is the integral of 4x dx plus the integral of 3y dy.

Everybody follow that step? I'm just wondering how you do your limits of integration for f. How do you write it? Because you have information for both. Yeah, so you need to write a new vector. You do ri rf, right?

Or you could explicitly write components here. You could say 0 comma 0 to 5 comma 0. Vector notation, remember, is just a shorthand way of writing several things at the same time.

So avoid having to write out x, y, blah blah blah. Also, mathematicians love writing things in a short way. So any time they find something interesting, they say, let's figure out a way to write that with only two scratches. Because mathematicians spend their life with pencil and paper, so they try to minimize the amount of pencil scratches they have to make or something. Also, it's like shorter equations are prettier.

So that can be confusing sometimes, because it's going to be sometimes hard to see into the notation to see what it really means. So that's why when I look at this, I always try to unpack it in my mind, remind myself what this really means physically, so you don't get confused and lost just in writing down the notation. Also, it's a good way to check that what you're writing down makes sense.

Okay, so now we have these integrals, 4x dx 3 plus 3y dy. These turn out to be pretty straightforward, right? And this is evaluated from 0, 0 to 5, 0.

Okay, and I get 50 joules. Okay, so that combined a lot of concepts we used. Varying forces, dot products, and work.

The y's drop out? That's right, there's no change in the y direction exactly. So this is just 2 times 5 squared. And we could have known that from the beginning, right? Because we knew that the motion was only an x.

So I just wanted to show you how you can do it for our calculation. Also, notice how they conveniently put x variation in the x force and y variation in the y force. Could have been the opposite. Could have been y variation in the x force and x variation in the y force. Would have made this integral simpler, because here we really had a y, we wouldn't have actually integrated it. We just pulled it out.

Other questions? Okay, so let's take our new concepts, energy, work, force, etc., and apply them to a very useful physical system, the spring.

So we have a spring, there's a block here. There's a spring attached to the wall. This is where the spring likes to rest. We call this x equals 0.

The thing we notice about a spring is that it resists motion from x equals 0. Looks like x equals 6. There is this motion from x equals 0.

Meaning that if I pull the spring so the block is here, then the spring applies a force this direction to return the block back to x equals 0. If I push the block over here,

then the spring applies a force this way to push the block back to x equals 0. And our model for the force applied by a spring is f equals negative kx.

So what does that mean? Well, x equals 0 is defined as the resting point of the spring, so plug in 0 here, you get 0 force. Makes sense. K is just a number associated with every spring. A strong spring has a large k, a weak spring has a soft k, a small k. Now, why the negative sign?

Well, let's think about what this force looks like in x. So positive x gives negative force. Negative x gives positive force. If our axis is in this direction here, plus x,

then that makes sense because if you are in the positive x region, it's going to apply a force this way, which is definitely negative. If you're in the negative x region, it's going to apply a force that way, which is positive. So a spring resists the motion away from x equals 0.

And k's is a constant you can measure for a spring. Depends what the spring is made out of, how tightly it's coiled, etc. So a lot of the complexity of how springs work is encapsulated in just a simple number k. Now, you might ask, how true is this? If I go out and I take a real spring in real life and I attach it to some, you know, some meter

and I measure the force of various distances, how well will that work? Well, that's a cool physics problem, right? To test this model, how well does this model work? In this class, we're going to assume the model works perfectly and we're living in a universe where we have ideal springs. But remember that this is a simple model, right? Remember in the first class I talked about how we build simple models to solve problems.

It's a simple, very effective model which is mostly going to give us the right answer. And springs actually do behave remarkably well according to this, according to this equation. Okay, it's remarkably accurate, okay? Now, if you want to think about the work done by a spring, work done by a spring,

well, the work is the integral of the force, dot over dr, but we'll just think about one dimension. So it's the integral of fx dx, right?

In the case of a spring, this is negative kx dx, right? I've put in this equation that we know for the force of the spring, right? So this is just one half kx squared evaluated from the final to the initial position squared.

So this is the work done by a spring. You notice the work goes as x squared. Why does it go as x squared? Because as you push further and further away from x equals zero,

the spring pushes harder and harder. Okay, so let's do an example from the book. So the example says an archer pulls her bowstring back 40 centimeters

by exerting a force that increases uniformly from zero to 230 newtons. Okay, so we have an archer, pulls her bowstring, hits the bow,

leaves the string at rest, pulls her bowstring back to some position here. This is 0.4 meters, okay?

And this we're obviously going to treat as a spring. By exerting a force that increases uniformly from zero to 230 newtons. What does that mean? Well, if we write the force as a function of distance, right, it means that force at x equals zero is zero.

Force at x equals 0.4 meters is 230 newtons. All right, but let's think carefully about our sign. All right, so let's define positive to be over here, plus x minus x, okay?

So then this is going to be negative 0.4 meters, all right? Okay, so the question says, A, what is the spring constant?

Okay, treat this bow as a spring and tell me what is the k value.

So how do we figure that one out? Well, force equals negative kx, right? And we know the force, 230 newtons, is negative k,

and we know the distance is negative 0.4 meters, right? So sometimes you look at one of these problems and you're like,

I have no idea what you're asking. What? And usually that's just because you haven't understood that the problem is much simpler than you might be imagining. Usually in these cases, the first part of the problem is just like, have you understood the setup? All right, you know this equation, I gave you this, I gave you this, find this one.

The first part, part A, is usually supposed to be like a confidence building step. All right, so if you're finding part A to be very complicated and tricky, then you've misunderstood something and you need to go back. Okay, questions about that? Does that make sense? Okay, part B says, how much work does she do?

What work does she do in pulling back the spring, pulling back and drawing the boat? The work we just discovered of a spring is one-half k delta x squared, right?

Which, if you didn't remember, you could very easily figure out just by integrating the force, negative kx, from zero to x, right? You can re-derive this in two seconds, so you don't have to memorize it.

But we already have it right there on the board, so we just plug it in. So it's one-half, we know what k is, we know what x is, 46 degrees.

Okay, so you see how solving this kind of problem requires applying the physics model to these books. I say, oh, I think this is definitely a spring problem. I'm going to model it with my, model the boat as a spring, and once you've made your decision, I have this physics problem, I'm going to use this mathematical model, you send all the interesting physics.

The rest is just figuring out which parts of the model you have, which parts of the model you need to derive, turning the crank on the equations. The physics is saying, I have this problem, which model will I use? Okay, alright, and then since we have a little more time,

I was going to just do a bunch more examples. Unless anybody else has a question. Because chapter six has a bunch of sort of crazy problems.

So here's chapter six, problem eleven. A four kilogram ball rotating in the horizontal plane supported by two strings. Alright, so this one has some sort of pole here, and then there's a ball, and the ball is being attached by one string,

and by a second string, and it's moving in the horizontal plane, like this. And so we're also told the mass...

here is four kilograms and that this is two meters, two meters, this height here is three meters and then it asks us to build a free body diagram and find the tension. It really

says find the tension in the strings, okay? Find the tensions in the string. Well the first thing you do is build a free body diagram, right? You have your picture of the problem which helps you understand what's going on, but we want to understand the motion of the ball, so let's focus on that. So what are the forces here? All right, let's start

with gravity, mg, and then tension, somebody said? Which direction does tension go here? Two tensions, right? T and T, and let's call this TA and TB, okay? Now these guys are at an angle,

so it'd be very handy to know what that angle is so we could break these guys into different points into our force balancing equations, right? How do we figure out what that angle is? Yeah,

we're told this is two meters, this is 1.5 meters, so we can figure it out. Sine of theta is 1.5 over 2, giving theta is 48.6, okay? And that means a radius of 2 times

cosine theta, so 1.32 meters. So just, this one is R here, because this thing is moving

in a circle, right? Okay, so what do we do with that? Well, we want to figure out what TA and TB are, so what we do is we write down the sum of the forces and we figure out what that relates to, right? So in Y, sum of the forces, all right,

and let's be careful again about our axes. This is positive Y, negative Y, we'll call this positive X. Okay, so in Y, what are the equations? Gravity, right,

which is down, negative MG, and then we have TA Y component, right? TA sine theta

minus TB sine theta, right? And what is this whole thing equal to? Zero, right? It's moving in a horizontal plane, it doesn't move off the plane, so it's an equilibrium in Y. Okay, so here we have the equation, we know theta, we know MG, so we have one equation with two unknowns, fortunately we have another component, right?

So let's do the X component. Okay, so in this case the sum of the forces in X, so what are the forces here in X? So we have two tensions, right, and they're both

pointing in the same direction? Yeah, they are. So it's two, we don't know that A and B are the same, so TA cosine theta, TB cosine theta, although these

really should be negative, because I define X to be negative to the left, right? And what is this equal to? Is it equal to zero? MB squared over R, right? And I forgot that it says up here, rotating velocity is six meters per second.

Okay, so now we have two equations, how many unknowns do we have? We know mass, we know velocity, we know radius, we know theta, we don't know TA or TB, which is the point of the problem, right? And now we have two equations, and two unknowns. So in principle we can solve this one, right? So how do you do it? Well, this one you can write as

TA sine theta minus TB sine theta equals mg, right? Which means TA minus TB

equals mg over sine theta, whereas this one you can reorganize to say TA plus TB equals

MB squared over R cosine theta, and then you have TA minus TB, TA plus TB, and then add these two together to get two. TB is MB squared over R cosine theta minus mg

sine theta, and so there's a solution for TB, you can plug in the numbers we know,

and then you can solve for TA given any of these guys. So I get TB is 56 and TA is 108. So does that make sense?

Why are they so imbalanced? That's right, TA is balancing gravity while TB is not, right? In fact, TA is balancing gravity and TB. TB has to pull in order to make it move in a circle,

because TA has to compensate for that as well. So TA is balancing gravity and the part of TB which is helping gravity out. Other questions? Okay. Okay, that's actually the only example I

had here. So we'll end a few minutes earlier. If somebody has a question, I'll stick around. Yeah, there is rotational energy. You can do rotational work. We'll talk about that later.