Okay, so we have one more topic to cover before we round out this bit about energy and conservation. So today we're going to talk about conservation of energy and potential energy diagrams.

And after that we'll be moving on to the fun things like collisions and momentum and stuff like that. So far we've mostly been dealing with single objects.

So today I sort of want to bring together some ideas, show you one or two new ideas and do a bunch of examples. And then next time we'll move on to a new topic. Okay, so we've been thinking about how things move. For example, if you have a projectile, you have an object here and you throw it up in the air and it comes back down.

It starts here with a lot of kinetic energy. So a lot of kinetic energy and no potential energy. Here where you're throwing it. Here at the top all that kinetic energy has been turned into potential energy.

And then back here you've got kinetic energy again and almost no potential energy.

So the energy just shifts from kinetic to potential and back. You guys know that well. Why is it that you end up with the same amount of kinetic energy here as you started? Gravity is conservative, right? All the potential energy that's created is turned back into kinetic energy.

Also in this simple model here there's no air resistance, there's no friction, right? This is an isolated system from the rest of the universe. And so that's why there's conservation of energy for this guy, right? If we added air resistance then you'd lose some of the kinetic energy.

You'd lose some of the, you wouldn't go as high, right? And on the way back down some of the potential energy would get turned into heat of the air around it or wind or heat of the object and you wouldn't get all this kinetic energy. Because then in order to have conservation of energy we'd have to include all of the air also in our system, right? We only consider the ball in this conservation of energy with no air resistance, okay?

And the interesting thing is that the ratio here, kinetic energy to potential energy, depends only on the height, right? It doesn't matter how far out an x you are, it only depends, the ratio only depends on the height, okay?

And we've seen other systems with things that are only in the y direction, right? Like we talked about if you have a block and it gets moved by some force to some other location, right?

And you have, on this block you have some force and you have gravity, right? Now gravity is only in y, right? To calculate the work done by gravity you only need to know this delta y, right?

So the work done by gravity is only affected by the difference in height, why? Well you might say because gravity is a conservative force or you might think about the motion of the object dotted with the direction of gravity, right? Gravity only points down and so the work only matters in the down direction, okay?

What about a slightly more subtle example? What if you had something like this? Alright, you have a skateboarder on a track, okay? Imagine that's a perfect semi-circle, alright? And he starts here and ends up over here of course.

Okay, so when he starts up here at the top, gravity is pulling him down, right? And if this is in fact totally vertical, what is the normal force on the skateboarder at this point?

Zero, right? Because he's not exerting any force on the circle, he's basically falling next to it, right? As soon as he, and then he falls onto this first part of the curve and then it starts to push back.

So what about over here? Here, alright, he's got gravity of course pulling him down and he's got a normal force which is pushing out this way, right? Right through his head. It's a normal force.

Now between here and here, there's some delta y, so of course there's some work done by gravity. What about the normal force? Does the normal force do any work? No, why not? Alright, the normal force is always perpendicular to the surface, he's moving along the surface.

So the normal force is doing no work in this case, right? So here, there's no friction, right? Again, there's no friction, there's conservation of energy, right? His potential energy turned into kinetic energy. It doesn't matter that he took a semi-circular path or if there was a wiggle in here, right? The normal force is always perpendicular to the surface that he's on.

So as long as his skateboard is small enough to stay on the surface, the normal force is going to do no work, okay? Now if, let's imagine another case, more realistic, when there is some friction here. So what's going to be the direction of the force due to friction?

Opposite of the motion, right? So along the surface, but the opposite direction that he's going, right? So is friction going to do some work? Yeah, negative work, right? It's always pointing in the opposite direction of his motion, so it's going to steal some of his potential energy into heat rather than into kinetic energy.

So friction here does some work, right? Okay, so similarly, if you have, here's another example they talk about in the book. You have a rock in a hemisphere, right?

It starts here and it says it's a frictionless bulb, okay? So no friction, right? It starts here. How far is it going to go if it comes down and goes back up? How far up is it going to go? Exactly the same height, and the reason is?

Conservation of energy, right? The potentials turn into kinetic and turn back into potential energy. Because there's no friction here, all the potential energy goes to kinetic energy and then back to potential energy. If there was friction, of course it would slide down, not come quite as far up, because some of the energy would be lost due to heat, right?

And then back, and then it would slowly oscillate in this damp oscillation. And you know that means if you take the differential equations, they eventually become the rest in the middle with all of the potential energy turned into heat, right? Okay, that's pretty simple stuff. So in this case we have conservative forces like gravity, right?

You can take the energy and retrieve it, and non-conservative forces like friction, right? So, non-conservative like friction, okay?

What if we took a different example, and we said a spring in a block, right? Is the force from the spring conservative or non-conservative? Somebody's shaking their head, does that mean non-conservative? I mean if an object hits the spring, it'll compress, and once it gets back to the point

of decompression where it originally hit, the amount of kinetic energy is not going to be the same.

Right, so it's a conservative force if it's reversible, right? If you can get back the energy that you've turned into potential energy. So if you take your kinetic energy, put it into this kind of potential energy, and you can get it back, then it's a conservative force. In the case of friction, you can't unless you have some way to capture the heat of friction and turn it back into kinetic energy, right?

But in the case of a spring, if you stretch this guy out, right, so he's over here, then he has potential energy. The potential energy is turned into kinetic energy, but then it's turned back into potential energy, right? So there's no loss of energy here, right?

So the spring force is also a conservative force. If you have a system with just a spring force and kinetic energy, there's no loss of energy, right? The spring force, similarly, depends on only the position, right?

Okay, so now there's a useful way to think about these things, to think about the connection between forces and energy. Okay, so let's draw the function of the forces. Alright, so let's do gravity first.

Okay, so in the case of gravity, so this is the force and this is the potential.

Okay, so in the case of gravity, I guess this should be y, what is the force of gravity as a function of y?

A function of height? You know, it's basically constant, right? I'm assuming you're close to the surface of the Earth, it's just constant. Okay, now, and what about the potential energy? What does that look like as a function of height? Linear and positive, right?

Because this is mgy, right? So it goes like this, right? Now, does it matter where I put the zero? Potential energy, I'm going to draw an axis here. Does it matter where I put the zero? No, why doesn't it matter?

Sorry? It's only the change that... The change that matters, what do you mean that matters? Well, it's going to have some potential energy, like you could say that this has potential energy, but if it's not moving anywhere, it's not relevant, so it doesn't change. It doesn't have any physical impact, right? Yeah. Sometimes we have mathematical models that have mathematical consequences, but no physical consequences, right?

For example, you've got negative solutions to some quadratic equation, it's not a physical solution. You've got mathematical answers that are not physical. So in this case, potential is not a physically measurable quantity, right? You can't measure potential.

What you can do, however, is measure force, because force is mass times acceleration. The relationship between these two guys is force is the negative derivative of the potential, and this makes explicitly mathematically clear why the value of this doesn't matter, because force is the physical quantity.

You add a constant to u, it doesn't change the force, right? Because the constant doesn't have any derivative with respect to x. So now take this potential, which is linear in y, take its derivative, and you get a constant, which is exactly the force of c. So a linear potential, potential energy function like this, generates a constant force.

Now there's an easy way to think about this using your intuition, where you say, well, if I imagine this potential shape to be the shape of a hill, right? Imagine rather than just being a function, say, this is a physical object and it's a hill,

and I put a ball on it, which way would it roll? Well, in this case, obviously, it would roll down, right? So the force is to negative y. So here, I should have put a zero somewhere, right?

The force is definitely negative. The gravitational force is negative y. So you can either just use your intuition and say, what would happen if I put a ball on this potential on this hill? Or you can do it mathematically. You could say, if I have a potential with this dependence on y,

I can derive the force on an object in that potential by just taking the negative derivative. Let's do a slightly more complicated case of a spring, and then we'll do some exercises to get the hang of this, because it might be a new concept. So here we have force as a function of x, potential as a function of x.

So what does the force look like in the case of a spring? It's a negative kx, right? So if this is zero, the force itself is linear.

So what does that mean about the potential? It's parabolic, right? I mean, if you know this relationship and you have the force, then you can get the potential by integrating, right? But we already know the potential energy of a spring is one-half kx squared just by doing the integral of the force up with the distance, right?

So this is the potential of a spring. Now similarly with thinking about it as earlier, what would happen if you put a ball here? Well, we just did this example, right?

You would roll down to the middle, picking up speed, and then come back up to the same place and then come back down. And in fact, what happens if you take a spring and you pull the spring back some distance? Well, it oscillates around the equilinear point, right? Because it receives a positive, sorry,

you guys totally let me draw this the wrong way. It receives a negative force when it has a positive value, pushing it back this way, just like a ball rolling up this hill would receive, right?

And it receives a positive force pushing it this way when it rolls up this hill. Okay, so the only point I wanted to make is the relationship between force and potential energy for conservative forces. So this guy is, but that's a good idea, let's think about that.

So the nice thing about this is it gives you a rule

to derive the force just from the potential. So let's say you had some physical system, I don't know what it would be, you had some physical system that had the potential that you just described. If this was the potential, what would the force be like?

Negative one-half kx squared, right? What would the force be like in this case?

Align with positive stuff. So two ways you can do this. One is you can say, well, you just told me to take the derivative with respect to x to get the force, so I'll just follow your instructions without thinking about it too carefully and say this is f equals kx, right? Align with the positive.

Or you could say, well, I'm not sure I'm going to use my physical intuition. Imagine this was a hill and it had a ball here, but what happens if I put the ball exactly on the very, very tippy top of the hill? Nothing, right? Feels no force. Force at x equals zero equals zero.

Now as soon as it gets moved, so if somebody comes along and displaces it over here slightly from the center, then what happens? Do I get pushed back towards equilibrium? It rolls off the side of the hill, right? Then the positive x gets a positive force. I mean, when negative x gets a negative force, there's no chance to ever get back to zero. So you can describe a completely different physical system using a different potential.

So the potential is the key, and often in physics, as you'll see as you get to higher and higher levels in physics, it's much easier to work with potential directly than with forces. You can have potentials in various things. You can have a spring plus gravity, add them together linearly, right?

It's not a vector quantity, and from that you can derive the force for the equation of motion. And these are just simple examples to show you the relationship between force and potential. Is there a question? The question is, can you just...

The directions... Like, you define positive x. I say that's negative x. Flip the graph. Sure, but this is true regardless of your definition of x, right? And if you flip the graph, you have to flip all the graphs. So it's like you're sitting on the opposite side of the board.

Everything I wrote would be true still. You just see it from the other perspective. Did I answer your question? Okay. So another way to think about this potential is sort of like a particle trapped in a hole, right?

So if you had your spring set up, you displaced it a little bit. If you draw the potential for this guy, it's a parabola, right?

Now, displacing it a little bit means you go over here. These are units of potential energy, right? So essentially, you've given it this much potential energy. You've given it this much potential energy at the intermediate state.

And it's never going to get more energy than that, right? No matter what happens to it, it's never going to get more potential energy. So it's sort of trapped. Trapped here, right? And that determines how far it's going to oscillate. I'm just saying that when you pull it back initially, you give it a certain amount of potential energy. That sets the upper limit for how far it can never go,

which you could also see just by looking at this graph and understanding that it's going to roll around here. It's never going to go up higher than its original position, right? And so, for example, say you had a complicated potential. You had a complicated potential that looked like it comes down and comes up and then it would.

Okay? So this is u of x goes to the x. Then if you had a particle that started here, that particle would never be able to escape this web, right?

Why does that mean? Well, you can use the physical model of a ball rolling down a hill. If you put a ball here and let it go, it's never going to get over this hill. Or you can use your physics knowledge and say, to be on top of this hill, for example, here, would require more potential energy than it has, so it'll never get there.

It's essentially trapped in this well. Whereas a particle, if you started it here, give it this much potential energy, it would be constrained to this larger well. It's capable of existing here, because here it has this much potential energy, so there's still room for it to have kinetic energy.

As you know, if you imagine what would happen if you did the experiment, the ball rolls down the hill, speeds up, it's going fast here, starts to slow down, but still has enough kinetic energy so that it's still moving at non-zero speed when it gets here, and then it rolls up here. But it can't go over this hill. Whereas if you started a particle up here,

then it could roll over both of these hills and escape. So you can also think of them as like energy level diagrams. And then, to be thorough about it, let's think about the direction of the forces. So remember that.

So here, the derivative of this guy is negative. So the force, so here, the derivative is negative.

The force is the negative derivative. So here, this is going to have a positive force,

which makes sense, because if it's on this hill, we push in this direction. So the force is positive here, all the way to here, when the force goes to zero. So let's put a zero here. The force is positive in this region. And then here, the derivative is positive. We want the negative derivative, so then the force is negative,

until here, when the force is zero again. And here, the force is zero. And here, the force is zero. I'm finding all the zero derivative points to anchor my curve. Now, from here to here, the derivative of the potential is negative,

so the force is positive. And from here to here, the derivative of the potential is positive, so the force is negative. And why do I get these curves? Well, I'm just, you know, quickly doing the derivative.

Here, it's zero. There, it's very slightly greater than zero. Here, it's probably the largest positive value. Then the slope decreases back to zero. So that's why I get some of these curves. There's that for a second. Make sure I did it right. Reach yourself. It's accurate.

Okay, let's do a couple of examples on the clicker, just to get yourself used to thinking about relationships between force and potential. So let's do some practice problems. So here, you have a graph showing potential energy.

Let's turn down the lights a bit so people can look in the back and see. Potential energy. The particles, so this is a potential energy drawing, right? Not a force diagram. The particle is initially at x equals d and then moving in the negative x direction. The question is, at which of these labeled coordinates, a, b, c, and d,

does the particle have its greatest speed? So it's giving you the potential energy function, from which I want you to think about the kinetic energy and also the forces on the object. Okay, so who wants to tell us what I answered?

That's right. So he says it starts at d, right? So it must have at least this much potential energy, right?

And because of conservation of energy, if it ever has less potential energy than that, it must have turned into kinetic energy. We're interested in the particle's speed, so we want the place with the most potential, most kinetic energy, which is going to be the place with the largest difference in potential energy. Another way to think about it is it starts here. It was moving in this direction.

It came down this hill, and there was a force on it the whole time, right? This force was because there was a slope in the potential. The slope in the potential means a force. And so it accelerated, accelerated, accelerated until it came here when it starts to decelerate.

So yes, it's at b. Okay, what about at which of these coordinates is the particle slowing down? By which I mean its magnitude of velocity is decreasing. Okay, so who thinks the answer is d?

How about the c? How about the c?

It's increasing. Well, it starts at d, remember, and it's going this direction. Up. Good. It is, in fact, a. I suspect some of you made the same mistake.

Can it go back from a to d since the force is going to go back, so isn't c also an option? So he's thinking in the future, he says what happens when it gets to a, it turns around and comes back, right? Well, do we know it's going to turn around and come back? Do we know? Yeah, because we started at d.

The particle is at d, but we don't know the initial kinetic energy, right? I see your point, but it's possible that it could just go off the rails and go over some hump and never come back. But you're right, if it changed direction, then it would be slowing down at c. Okay, what about at which coordinates is the force on the particle zero?

Okay, so some people think d, some people think e. Why don't you discuss with your neighbor if you've got a different answer, argue about it, and then we'll vote again.

Okay, so some convergence. So let's hear from somebody who didn't vote d last time but was persuaded to vote e.

Yeah, so at the point where the slope is zero, there's no force, right? Anybody disagree with that? Well, we don't know if there's gravity here. It could be an outer space, something you're getting the potential, right?

Any other questions? Don't overthink the problem, right? There could be supernatural forces. All right, so it's here is the slope is zero and here the slope is zero. Okay, let's do another one.

Now we're looking at a force diagram, okay? This is not a potential energy diagram, it's a force function. The graph shows a conservative force f of x as a function of x in the vicinity of x equals a. As the graph shows, f is equal to zero at x equals a. Which statement about the potential energy function of x is correct?

Is it zero at x equals a? Is it maximum, minimum? Or d, which refuses to really say anything? Okay, so b is winning, but there's some people who still believe in c. So who thinks c is correct? Who thinks the potential energy is a minimum at x equals a?

You wouldn't listen to your neighbor, you voted twice, but you voted mid. Nobody? Anybody? Since force is the slope of the potential energy graph, it goes from negative to positive.

Okay, so imagine the force looks just like this one, right? So what does the potential energy look like? It looks like a hill, right?

So one way to do it is to take the force, think about the equation. This is going to be something like f equals kx, integrate that, right? And flip it, right? You integrate it and you get a minus sign, you get negative one-half kx squared. So the potential energy is going to be a maximum at x equals a.

Questions about that? Another way to think about that is this is the kind of force that's pushing the particle, pushing the object away from point a, right?

And so it's unstable equilibrium. If it's at a, it's zero, but if it's anywhere deviated from it, it gets pushed away. So that's, you know, potential energy of this shape has that effect.

Okay? Well, what about this one? So this point here is a. This is a, this line here. Okay, right?

So it's a minimum here for exactly the same arguments. And in this case, the force function would be something like negative kx. And so if you, and so if you integrate, if you integrate that to get the potential, then you'll see that it looks just like a spring, right?

It's negative one-half kx squared. So force is the negative derivative of potential. You're asking why is there a minus sign here? So another question might be why is there a minus sign here?

Is it? In the case of a spring, the force is pulling you towards it. In the case of a hill, you're rolling away from it.

So I feel like you've given me a couple of examples which demonstrate this is true. It doesn't really tell me why it's true, though. Why?

Why does one fall from the other? Why does a reversible force determine this relationship? Still thinking?

Let's just say the force does work on the system, and then the system does work. It doesn't do work, that's what I'm trying to figure out the wording. But it, you're doing work on the system, and then that's just, you're using energy. Okay, anybody else?

Let's think about what is potential? What is potential energy, right? Potential energy really a physical quantity? I mean, I can have an infinite amount of potential energy, right? I can just define myself to have it. Say, yeah, say, you know, the potential is mgh, and I define my h equals zero to be an alpha centauri.

I have a mass of h. Look how strong I am. Right, so is potential energy really a physical quantity? Not really, right? It's a construct that we use that's convenient mathematically. You'll see this later on in physics, for example, when you talk about particle physics, you talk about particles having a wave function. Wave function is not something that's physical, or it can never be measured.

It's square can be measured, right? Wave function squared gives you the probability distribution for a particle. So not every quantity you can calculate necessarily is physical. Anyway, think about that one some more. Think about your best answers. Send them to me and we'll see what people come up with. It's a bit of a philosophical point. Okay, so let's do some example problems.

So here's one that looks nasty in the book. Okay, it says a three kilogram block is connected to two springs, like this, like this,

having force constants k1 and k2, where k1 is 25, k2 is 20.

The system is initially in equilibrium on a horizontal frictionless surface. The block is now pushed 15 centimeters to the right, displaced 15 centimeters to the right, and released from rest.

What is the maximum speed of the block? First, let's just think about what is going to be the motion of this guy, right?

Are the springs going to be acting together, or are they going to be fighting each other? Who thinks they're going to be fighting? Why? Well, imagine what happens. What's the direction of the force here from this grid? It's going to be pulling it back to equilibrium, right?

And from k2, it's going to be pushing it back to equilibrium, right? So they're reacting together. How can they be acting together when it's f equals negative kx, right? It should be negative kx and you have the same x. The x's would be negative because you're stretching one and you're compressing the other.

You have to be very careful about your axes here, right? The kx is relative to the equilibrium position, right? And so one of them has an x starting from here, the other one has an x going in the other direction. And let's think about the potential. What does the potential energy in this thing look like?

Is it a zero? What does the potential energy of one spring system look like? It's a parabola, right? What happens if you add another spring to it? You have another parabola, right?

Slightly different slope because it's a different k, right? So it's going to be a tighter or a wider parabola. Just add the two together. That's the potential energy of this system. What happens if you add two parabolas? You get a funny shaped parabola, right? So the sum of them is going to look something like this.

So already let's just look at the potential and understand what's going to be the motion of this system. Is it going to be some weird, crazy thing like when you have a pendulum on a pendulum and it's all chaotic? No, it's going to be pretty simple. It's just going to look like one spring, right? In fact, you could probably find some way to combine these two k's into an effective k for a single overall spring.

You look at the potential shape, you can get a sense already. What are they going to be, the dynamics of the system? You know you're not going to have to go crazy calculating weird forces on it. You just have to think about these two springs acting together. So the potential, drawing a potential diagram can be helpful just in getting a sense for what's going to be the motion of this system.

So let's get back to the question. How do we answer the question? What is the maximum speed of the block? The place where it has no potential energy? So we're giving a potential energy by displacing it.

Now we're asking where is it going to have the maximum speed which means the maximum kinetic energy. It's going to have the maximum kinetic energy when it has the least potential energy because of conservation of energy, right? At the equilibrium position. Just like in a simple spring question, right? Take any spring, you pull it apart, you stretch it and then you ask

where is it going to have its maximum velocity? It's going to have maximum velocity at the point of minimum potential energy because that's where most of the energy is turned into kinetic, right? Just like the little diagram we thought about where the ball is rolling down the potential shape. Or if you just look at this guy and say, well, it's going to be here, right?

Through the place of minimum potential. It's zooming down the hill, it's going to be going faster here and then slowed back up. Okay, so let's do energy conservation. Initially we have only potential energy so we can not even write the kinetic term, right? There's no friction here so it's a simple setup, right?

Now the maximum speed is going to be when we have minimum potential energy so we'll set that to zero, right? We'll say when UF equals zero, okay? And that we know is going to be at X equals zero. So what is the potential energy in the initial state?

Well, it's one half k one X one initial squared plus one half k two X two initial squared equals the kinetic energy, right?

One half mV final squared. Okay, now we have to think, so this is what we want, right? We're interested in this speed. We know the mass, not a problem. We know the k's, not a problem. What about the X's? We have two X's.

Well, it's really a one dimensional problem, right? And it's a single block so you can think about it as just two frames, right? Obviously there's going to be a very tight connection between X one and X two, right? This is positive X one and this is positive X two.

So what can we do? Well, first, are we sensitive to the signs? X one squared, X two squared. We don't even really care in the direction, right? So what can we say?

Let's just treat them as one, right? Because a displacement of one centimeter in one direction affects both springs by the same amount, right? So we can just say, see, in fact, we can treat it as a single effective spring

and the effective constant is just the sum of the two. k's are newtons per meter.

So let me ask you a question. Does it matter that I drew this other spring on that side? What if I had said two springs, identical length, different spring constants? The problem would be exactly the same, right? But it's trickier.

It looks trickier when it's attached to the other wall, right? Because you're imagining some potential craziness, okay? So from here, it's pretty easy because we know how much it's been displaced. They told us 15 centimeters. We know k one, we know k two, so we can just solve, okay?

So just plug in the numbers we get, right? And we get six units per second. Now, there's a second part of the problem which says,

what is the maximum compression in distance of spring one? All right, so we started out compressing spring two, and now it's going to get kinetic energy, go through the equilibrium point,

and then start to compress spring one. The question is basically how far is it going to squish spring one before it stops? 15 centimeters. He says 15. He's trying to make an argument from symmetry. That would be valid if it had the same spring constant, like a dump, all right? So, like, wouldn't it matter where you started?

It does matter where you started, yeah, but we started at 15 centimeters on k two. So what do we know about the point when it has the maximum compression? What's its speed at that point? Zero. Zero. Zero kinetic energy, right?

So it has zero initial kinetic energy and zero final kinetic energy. So the equation of energy conservation is pretty simple, right? So you say, well, one-half k one x squared is one-half k two x squared, right?

Not quite. You might be tempted to do it this way, but remember, in the initial configuration, both springs are extended and both springs are applied of course. Also in the final configuration, when it's come and squished k one,

both springs are extended and both are applying a force, right? So you have to account for both in both cases. So make this initial x final.

Both springs give potential energy in both cases. So what do you notice here?

All the halves cancel, right? So in fact, there is a symmetry, right? There is a symmetry and you were right.

It's 15 centimeters here, 15 centimeters the other direction because both springs are acting on it in both cases. We're not passing it off from one spring to the other. So in fact, it is x initial equals x final. You were right and I persuaded you you were wrong. Stick to your intuition.

So it doesn't matter which side you start on. What's that? It doesn't matter which side you start on. It doesn't matter the ratio of the spring constants. One can be very weak and one can be very strong or they can be equal. Put a technically negative. I'm sorry? Put a technically negative.

Yeah, you have to be careful how you define it. Yes, thank you. Is this all under the assumption that they both have the same equilibrium? Yes, they have the same equilibrium position, exactly. Okay, let's do another one.

A sled with a rider having a combined mass of 125 kilograms,

125 kilograms. That seems like a lot for a sled and a rider. It travels over the smooth icy hill shown here. How far does a sled land from the foot of the hill?

The question is, what is this? Oh, and its initial velocity, 22.5 meters per second. How fast is 22.5 meters per second? Pretty fast.

Is that a running speed? How fast does the interesting bull run when he wins the bull belt? He runs 100 meters in 10 seconds, right? So that's 10 meters per second. This is blisteringly fast for a little fatty on a sled. Somebody must have loaded up a big slingshot.

Okay, so this is an energy problem, but it's also a kinematics problem, right? So let's think about how we're going to do this. We can use conservation of energy to say how much kinetic energy does he have left at the top of the hill, right?

And then we need to know how far is he going to go. If we have kinetic energy here, we know his horizontal velocity. And then we just need to figure out from kinematics how far he's going to travel. We can do that by figuring out how long it'll take him to fall and how far his velocity will take him in that time, right? So step one is find the x.

Step two is find the time of flight, right? Because step three, the distance is the velocity times the time, right? So how do we do this? Well, how do we find the velocity? Do we have how high the money is?

Yes, 11 meters, thanks. 11 meters high, which is a huge hill, right? So, initial potential plus initial kinetic is final potential plus final kinetic.

This guy is zero. One half mv squared is the number we know, right? So we're given the initial velocity. The final potential energy is mgh.

We know mass, we know g, we know h. So this is looking pretty straightforward, right? So we know everything in this equation except for the thing we're interested in, which is the final velocity, final being up here at the top of the hill, right? So vf squared, so the masses cancel, bring the two up.

vf squared equals vi squared minus 2gh, so I get 17.1 meters per second. So it doesn't slow down a whole lot, right? Now, is it important that we know the shape of this guy at the top, the shape of the hill?

Well, what did we just calculate? We just calculated basically the magnitude of the velocity, right? To know how far he's going to land, it's also important to know the direction. If, for example, the hill looked like this, he would have a different trajectory than if the hill looked like this, right?

So in this case, the top of the hill is flat, which means we can assume that he has zero y velocity. And when he's up here, then he starts to fall. And all of this goes into his x velocity, right?

So if he has zero y velocity, all we need to do to find the time of flight is to calculate how long it takes to fall from rest, 11 meters, right? Which is not hard, right? And it's just one half a t squared, right?

Where we know y, it's 11 meters, we know a half, we know g, so we can just solve for t, right? So I get 1.5 seconds. So he's got the thrill of his life for a second and a half.

And then for the third part, the distance he travels. Here this is easy because once he leaves the top of the hill, he's no longer losing velocity, right? His velocity in the x direction is constant. So this is 1.5 seconds times, thank you, 17.1 meters per second.

So I get 25.6 meters.

So why doesn't his velocity, so here his x velocity is basically constant, right? But from here to here, he's gaining y velocity, right? What's going to be the magnitude of his y velocity just before he hits?

What's that? And why is that? Because conservation is energy. Yeah, exactly. Good, okay. His magnitude, his total velocity here, right? His x velocity, just before he hits, his x velocity is still going to be 17.1.

Nothing is going to change that until he splats, but his y velocity is going to increase until it makes up for the potential energy that he's gaining here, turning into kinetic energy.