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>> All right, let's ask the following question. Let's say we have a skier going down a slope, and we're going to start and say it is a frictionless slope. Okay, and we want to calculate the speed at the bottom. All right. Let's draw a picture of what we're talking about. Here is our slope. We've got an incline here of angle theta. And now we've got a skier that is going down the slope, and we'll say that they start at a height h. Now, we know a little bit how to solve these problems, right. We've done these problems before with kinematic equations. Let's see if we can attack it from that point of view to start. So if I think about the kinematic equations, what can I say? Well, any time we're dealing with a slope like this, we usually want to go to a rotated coordinate system. And if I go to a rotated coordinate system like this, I'm not really worried about the y direction. I'm only worried about the x direction. And so I can simplify this to one simple equation: x final equals x initial plus (vx initial times t) plus 1/2a sub x t squared. That looks like it might help us, except if we're looking for v at the bottom, we don't really have v at the bottom in here. We have v initial, but we don't have v final, okay. So let's go back to our kinematic equations and let's look at a different one. vx final squared equals vx initial squared plus ((2 a sub x) times (x final minus x initial)). That one looks like it might help us because if you're starting from rest, we know vx initial. If we know how far you go, we know this. a sub x, we might remember what that is for a slope. Can you hand the mic to Kevin? Let me have a chat with Kevin real quick. It was Kevin, right? >> (student speaking) Yes. >> Okay, so if I'm on a slope like this and I'm thinking about my acceleration in this direction, do you happen to remember what I should put here? What is my acceleration down the slope? >> (student speaking) Should be mg cosine theta. >> Okay, very close, except I think we want a sine theta, right? If I think about the force diagram like this, then I break mg into components, and it turns out cosine theta ends up in that direction. Sine theta ends up in that direction. That is, of course, our normal force, and if there's no friction, then there's no force going back up that way. So the acceleration down the x direction is going to be g sine theta. The force down the x direction is mg sine theta. Okay, so in this case, we've got this. a sub x equals g sine theta. Now, how do you know that? How do you know it's sine theta and not cosine theta? Because if it goes flat and theta goes to zero, I had better not have any acceleration horizontally. Sine of zero goes to zero. Cosine of zero goes to one. So that's how you can double-check that. All right, what about this x final minus x initial? Well, that's just how far I go. So let's call this hypotenuse, how about we'll call it L. So this thing right here is just going to be L, and vx initial, we'll say we start from rest up there, okay. And now we can calculate vx final. We get vx final squared is equal to 2g sine theta, all that times L. All right. That looks pretty good, but we don't know maybe what L is. Maybe we were only giving, we were only given h. But of course, there is a relationship between h and L. It's that angle theta. And what we can say is sine theta equals h over L. And so I can solve this for L. L is equal to h over sine theta. And so now when I stick that in here, look what happens. I get 2g sine theta times L, which is h over sine theta, and the sines cancel out. And so that whole thing just becomes 2gh, and now I can calculate this final speed. vxf is the square root of 2gh. Okay. That's your speed when you get to the bottom. There's our person at the bottom, and they're moving at vx final. Remember, we're in this rotating coordinate system, so that is their speed at the bottom. So this is how you do it with the kinematic equations. But there must be an easier way. Kevin, is there an easier way to do this problem? >> (student speaking) Using energy. >> Yeah. >> (student speaking) Conservation of energy. >> Exactly. Let's do it again using conservation of energy, and let's make sure that we get to the same result. So we will call this v initial was zero. They started at rest. And v final is what we're looking for. Okay, so Kevin suggested we use conservation of energy. Conservation of energy is just that. Energy is conserved. Whatever energy we have initially has to be equal to whatever energy is in our system finally. And initially, we have two things. We have gravitational initially. We maybe have some kinetic initially. In the final picture, we're going to have gravitational final plus kinetic final. Now, gravitational we said was just, how high are you above the surface of the earth? If we start up at h, that just becomes mgh. Kinetic initial was 1/2mv squared, but we said that you're starting from rest up there, so that turned to zero. Gravitational final is down here when you're at the bottom of a hill, and now you're at a height of zero, so that term is zero. And you're moving at some speed 1/2mvf squared is therefore your kinetic energy. And now look how simple this is. mgh equals 1/2mvf squared, and you can solve this for vf. Cross out the m's. Multiply by 2. Take the square root. And you get exactly what we had before in a lot fewer steps, okay. Just straightforward. This is the great thing about conservation of energy. A lot of problems that were difficult to do using the kinematic equations are suddenly really easy to do using conservation of energy. And this is kind of a nice result, okay. This v is square root of 2gh. Why is it an interesting result? Because we did not take into account the angle theta here, right? We said the skier's going to go down this hill of angle theta, but theta never fits into this problem. So are we saying that the person that goes down a ramp like that is going to have the exact same speed as somebody that goes down a much steeper ramp? And the answer is yes. If they're starting from the same height and we're ignoring any friction, then that person going down the steep hill will have the same speed at the bottom as somebody going down this relatively mild hill. In fact, the hill is irrelevant in this case. If you dropped the skier straight down from the top of the hill, how fast would they be going right before they hit the ground? This right here -- square root of 2gh, okay. Now, in the real world, of course, friction comes into it. And friction is therefore important in this problem. So let's take a look at it again, but let's add some friction. And let's do one more thing. We're going to say that this person going down the hill is moving along at constant speed, okay. You've all been skiing or snowboarding. You know that after a while, you just move along with constant speed. Why are you doing that? Because there's friction between your skis and the snow. There's also air resistance that you have to worry about. And so typically, you're just moving down at constant speed. So let's say that that's what's happening. v initial is just some speed v. v final is just some speed v. And let's see if we can calculate how much energy is lost to thermal energy. How much energy goes into heat, all right? The energy transformation that we're worried about now is the following. When I start out, I have gravitational potential energy, but I also have kinetic energy. At the end, I get rid of the gravitational energy, but I still have potential energy. And I have generated some heat. This is our thermal energy, okay. But energy has to be conserved. If I can keep track of all the energy, then it has to be conserved. Conservation of energy says the following. Energy initially has to equal energy finally. And we know what it is. Energy initially we said is gravitational plus kinetic. Energy finally is kinetic plus thermal. Ugi in this case is just mgh. How high did you start? You start at height h. That's your gravitational potential energy. At that height, we were moving at speed v. So this is 1/2mv squared. Down at the bottom, we're still moving at speed v, so this is 1/2mv squared. And we have generated heat. Energy thermal. So in fact, in this problem, look what happens. We can cross out the 1/2mv squared from both sides. I subtract those, and how much heat have I generated? It's just how much potential energy I started from. All of the potential energy that you had initially went into heat. It heated up the snow. It heated up the skis. It heated up the air as you pushed it out of the way, okay. All of that energy is just equal to mgh. Now, you maybe don't know exactly where it went and how much went into heating of the snow and the skis versus how much went into air resistance, but that's okay. You still know that it has to add up. It has to go somewhere.

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