 Okay, so here is a work energy example with friction. Suppose you have a car and let's say it has an initial velocity in this direction of let's say making up something as usual 20 meters per second and you're going to stop, okay. Let's just see. Wait, let me make sure. Real quick here. I should have done this beforehand but I'm not prepared. Okay, that's fine. Um, and so let's say it has a mass M and let's say it has a coefficient of let's say it's let's say you're slamming on the brakes and the difference between slamming and not slamming is slamming the wheel stop if you have good enough brakes and the tires slide. Not slamming with any of the any lot brakes do this. They roll. Okay, so you're using static friction for any lot brakes and kinetic friction for sliding. You could really do the problem either way but I'm just going to randomly choose sliding mu K value and I'm making up a value of 0.5, okay. So that's the coefficient of static friction. And so you want to stop this car and the question is how far does it take you to stop? Right there when it says far you should say, okay, you have two tools. The momentum principle and the work energy principle. But since we're trying to say how far does it take to stop, here's my car again and we'll call this distance S, then we should think oh it'd be better to do that with the work energy principle because that uses distance. You can do this with the momentum principle and you should try that but it's just a little bit easier with the work energy principle. Okay, so first let me draw a diagram, a force diagram while it's stopping. What force are acting on it? Well I have gravity, I have the normal force and then I have the frictional force. So in this case I do need to look at the net forces in the y direction because I need to find the normal force, I need the normal force so I can find the friction force. So in this case it's not accelerating up or down, so N minus mg equals zero, N equals mg. So the model for friction says the frictional force, the magnitude is equal to mu k times N equals mu k mg. Okay, but I'm not going to find the acceleration in the x direction, that's not what I want to do. I want to find out how far it takes to stop. Okay, so I can use the work energy principle so I can say the work done in the car is a change in energy in the car. And now the next thing is, well what's my system? This is an important part that a lot of people skip. I could choose, in this case let me choose just the car as my system. If you wanted to you could choose the car plus the earth but I think in that case it may be kind of difficult to say whether friction is a force acting on the outside of the system or inside. So I think it's better to do it this way. Okay so now I need to do the work done on the car. Let me just say that, write that as work done by gravity plus work done by friction plus work done by the normal force equals just a change in kinetic energy. If it's just the car it can't have potential energy, it only has kinetic. So I need to look at these three works. So first let me look at the work done by gravity. The work done by gravity is going to be the gravitational force times the distance s times the angle between the direction the car moves and the gravitational force. In that case that's 90 degrees so it's going to be cosine of 90 degrees which is zero. So gravity does no work on this as it moves along that way. Now let's look at the normal force. Again we'll have ns cosine 90 again equals zero. So neither the floor put the ground pushing up or gravity pulling down do any work because the car is moving perpendicular to that. So no work done. So now we just have the work done by friction. So that's going to be the frictional force mu k, mg times the distance s times the cosine of the angle between them and it's not zero. S is this way, f it's really delta s the vector that way and f is that way the frictional force is that way. So the cosine is the cosine of 180 degrees which is negative one. So I get the work done by frictions negative mu k, mg, s. So what does that mean? Can you do negative work? You can. Okay. If you do negative work that means that the energy decreases so we're okay there. Okay. Let me erase this because I need more room. And so that's where I'm going to start. I already have that work equals negative mu k, mg, s. And that's going to be the change in kinetic energy. So that's going to be k2 the final minus k1. So if I stop this car what's the final kinetic energy? So k2, one-half mv2 squared, but the final velocity is zero so that's zero. And then the initial velocity is, this is v1. So I have negative mu k, mg, s equals negative one-half mv1 squared, zero minus that. Okay. The negatives cancel. The mass cancels too. Okay. And then I can solve for s. S is going to be equal to v1 squared over 2 mu k, mg. Okay. Check the units here and check reasonableness. The units, meters squared per second squared over, this has no units, meters per second squared does give me meters. That's good. The faster I'm going, the longer it takes to stop. The farther it takes to stop. The greater the coefficient of friction, the shorter you stop. Those all make sense. So that's good. So let's just put in some values here because I know how happy that makes people. So this is going to be 20 meters per second squared over 2 times .5, this is going to be 1 divided by and times 9.8 meters per second squared. So that's going to be about, this is 400 divided by 10, so 40 meters, approximately 40 meters. That seems reasonable. 40 meters to stop. Yeah. Yeah. Maybe your coefficient higher than that probably is. I think in the previous problem I had a coefficient of .4 on wet roads, so okay. Now let's take this to one other example. What if V1 equals 2 times, I'm sorry, 2 times 20 meters per second squared, meters per second. So 40. What if you're going twice as fast? And nothing changes except in the end, I can calculate, let me calculate S2. It's just going to be, I'm going to put this as 2 times 20 squared over 9.8. So that's going to be equal to 4 times what we had before, so it's 160 meters. So when you double the speed, it doesn't take twice the stopping distance to stop. It takes 4 times the stopping distance to stop. Okay. Okay.