 Okay, another example with the work energy, in this case I want to do an example with the spring. If you take a lot of these toy guns that shoot something, almost all of them have a spring in there. So how it works is you take, here's your little toy gun, you have a spring in there and you compress it and then you have a ball and then it uncompresses and then it shoots out. Let's say that this has a spring constant K equals 10 newtons per meter and the compression distance S right there is going to be 10 centimeters, how about that? And I want to shoot this at a, let's say, 35 degree angle. And I have a mass, a ball, it's a little plastic ball and let's say it has a mass of 100 grams, so 0.1 kilograms. Okay, so I shoot it and the question is, I want to know if I shoot at this angle, how high does it go, so I want to know how high and how fast at the highest point, okay, because it's going to do this. So at the highest point it's still going to be moving. Now there's two ways to do this problem. The first way is to say, okay, I have a ball and a spring and I can find out how fast it's going right here and then I can use projectile motion to find out how high it goes. That is completely 100% valid, but I'm not going to do it that way. Instead, here I'm saying how high? So your little work energy light should go off saying, oh, use work energy because it's a distance, okay, so I'm going to use work energy. And I'm going to call this a starting point, y equals 0. You've got to, you know, there's a couple things when you're doing projectile motion and work energy, you need to say where y equals 0 is so that you can use that later. The next thing I need to say is, what's my system? In this case, my system's going to be the spring plus the ball plus the earth. Well, if I don't do that, what if I just had the ball? First, I'd have the work done by the spring during this part because the spring exerts a force on it. Not easy because that force changes as it gets further and further away. So that would be very difficult to do. Second, as it moves up and down, I'd have to include work done by gravity. It turns out that if I use all these things, if I have the spring in my system, then I can have spring potential energy. If I have the earth and the ball, I can have gravitational potential energy. So, if that's the case, what's the work done? Well, what's acting on the, during this interval right here, the whole time I have gravity, but gravity doesn't do any work because it's part of the system. The gravitational force from the earth is part of the system. The spring exerts a force on it, but that's part of the system. So there's nothing that does any work on the system, so there's no work done. So my change in energy, I have change in kinetic, change in spring potential, and change in gravitational potential equals zero. So it's called this position one, and I'm going to skip this. I don't even care about this position right here. I just want to find up here, moving that way, and this is position two. And that's going to be y2, which is the site. Okay, so let's just start putting in our stuff. We already had, the work is zero. The change in kinetic is going to be k2 minus k1. I'm just going to write out like that. Spring potential energy is going to be us2 minus us1 plus ug2 minus ug1. Okay, are any of these things zero? What about kinetic energy? Start from rest. So the initial velocity is zero, the initial kinetic energy, zero. What about this, the final spring potential energy? When it's up here, the spring's not compressed or stripped. If you remember, let me write down here, k equals 1 half mv squared. U spring is 1 half ks squared, and then U gravity is going to be mg y. So when this is up here, the spring's no longer compressed or stretched, so s is zero. So the spring potential energy right there is zero. What about the potential, gravitational potential energy at the top? Well, it has a y value, so that's not zero. But since I called this my y equals zero right there, then my initial gravitational potential energy has a y value of zero, so that's zero. So it doesn't look so bad anymore, I only have three terms. So I have zero equals 1 half, okay, I made a trick, I tricked myself. Okay, let me just write it down anyway. 1 half mv2 squared minus 1 half ks squared, where s is the amount stretched. And this k is little k for the spring constant. And then plus mg y2, okay. So I made a little mistake here because I don't, I mean, I'm in trouble because I know k, I know s, I know g, I know m. I don't know v2 and I don't know y2, so I have one equation and I have two unknowns. But I can use my knowledge about projectile motion and I can find v2. Okay, so that's what I'm going to do. Because at the highest point, it only has motion in the x direction. And the velocity in the x direction doesn't change. So if I know the velocity right here, which makes it a little more complicated than I want it to be, because I'm a dummy. Then I can find the kinetic energy up there and then solve for y2. I'm going to do it anyway and I don't care if this thing takes 20 minutes. I've used seven already. Okay, so let's do this problem. Now I'm actually going to have to find the velocity right here at the beginning. Okay, so I have another problem. It looks just like this, except now I do know how high it is. It looks exactly the same. This is s, so that's going to be s sine 35. Okay, so now it's the exact same problem, but I know that. I know I'm skipping some steps that I probably shouldn't have. But I'm just trying to finish this before the whole camera runs out of batteries. So I have zero equals, zero equals one half m. I'll call this v a, because it's in the middle. And I'll call everything a. So one half m v a squared minus one half k s squared, still that's the same. And then plus m g s sine 35. So now this I can solve for v, right, because I know everything. So I get v a squared equals k over m. I'm skipping some steps because I'm going fast, s squared, minus 2 g s sine 35. And then I can take the square root of both sides. So let me get a numerical value for that. Okay, so I get 10 and the mass was one, you know what? My spring constant is not large enough to move that mass up that high. Okay, so that's not going to work. Let me increase this to, let's say, 20 Newtons per meter. And let's see what happens. I'm wasting everyone's time because I didn't prepare beforehand. This is what happens when you don't prepare beforehand. You waste everyone's time. Oops. Okay, so 20. Let's train it up to 20. Okay, so 0.86 meters per second. Oh, I didn't take the square root. 0.94. So the velocity right here, v a, 0.94 meters per second. So now what's the velocity at the top? Well, what's the x component of this velocity? Well, that's going to be v a, x equals v two equals v a cosine 35 degrees. So I can get the speed at the top. So that's going to be, let me just plug that in, 0.77. So that's the velocity at the top. And then going back down here, now I know v two, I know m, I know k, I know s, I know m, I know g. I can solve for y two and I'll let you do that. Okay, so it looks like a complicated problem, but the key thing here is I can still use ideas from productile motion and I can use those with my work energy principle to find things for a more complicated case. I do like this problem because it involves a spring, spring potential energy and gravitational potential energy.