Hey guys. So for this video, I want to introduce you to a type of problem I like to call the curved path problem. Let's check out the example and we'll come back here and fill out the rest. I've got this block that is traveling with some speed, and I want to figure out what is the minimum speed, what's this minimum v here, so that this block can travel up this curvy path and reach the top of the hill, just barely reach the top. So how do we solve this?

We have the height of the hill. We know this height is equal to 20. So how would I solve this? Do I use forces? Do I use kinematics? Well, there are a couple of reasons why those approaches aren't going to work. For one, we actually don't have the angle of this slope because it's constantly changing all the time. And even if we did actually know the angle, what's going to happen is imagine you had a box that was sort of at this point right here, the mg would point downwards and your component of mg that's pulling you down would point in this direction. But then when you reach the steeper part right here, the component that's pulling you down is actually in a different direction. So along the curvy path, your acceleration is never going to be constant. So you can't use forces or kinematics. What I want to point out here is that you're always going to solve these curvy path problems by using energy conservation. There's no other way that you can solve these kinds of problems. Now the reason I say only is that there's technically, in some very rare cases, you may be able to solve these using calculus. But almost always, you're going to use them using energy conservation. So let's go ahead and get to this.

So we have our diagram here. Basically, we're trying to figure out from point A to point B what is the speed that we need so we can reach the top of this 20-meter hill. So now we're just going to go ahead and write our energy conservation equation. So this is going to be Ki+Ui+workdonebynonconservative=Kf+Uf. So let's go ahead and eliminate and expand all the terms. We do have some kinetic energy initially. That's really just going to be related to this speed here. So do we have any gravitational potential? Well, here what happens is if your height is 20, then you can say that your height's initial is equal to 0 when you're at the ground here, so you don't have any gravitational potential.

So do we have any work done by non-conservative forces? Remember, that is work done by applied forces or work done by you or work done by friction. And you actually have neither one of those because you're basically just, you know, this block is just traveling, and there's going to be no friction because it's on a smooth hill. So there's going to be no work done by non-conservative forces. Now do we have any kinetic energy final? The whole idea is that you might be thinking that there might be some kinetic energy because when it reaches the top just barely, it might be moving with some tiny velocity like this. But the whole idea behind these problems is the final velocity and the final kinetic energy is going to be 0 because it's kind of related to the idea of the minimum speed.

So, for example, imagine we had a velocity here of 20 or a minimum speed of 20, and then by the time we reach the top, the block had some speed v_{b} and that was equal to 5. Well, if it still has some speed here, then that means that you could have actually had a smaller initial speed. This could have been less than 20 and you would have, you know, and you still would have made it to the top. Now imagine I had this as 18 and then v_{b} is equal to 2. You still could have gone a little bit slower to reach the top of the hill. Eventually, what happens is you're going to reach some number here and that's going to be the minimum speed. If you go any lower or any slower than that, you're actually not going to reach the top of the hill and then you're going to come crashing right back down again. So that's the whole idea, is at the top of the hill, we're going to have 0 kinetic energy.

Finally, we're also going to have some gravitational potential energy because we're at some height y_{b}. So really, we're just going to expand these last two terms here. So we have ¼m v_{a}^{2}=mgy_{b}. What we can do is we can cancel out these masses here, and we can go ahead and solve for this v_{a}. So if you go ahead and rearrange for this, you're going to get v_{a}=2g y_{b}. This is really just the square root of 2 9.8 × 20. If you go ahead and work this out on your calculator, you're going to get 19.8 meters per second.

So there's no other way you could have solved this instead of using energy conservation. That's it for this one, guys. Let me know if you have any questions.