So occasionally, you run into these kinds of problems where you're going to have to solve these connected systems of objects problems. Kind of like this Atwood machine that we have down here, that we've talked about before when we talked about forces. Except you're not going to solve them by using forces, you're going to have to solve them by using energy conservation. So I'm going to show you how to do that in this video. It's pretty straightforward. So we're actually going to come back to this in just a second here because I want to start the example. So the whole idea here is that you have these blocks that are connected by this pulley. I'll call this one a and this one b, and you're going to release this system from rest. So the initial velocity is equal to 0. Now what we want to do is we want to figure out the speed of this 5 kilogram block when it hits the floor, right before it hits the floor. So this 5 kilogram block starts off with a height of 3 meters, which means its initial height is 3. And what happens is it's going to get to the bottom right before it hits the ground. It's going to have some final speed here. So we know that this y final is going to be 0. So what we want to do is we want to calculate this speed here, but not by using f equals ma forces, but by using energy conservation. How do we actually do that? We're just going to go ahead and stick to the steps. Right? We have our diagram. Now we just want to write our big conservation of energy equation. So let's go ahead and do that. The whole idea here, guys, is that you can often solve these kinds of problems even if there are multiple objects by using your energy conservation equation. So you're going to use Kinitial + Uinitial + work done by non-conservative forces = Kfinal + Ufinal. Right? So you're still just going to use this one equation here. Before, when we solved this by using forces, we had to draw the diagrams for both objects. We had to write f equals ma for both objects. We'd come up with equations and yada yada yada. Here we can actually just solve this by using one energy conservation equation. However, what happens is we're going to have to consider the energies of each individual object because if you have these things that are connected, you actually have both objects that are changing heights and speeds. So what this means here is that when we get to step 3 and we actually start expanding each one of our terms, we're going to have to consider the kinetic energies, the initial potential energies, and all that stuff for each individual object. So let's go ahead and do that. This initial kinetic energy for the whole entire system is actually going to be the initial kinetic energy for a plus the initial kinetic energy for b. Do I have both of those? Well, remember what happens is that both objects are going to start from rest, so the initial speeds for both of them is going to be 0. And if that's the case, there is no initial kinetic energy for either one of them. And basically, there is no kinetic energy for the whole system. What about potential energy? So potential energy is going to be Uginitial for a + Uginitial for b. So let's take a look. For a, we have an initial height. This y initial is going to be 0. Right? Because block a is actually on the floor. So there is no gravitational potential for Ug for a. What about b? Well, b actually starts at a height of 3, so we know there's definitely going to be some gravitational potential there. Alright. So what about work done by non-conservative forces? Well, actually, we don't have to look at both objects for that. Remember work done by non-conservative forces is any work done by you plus any work done by friction. So once you release this system, you're not doing anything anymore. There are no applied forces and there's also no friction here because we're told to ignore the effects of friction and air resistance and all that stuff. So there's no work done. So what about the kinetic energy final? Well, the kinetic energy final is going to be your Kfinal for a + Kfinal for b. So let's take a look. What happens is this b block is actually going to come down to the floor like this, and it's going to be traveling with some speed, so we know there's definitely going to be some kinetic energy for b. But what about for a? Well, what you have to remember here, guys, is that these connected objects, if they're connected via this pulley in the string here, they're going to have to move together, and they move together with the same acceleration and speed. So what happens is as block b pulls down, block a has to go up because these things are connected by the string. So what ends up happening is that a basically does the reverse of b. So b falls down like this, but a actually goes up to some height like this, which actually is going to be a height of 3, and it's going to be traveling with some speed, which is going to be va final. Now what we just said is that both objects are going to have the same speed, so instead of actually writing va final and vb final, we're just going to write vfinal for both of them. Remember, that's the velocity for both of these objects and they're going to be the same. Alright. So what that means here is that we definitely have kinetic energies for both. What about gravitational potentials? Well, this Ufinal is going to be plus, Ug this is going to be Ugfinal for a, and then Ugfinal for b. So what happens here is what we just said is that a is going to go up and b is going to come down to the floor. So b is now the one that's on the floor, so it has no gravitational potential here, but a is actually going to have some gravitational potential because now it's at a height of 3. Alright. So those are all our terms. Now we're going to go ahead and start plugging in our expressions. So Uginitial for b is going to be mb g yinitial, and this is going to equal now our K's, this is going to be 12 ma vfinal2, plus 12 mb vfinal2, and then plus our Ugfinal for a. So this is going to be, ma g yfinal. Alright? So now we're going to go ahead and start plugging in our numbers. One thing I want to warn you against is not to actually cancel out these masses because now we actually have 2 objects here. You can only cancel them if they appear in all sides of the equations. We actually have na's and mb's, so we can't cancel out those masses. Alright. So now we're just going to start plugging everything in. Mass for b is 5. This is 5 times 9.8 and the initial height is 3. So this is going to be 12. Mass of a is 4. This is vfinal2. Remember this is actually what you're trying to solve for. This is your target variable. Plus one half of 5 times vfinal2. And then we've got plus the mass of a which is 4 times 9.8 times the final height of 3. Alright. So basically, what happens is that the rest of these things are numbers and the only variables are this vfinal here, which you end up getting when you plug everything in. You're going to get a 147 equals 2 vfinal2 plus 2.5 vfinal2 plus and this is going to be a 117.6. Alright. So what you end up getting here is when you rearrange everything, you can actually combine these two terms here because, there's vfinal2 in both of them. And I'm actually going to flip around the equation. I'm going to get 4.5 vfinal2 is equal to and then when you subtract these two things together, you're going to get 29.4. So if you go ahead and work this out, what you're going to get is the square root of 29.4 divided by 4.5. You're getting a speed of 2.56 meters per second. And that's your answer. So that is the speed actually of both of these objects. So B is going to be going down with 2.56 meters per second, and A is going to be going up with 2.56 meters per second. They're both going to have the same speeds, and so they're both going to have some kinetic energy over here. Alright? So that's how you do these kinds of problems, guys. Let me know if you have any questions.
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10. Conservation of Energy
Energy in Connected Objects (Systems)
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