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Systems Of Objects with Energy Conservation

Patrick Ford
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Hey guys, so occasionally you'll run into these kinds of problems. We're gonna have to solve these connected systems of objects problems kind of like this uh Atwood Machine that we have down here that we've talked about before when we talk about forces. Except you're not going to solve them by using forces, You're gonna have to solve them by using energy conservation. So I'm gonna 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 gonna release the system from rest. So the initial velocity is equal to zero. Now, what we want to do is we want to figure out the speed of this five kg block when it hits the floor, right before it hits the floor. So there's five kg block, starts off with a height of three m, which means its initial height is three. And what happens is it's going to get to the bottom right, right before it hits the ground. It's gonna have some final speed here. So we know that this why final is going to be zero. So we want to do is we want to calculate the speed here, but not by using F equals a man forces, but by using energy conservation. So how do we actually do that? We're just gonna go ahead and stick to the steps, right? We have our diagram. Now we just run right are 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 is multiple objects by using your energy conservation equation. So you're going to use K. I plus you, I plus work done by non conservative equals K. Final plus you. Final. Right. So you're still just going to use this one equation here. Before. When we solve this by using force is we have to draw the diagrams for both objects. We had to write F equals M. A. For both objects were to come up with equations and gary yada yada here, we can actually just solve this by using one energy conservation equation. However, what happens is we're gonna 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 when we get to step three and we actually start expanding each one of our terms. We're gonna have to consider the kinetic energy is the initial potential energy and all that stuff for each individual objects. So let's go ahead and do that. This initial kinetic energy for the whole entire system. It's actually gonna 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 zero. 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 UG initial for A plus Yugi initial for B. So let's take a look for a. We have an initial heights. That's why initial is going to be zero. Right? Because the block is actually on the floor. So there is no gravitational potential for Yugi for a What about B. Will be actually starts at a height of three. So, we know there's definitely gonna be some gravitational potential there. All right. So, what about work done by non conservative? Well, actually, we don't have to look at both objects for that. Remember work done by non conservatives. Any work done by U. Plus. Any work done by friction. So once you release the system, you're not doing anything anymore. There's 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 K. Final for a plus K. Final four B. So, let's take a look what happens is this B. Block is actually gonna come down to the floor like this and it's gonna be traveling with some speed. Something that is definitely gonna be some kinetic energy for B. But what about for a? Well, we have to remember here guys is that these connected objects that they're connected via this pulley in the string here, they're gonna have to move together and they move together with the same acceleration and speed. So what happens is is 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 heights like this, which actually is going to be a height of three and it's going to be traveling with some speed, which is going to be V. A. Final. Now, what we just said is that both objects are going to have the same speed. So instead of actually writing V. A. Final and VB final, we're just going to write the final for both of them. Remember that's the velocity for both of these objects and they're gonna be the same. All right. So, what that means here is that we definitely have kinetic energies for both. What about gravitational potentials? Well, this you final is going to be plus uh you G. Is gonna be Yugi final for A. And then you G. Final four B. So, what happens here is what we just said is that A. Is going to go up and be is gonna come down to the floor? So B is now the one that's on the floor. So, it has no gravitational potential here. But you G. F. But is actually gonna have some gravitational potential because now it's at height of three. All right, So those are over terms. Now we're gonna go ahead and start plugging in our expressions. So you G. Initial for B is going to be M. B. G. Y. Initial. And this is going to equal. Now our case, this is gonna be one half M. A. V. Final squared plus one half A. M. B. V. Final squared and then plus R. U. G. Final for A. So this is gonna be uh M. A. G. Y. Final. Alright. So now we're gonna 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 two objects here. You can only cancel them if they appear in all sides of the equation. So we actually have any S. And M. B. S. So we can't cancel out those masses. Alright, So now we're just gonna start plugging everything in mass for B is five. This is 9.8 and the initial height is three. So this is gonna be one half massive is four. This is the final square. Remember this is actually you're trying to solve for this is your target variable Plus 1/2 of five times v final squared. And then we've got plus the massive A. Which is four times 9.8 times the final height of three. All right. So basically what happens is that the rest of these things are numbers And you're only variables are this V final here, which you end up getting when you plug everything is you're gonna get 100 and 47 equals to the final squared plus, 25. The final squared Plus. And this is going to be 117.6. All right. So what you end up getting here is when you rearrange everything, you can actually combine these two terms here because um there's the final squares in both of them. And I'm actually gonna flip around the equation, I'm gonna get 4.5 V. Squared is equal to And then when you subtract these two things together, Um you're gonna get 29.4. So if you go ahead and work this out, what you're gonna get is square root of 29.4, divided by 4.5. And you're gonna speed of 2.56 m/s. 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 2.5 m per second and is going to be going up with 2.56 m per second. They're both going to have the same speeds. And so they're both going to have some kinetic energy over here. All right, so that's how you do these kinds of problems, Guys, let me know if you have any questions.