10. Conservation of Energy

Energy in Connected Objects (Systems)

# Solving System of Blocks on Incline Plane with Energy

Patrick Ford

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Hey everybody. So this problem is a little bit tricky, but we're gonna work it out together. So we have the system of blocks, One of those un inclined, the other one is hanging, we're gonna release the system from rest and we're gonna calculate the speed of the system of the blocks when the hanging block, the 30 kg one has fallen by a distance of one m. So here's what's going on, I'm gonna go ahead and start drawing my diagram here. This is sort of like the initial situation. The 30 kg block is going to fall by a distance of one m. So when it ends up over here, this is gonna be a delta Y of 30 sorry, of one, right? It's going to fall a distance of one m. Right? And because of that, the 20 kg block that's on the incline gets pulled up the incline by the same amounts. Remember if the 30 kg falls by one m, then because of the string, the 20 kg has to go up the incline by one, so it's going to end up over here and this distance here, I'm gonna call d is also just equal to one m. So that's really what's going on here. So this is sort of the initial and then this is going to be sort of like the final, I'm gonna call this block, this one block A and this one's gonna be blocked, be okay. And what happens is when we get down here, the blocks are gonna have some speeds, we have VB final and then this one is gonna have V. A final. But again, because the system is connected than whatever the speeds are for both of the objects, it has to be the same thing. So what I'm gonna do is I'm gonna write here that this V be final and V A final are actually going to be the same. So I'm gonna write that VB final is V. A final and that's equal to just the final for the whole entire system. Right? And that's why it says the speed of the system here. So let's go ahead and write out our energy conservation equation. So we have that kinetic initial plus potential initial and the work done by non conservative equals kinetic final plus potential final. Alright, so we're gonna go through and start limiting the terms. You don't have to write everything out because we can sort of just shortcut some of these things. Right. So remember um So for example the kinetic energy initial. Right, Both of the objects are gonna be at rest. So the whole system is released from rest. So when you write out the one half mv squared for both of them, you're gonna see that there actually is no kinetic energy because the V. Is for both of them are zero. What about any potential energy? Well, we have some things that are changing heights. So we're gonna have to write out some potential energies. What about any work done by non conservative forces. Well, there's no work done by you or friction because this is a smooth inclined plane? So there's no work done by your friction. And so that just means that everything here is really just changing kinetic and potential energies, there's definitely gonna be some kinetic energies because once you release the system the blocks start moving and then you're also gonna have some some potential energies as well that change because you have some changing heights. So let's go ahead and write out and expand some of these terms. The initial potential energy remember is going to be the potential for both of the objects. So what I'm gonna do here is I'm gonna write that, this is M A G times Y A initial plus mbG times Y B initial. Right, just MG wise for both of the blocks, that's what this sort of term works out to. This is going to equal the kinetic energy final. So this is gonna be one half of M A v A final squared plus one half M B V B final squared. But remember you can just sort of take this because both of these speeds are gonna be the same. We can actually just rewrite this and we can say that this whole kinetic energy term is gonna be one half of em a plus MB times v final squared. Right? So that's really what the kinetic energy works out to. And remember it's because they sort of share the same velocity. Okay, And then finally the potential energy is is just gonna be M A G Y final. Why a final plus M b g y B final. Alright, so let's go ahead and figure this out. Right. We're ultimately trying to figure out what this V final is. If you look through this problem, if you look through all the numbers, we have all the masses, we have M A. S and the M B s and all that stuff and then G is just the constant. The only thing that we're kind of missing, the information that we're missing in this problem is all of the initial and final heights. We have the Y A and the yb initial and final and things like that. Right? So we don't know what any of those numbers are. So, because these sort of things are unknown here. How do we actually go ahead and solve this problem? Well, if you look at this, the only information that we're given about distance is the fact that the hanging block is falling through a distance of one m. And that's the important part here. Remember that in conservation of energy problems and potential energies. The only thing that matters is the change in the heights not the actual initial and final. So what we're gonna do here is we're actually going to take these terms are going to move them to the other side. So basically going to combine them with their counterparts on the right side and this is what you end up with. When you subtract these from the left side, you just end up with zero over here, this is going to equal one half of uh This is I'm going to actually start plugging in some numbers. Uh This is going to be, let's see, this is Emma, which is 20 this is the 30 and this is going to be the final squared plus and this is what you end up with, right? So when you have M A G Y A final minus M A G Y A initial. This just ends up being M A G times Y A final minus Y A initial. And then the same thing happens for the B term M B G Y B final minus Y B initial. Alright, so all that's happened here is you've moved both of the left terms, these ones over to the right side and you sort of combine them with their counterparts on the right side. And what happens is you end up with these more simple expressions and the whole idea, the reason this is important is because remember that Y final minus Y initial, it's just the definition of delta Y. So this is just delta Y. A. And this is just delta Y for B. Okay, so now what we're gonna do is we're gonna move these terms back over to the left side because really we want to isolate and solve for this V final squared. So when you move these terms back to the left side again, what happens is they become negative M A G, delta Y A minus M B. G delta Y B. And this is gonna equal when you work this out, this is going to equal 25 V final squared. Ok, so remember we have the masses, all we have to do is just figure out what is the delta Y. A. And delta Y B. What are the changes in the heights for each one of these blocks? And they're gonna be different remember because one of them sort of going up the incline and the other one's sort of falling like this. So they're not necessarily going to be the same. The easiest one to look at is probably going to be this delta Y B first because if you look at it, the problem actually tells us exactly what that number is. The block. The block B is falling a distance of one m. So in other words, the delta Y. Here that we actually listed or that we labeled is actually the delta Y. For B. But we have to insert a negative sign here because it's falling. So in other words the final minus initial is actually negative. Okay, because you're falling downwards like this so we actually already have what this is, this is just going to be negative one. Alright, so I'm just gonna go ahead and start writing this out, I'm gonna skip this one for just a second here. So this is going to be 30 minus 9.8 times negative one. Ok now what about this one? This delta Y. A. So what I'm gonna do here is I'm gonna sort of go over here for a second and we can look at this triangle here. So in other words we have to figure out how high in other words the block is sort of going up the incline like this but that doesn't necessarily mean it's changing by one m because this is along the diagonal. So what you have to do here is you kind of have to sort of construct a little triangle to see what is the delta Y. For a. It's not necessarily going to be one and we're basically just going to use some trigonometry here. So I'm just gonna sort of I'm gonna go over here and I'm gonna draw this triangle again. So this is D. Equals one. We've got the angle of 53 degrees, this is just gonna be the delta X. And this is gonna be the delta Y. For a. This is really what I'm interested in here is delta Y. So how do we get it? Well it's pretty simple relationship? Right, it's just the opposite over hypotenuse. So basically what happens here is that the sign of 53 is going to equal? Remember opposite over hypotenuse. So it's gonna be delta Y. A. Over the deep, remember G. Is just one and that's just to make you know the numbers a little bit simpler here. But basically what you're gonna get here is that delta Y. Is equal to one Times the sine of 53. And what you should get here is that it's 0.8. Alright. So basically what happens is the hanging block falls by one. But because of the incline, the block goes up by not one but 0.8 m. That's the actual change in the height here. So that's what that sort of term, that's the delta Y. A. So what happens is this becomes 20 times 9.8 and this is going to be 0.8. And remember it's gonna be positive because it's actually going up in heights. The Y the final minus initial is going to be positive. So this is just going to be 25. Uh the final squared, so just go ahead and plug in some numbers here. What you get is that this is equal to negative .9 or sorry .8 and this is going to be positive. Uh 294. Remember what happens here is that you're subtracting a negative number over here. So you just gotta be careful with the negative signs and this is equal to 25 V. Finals squared. Ok so now all we have to do is just sort of tidy things up and get this V final squared and isolated. Um So what you end up with, when you sort of add these numbers and divide by the 25 is that your V final is going to be the square root of 5.49? And what you end up with as the final answer is 2.34 m per second. So 2.34 meters per second is the final speed of the system. Alright, So hopefully that makes sense guys, and it was kind of tricky, but hopefully you suck it, suck it out with me and I'll see you the next one.

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