Energy in Connected Objects (Systems)

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.

everybody. So let's get started with our problems. So we have a system of objects and we want to calculate the speed of the three kg block just before it hits the ground. So I'm gonna start things off by drawing are sort of diagram here. So we've got this sort of, my initial, I've got these two boxes, I'm gonna call this one A and this one be What happens is once you release the system this three kg Block is gonna fall like this and then eventually right before it hits the ground it's going to be traveling with some speed. I'm going to call this V be final. So that's really what we're interested in now. As a result, the A block is also gonna be sliding to the right because these things are connected as one falls, it pulls the other one to the right like this. So it's gonna be somewhere over here like this and it's gonna have its own speed which is V. A. Final. But remember that from systems of objects, these speeds are actually gonna be the same because they're connected. So the whole system here has the same final speed. So in other words VB final is equal to V A. Final. So really what we're interested in is just the v final of the whole entire system. Alright, so how do we do this? We're gonna write out our energy conservation equation. So I'm going to write this out, this is going to be K initial plus you initial plus work. Non conservative equals. Okay, final plus you. Final And a written out sort of spaced out like this because remember we have to consider all of the individual energies from each object. Alright, so we're gonna have lots of terms here. So what's the initial kinetic energy? Well, it's gonna be the kinetic energy from A. And B. So in other words it's going to be one half M A. V. A. Initial squared Plus 1/2 and be vb initial squared. And now with the potential energy, remember that includes things like potential spring energy and gravitational potential. There's no springs involved but that we do have some changing heights for the B. Block. So we're gonna have some gravitational potential but the initial potential energy for both objects is going to be just MG. Y. Right? So it's going to be a um So it's going to be M A. G. And then Y. A initial plus M B. G, Y B. Initial. Alright, and finally we have any work done by non conservative forces. We do have a non conservative force which is friction because we're told the coefficient of friction here. So what happens here is that friction is going to be acting not on both the blocks, just on one of them because one of them is gonna be sliding across the tabletop that's sort of rough, like this? So here's what happens if you have some force of friction that's going to be sort of pulling back against the block and that's going to do work. So this is really just gonna be friction. The work done by friction on block. A So in other words, this is just gonna be I'm gonna call this negative F. K. Um times and we're gonna call this deed that distance. All right. And then finally I have my my kinetic energy final, which is gonna be one half of M A V. A final squared Plus 1/2 MB VB final squared. And then the final potential energy which is going to be M A. G. Why a final plus MB G Y B final. Alright, lots of terms there, but we're gonna see that some of them actually are going to cancel out. All right. So, let's see what's going on here. Is there any kinetic energy? No, there's not. There's no kinetic energy. Is there any potential energy? Well, what happens is that really depends on how you sort of set your zero points. So, because we have some of the distances, the heights involved like this block here is let's say we know this is two m. This y here that we can do here that this is going to call this delta Y. We can actually just set the ground to B. R. Y. Equals zero. Right? We usually want to set the lower points like that. All right. So, what does that mean then for why a initial So by the way, um Yeah. What does that mean for this block? Because this isn't two m, this is much higher. Well, it turns out that it actually doesn't matter because for this block, this four kg block, the block A. You have the MG. Y. And the M. G. Y. On the other side. But the block is only moving horizontally. So what happens here is there actually is no change in the potential energy because these two terms are going to be the same. So you have the same two terms over there and you can basically just cancel it at whatever height that is, it doesn't change. Alright, so then um we do have some initial potential energy for the B. Block, right? Because it's some height above the ground or zero points, there is some friction and there is going to be some kinetic energy because after the system starts moving, both the blocks are gonna be moving like this. Alright, so that's definitely there. And then what about this final potential energy for be? What happens is B is gonna be at the ground at this point? So therefore there's gonna be no potential energy. All right, so you kind of simplify all these terms there to really just maybe three or four of them. Okay, so here's what happens. Right? So we've got MB G times Y. B initial um Plus and then this is going to be the friction times the distance. Okay, so this friction force. Remember the friction force on this guy is actually just going to be um mu K. Times M A G. Right? So in other words it's going to be negative mu K M A G. Now, what about this distance here that it travels? Remember this distance here is D. But if you think about this, remember because these systems are sort of connected like this, whatever distance the B block falls is the same distance the A block travels. In other words it goes like this because the string has to remain sort of the same length. So in other words, these two things are actually equal to each other and see what we can do here is we can actually replace this D. Term and actually just becomes delta Y. So that's sort of what happens here, that actually just equal to the same thing. And then finally what happens is um you've got these two terms, Right, So the two kinetic energies, now, what we said here was that the initial the final energies or final speeds for both of them are gonna be the same. So we can do is we can say well, if these two terms are equal to each other the same, Then what happens here is this whole entire term can just become 1/2 and then you have M A plus MB times v final squared, right? So basically they're the same exact v finals, you can sort of group those terms together and that's really what we're after. Okay, all right, so we're almost done here. Let's just go ahead and start plugging in some stuff. Um So you've got the massive B. Which is going to be three times 9.8 times yB initial. So in other words why be initial is actually just going to be uh this term over here? So this is gonna be my Y. B. Initial. So this is gonna be two plus and then we have negative and then this is going to be 0.5. That's the coefficient times M. A. Which is gonna be the four times G. Which is 9.8 and then times delta Y. So in other words, the delta Y. Is just the two again like this, right? Because it travels at distance, this equals one half. And then we have this is gonna be four plus three and this is gonna be the final squared. Alright. So if you plug some stuff into your calculators, I'm just going to sort of simplify these into into numbers which you get over here, Is you just end up with 58.8 minus And then this is gonna be this works out to 39 points to and this is going to work out too 3.5, 3.5. The final squared. Alright. And now what happens is um If you work this out, what you're gonna end up with, is that the final is equal to um to .37 m/s. And that's gonna be your answer when you finally work all that stuff out. All right, so that's it for this one, guys, let me know if you have any questions.

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.