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 K_{initial} + U_{initial} + work done by non-conservative forces = K_{final} + U_{final}. 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 U_{g}_{initial} for a + U_{g}_{initial} 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 U_{g} 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 K_{final} for a + K_{final} 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 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 v_{b final}, we're just going to write v_{final} 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 U_{final} is going to be plus, U_{g} this is going to be U_{g}_{final} for a, and then U_{g}_{final} 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 U_{g}_{initial} for b is going to be m_{b} g y_{initial}, and this is going to equal now our K's, this is going to be 12 m_{a} v_{final}^{2}, plus 12 m_{b} v_{final}^{2}, and then plus our U_{g}_{final} for a. So this is going to be, m_{a} g y_{final}. 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 n_{a}'s and m_{b}'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 v_{final}^{2}. Remember this is actually what you're trying to solve for. This is your target variable. Plus one half of 5 times v_{final}^{2}. 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 v_{final} here, which you end up getting when you plug everything in. You're going to get a 147 equals 2 v_{final}^{2} plus 2.5 v_{final}^{2} 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 v_{final}^{2} in both of them. And I'm actually going to flip around the equation. I'm going to get 4.5 v_{final}^{2} 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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# Energy in Connected Objects (Systems) - Online Tutor, Practice Problems & Exam Prep

In solving connected systems of objects, such as an Atwood machine, energy conservation is key. The equation for conservation of energy is $K{i}_{\mathrm{total}}+U{i}_{\mathrm{total}}+W=K{f}_{\mathrm{total}}+U{f}_{\mathrm{total}}$. By analyzing the kinetic and potential energies of each block, students can determine the final speed of the system, illustrating the principle of conservation of energy in mechanical systems.

### Systems Of Objects with Energy Conservation

#### Video transcript

### Solving Systems Problems with Friction using Energy

#### Video transcript

Hey, everybody. So let's get started with our problem. We have a system of objects and we want to calculate the speed of the 3-kilogram block just before it hits the ground. I'm going to start things off by drawing our diagram here. This is my initial setup. I've got these two boxes. I'm going to call this one A and this one B. What happens is once you release the system, this 3-kilogram block is going to 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_{B} final. That's really what we're interested in. Now, as a result, the A block is also going to 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 going to be somewhere over here like this and it's going to have its own speed, which is v_{A} final. But remember that from systems of objects these speeds are actually going to be the same because they're connected. So the whole system here has the same final speed. In other words, v_{B} 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 going to write out our energy conservation equation. So I'm going to write this out. This is going to be K_{initial} + U_{initial} + Work_{nonconservative} = K_{final} + U_{final}. I'm writing it out spaced out like this because, remember, we have to consider all of the individual energies from each object. Alright? So we're going to have lots of terms here. So what's the initial kinetic energy? Well, it's going to be the kinetic energy from A and B. So in other words, it's going to be 1/2 m_{A} v_{A} initial^{2} + 1/2 m_{B} v_{B} initial^{2}. And what's the potential energy? Remember that includes things like potential, spring energy, and gravitational potential. There's no springs involved, but we do have some changing heights for the B block. So we're going to have some gravitational potential. But the initial potential energy for both objects is going to be just mgy. Right? 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 on just one of the blocks because one of them is sliding across a tabletop that's sort of rough like this. So here's what happens: you have some force of friction that's going to be pulling back against the block, and that's going to do work. So this is really just going to be, I'm going to call this, negative f_{k}, times, and we're going to call this D, that distance. Alright?

And then finally, my kinetic energy final, which is going to be 1/2 m_{A} v_{A} final^{2} + 1/2 m_{B} v_{B} final^{2}, and then the final potential energy, which is going to be m_{A} g y_{A} final + m_{B} g y_{B} final. Alright. So lots of terms there, but we're going to see that some of them actually are going to cancel out. Alright?

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 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 2 meters, this y here, then we can do here this I'm going to call this Δy. We can actually just set the ground to be our y = 0. Right?

We usually want to set the lower point like that. So what does that mean for y_{A} initial? So by the way, yeah. What does that mean for this block? Because this isn't 2 meters. This is much higher. Well, it turns out that it actually doesn't matter, because for this block, the 4-kilogram block, the block A, you have the mgy and the MGY 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 out. Whatever height that is, it doesn't change. Alright?

So then, we do have some initial potential energy for the B block. Right? Because some height above the ground are 0 points. There is some friction and there is going to be some kinetic energy because after the system starts moving, both the blocks are going to be moving like this. Alright? So that's definitely there. And then what about this final potential energy for B? Well, what happens is B is going to be at the ground at this point, so therefore there's going to be no potential energy. Alright? So we've kind of simplified all these terms there to really just maybe 3 or 4 of them. Okay?

So here's what happens, right? So we've got m_{B} g times y_{B} initial, 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, μ_{k} times m_{A} g. Right? So in other words, it's going to be negative μ_{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. So 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. So what we can do here is we can actually replace this D term and actually just becomes Δy. So that's sort of what happens here. They're actually just equal to the same thing.

And then finally, what happens is, you've got these two terms. Right? So the 2 kinetic energies. Now what we said here was that the initial the final energies or final speeds for both of them are going to be the same. So what 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} + m_{B} times v final^{2}. Right? So basically, they're the same exact V finals, so you can sort of group those terms together, and that's really what we're after. Okay. Alright. So we're almost done here. Let's just go ahead and start plugging in some stuff. So you got the mass of B, which is going to be 3 times 9.8 and times y_{B} initial. So in other words, y_{B} initial is actually just going to be, this term over here. So this is going to be my y_{B} initial. So this is going to be 2 plus and then we have negative, and this is going to be 0.5, that's the coefficient, times m_{A}, which is going to be the 4, times g, which is 9.8, and then times Δy. So in other words, the Δy is just the 2 again like this. Right?

Because it travels that distance. This equals, one-half and then we have, this is going to be 4 plus 3, and this is going to be V final^{2}. Alright. So if you plug some stuff into your calculators, I'm just going to sort of simplify these into numbers. What you get over here is you just end up with 58.8 minus, then this is going to be this works out to 39.2, and then this is going to work out to 3.5 v_{final}^{2}. Alright?

And now what happens is, if you work this out, what you're going to end up with is that v final is equal to 2.37 meters per second, and that's going to be your answer when you finally work all that stuff out. Alright? So that's it for this one, guys. Let me know if you have any questions.

### Solving System of Blocks on Incline Plane with Energy

#### Video transcript

Hey, everybody. So this problem is a little bit tricky, but we're going to work it out together. So we have the system of blocks. One of them is on an inclined plane, the other one is hanging. We're going to release this system from rest, and we're going to calculate the speed of the system of the blocks when the hanging block, the 30 kilogram one, has fallen by a distance of 1 meter. So here's what's going on. I'm going to go ahead and start drawing my diagram here. This is sort of like the initial situation. The 30 kilogram block is going to fall by a distance of 1 meter. So when it ends up over here, this is going to be a delta y of 1. Right? It's going to fall a distance of 1 meter. Right? And because of that, the 20 kilogram block that's on the incline gets pulled up the incline by the same amount. Remember, if the 30 kilogram falls by 1 meter, then because of the string, the 20 kilogram has to go up the incline by one. So it's going to end up over here and this distance here I'm going to call d is also just equal to 1 meter. 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 going to call this one block a and this one's going to be block b. Okay? And what happens is when we get down here, the blocks are going to have some speeds. We have v_{B} final and then this one is going to have v_{A} final. But again, because this system is connected, then whatever the speeds are for both of the objects, it has to be the same thing. So what I'm going to do is I'm going to write here that this v_{B} final and v_{A} final are actually going to be the same. So I'm going to write that v_{B} final is v_{A} final, and that's equal to just v_{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, any work done by non-conservative, equals kinetic final plus potential final. Alright. So we're going to go through and start eliminating the terms. You don't have to write everything out, because we can sort of just shortcut some of these things. Right? So remember, so for example, the kinetic energy initial. Right? Both of the objects are going to be at rest, so the whole system is released from rest. So when you write out the one half m v squareds for both of them, you're going to see that there actually is no kinetic energy because the v's for both of them are 0. What about any potential energy? Well, we have some things that are changing heights, so we're going to have to write out some potential energies. What about any work done by nonconservative 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 you or friction. And so that just means that everything here is really just changing kinetic and potential energies. There's definitely going to be some kinetic energies because once you release the system, the blocks start moving, and then you are also going to 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 going to do here is I'm going to write that this is m_{A}g times Y_{initial} plus m_{B}g times Y_{B} initial. Right? It's just mgy 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 going to 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 these speeds are going to be the same, we can actually just rewrite this and we can say that this whole kinetic energy term is going to be 1 half of m_{A} + m_{B} 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 energies are just going to be m_{A}gy_{A} final, plus m_{B}gy_{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 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_{initial} and the Y_{B} 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 1 meter, 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 going to do here is we're actually going to take these terms and we're 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 0 over here. This is going to equal 1 half of, this I'm going to screw up actually start plugging in some numbers. This is going to be let's see. This is m_{A}, which is 20. This is the 30, and this is going to be v_{final} squared. Plus And this is what you end up with. Right? So when you have m_{A}gy_{A} final minus m_{A}gy_{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}gy_{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 sort of end up with these more simple expressions. And the whole idea, the reason this is important is that y_{final} minus y_{initial} is just the definition of delta y. So this is just delta y_{A} and this is just delta y _{B}. Okay? So now what we're going to do is we're going to 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 going to equal, when you work this out, this is going to equal 25 v_{final} squared. Okay? So remember, we have the masses. All we have to do is just figure out what the delta y_{A} and delta y_{B} are. What are the changes in the heights for each one of these blocks? And they're going to be different, remember, because one of them is 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 1 meter. 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 going to go ahead and start writing this out. I'm going to skip this one for just a second here. So this is going to be 30 minus 9.8 times negative one. Okay? Now what about this 1, this delta y_{A}? So what I'm going to do here is I'm going to 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 the block is sort of going up the incline like this, but that doesn't necessarily mean it's changing by 1 meter because this is along the diagonal. So what you have to do here is you kinda have to sort of construct a little triangle to see what the delta y for a is. It's not necessarily going to be 1. And we're basically just gonna use some trigonometry here. So I'm just going to sort of I'm going to go over here, and I'm going to draw this triangle again. So this is d equals 1. We've got the angle of 53 degrees. This is just going to be the delta x and this is going to be the delta y for a. This is really what I'm interested in here, this delta y. So how do we get it? Well, it's a pretty simple relationship. Right? It's just opposite over hypotenuse. So basically what happens here is that the sine of 53 is going to equal, remember, opposite over hypotenuse. So it's going to be delta y_{A} over the d. But remember, g is just 1, that's just to make, you know, the numbers a little bit simpler here. But basically what you're going to get here is that delta y is equal to 1 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 1, but because of the incline, the block goes up by not 1 but 0.8 meters. That's the actual change in the height here. So that's what that sort of term, that's the delta You. So what happens is this becomes 20 times 9.8, and and this is going to be 0.8. And remember, it's going to 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, v_{final} squared. So just going to go ahead and plug in some numbers here. What you get is that this is equal to negative 156.9 or sorry. Point 8. And then this is going to be positive, 294. And remember, what happens here is that you're subtracting a negative number over here, so you just got to be careful with the negative signs. And this is equal to 25 v_{final} squared. Okay? So now all we have to do is just sort of tidy things up and get this v_{final} squared and isolate it. So what you end up with when you sort of add these numbers and then 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 a final answer is 2.34 meters per second. So 2.34 meters per second is the final speed of the system. Alright? So, hopefully, that makes sense, guys. I know it was kinda tricky. Hopefully, you stuck it out with me, and I'll see you in the next one.

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More sets### Here’s what students ask on this topic:

How do you solve an Atwood machine problem using energy conservation?

To solve an Atwood machine problem using energy conservation, you start by writing the conservation of energy equation: ${K}_{i}+{U}_{i}+W={K}_{f}+{U}_{f}$. Identify the initial and final kinetic and potential energies of each block. For example, if block A is on the floor and block B is at height h, the initial potential energy of block B is ${U}_{i}=\; m\_bgh$. Set up the equation considering both blocks' energies and solve for the final speed.

What is the principle of conservation of energy in mechanical systems?

The principle of conservation of energy in mechanical systems states that the total energy in an isolated system remains constant. This means that the sum of kinetic energy (K) and potential energy (U) at the initial state plus any work done by non-conservative forces (W) equals the sum of kinetic and potential energy at the final state: ${K}_{i}+{U}_{i}+W={K}_{f}+{U}_{f}$. This principle helps in solving problems involving connected objects by analyzing their energy changes.

How do you calculate the final speed of connected objects using energy conservation?

To calculate the final speed of connected objects using energy conservation, write the conservation of energy equation: ${K}_{i}+{U}_{i}+W={K}_{f}+{U}_{f}$. Identify the initial and final kinetic and potential energies for each object. For example, if block A is on the floor and block B is at height h, the initial potential energy of block B is ${U}_{i}=\; m\_bgh$. Set up the equation considering both blocks' energies and solve for the final speed.

What are the steps to solve a connected system problem using energy conservation?

To solve a connected system problem using energy conservation, follow these steps: 1) Draw a diagram of the system. 2) Write the conservation of energy equation: ${K}_{i}+{U}_{i}+W={K}_{f}+{U}_{f}$. 3) Identify and calculate the initial and final kinetic and potential energies for each object. 4) Substitute the known values into the equation. 5) Solve for the unknown variable, such as the final speed.

Why is it important to consider the energies of each individual object in a connected system?

It is important to consider the energies of each individual object in a connected system because each object may have different initial and final kinetic and potential energies. For example, in an Atwood machine, one block may start on the floor while the other is at a height. By analyzing the energies of each object separately, you can accurately apply the conservation of energy principle to determine the system's behavior, such as the final speed of the objects.