Hey, guys. So for this video, I want to cover the conservation of mechanical energy, which is probably one of the most important principles that we'll learn in all of physics. So you're going to use this whenever you're asked to calculate the mechanical energy as this object or system, which we'll talk about in just a second, moves between 2 different points like we have in this example right here. So let's check this out. The whole idea is that we have this ball that's at 100 meters and it's going to fall down to the ground like this. So the y final is going to be 0 and we want to calculate the mechanical energy before and after. So let's take a look here. For part a, we're going to calculate the mechanical energy at the top of the building. Remember the mechanical energy of any object or system is really just going to be the sum of kinetic and potential energies, k plus u. We're going to talk more about systems in a later video. For now, a system could be as simple as our ball right here. Okay? So really the mechanical energy initial is just going to be the kinetic initial plus the potential initial. And we can expand out both of these terms because we know what those equations are. So this is going to be 1/2 mv_{initial}^{2} plus mgy_{initial}.k+u Alright? So we can see here is that if we're dropping this ball from this height, the initial velocity is going to be 0 which means that we can actually eliminate our kinetic energy initial term. So this v initial is going to be 0. That whole term goes away. So all of our mechanical energy initial is actually gravitational potential. And we can calculate this because we have all the numbers. This is 29.8 and this is 100. So you get an initial gravitational potential energy of 1960, which means that your total mechanical energy initial is 1,960 joules. Alright. Let's move on to the second part. For the second part, we want to calculate now the mechanical energy final once it reaches the bottom. So when it reaches the bottom here right before it hits the ground, it has all of this velocity here, which means it has some kinetic energy. And our final height is going to be 0. Right? So we're just going to expand the terms exactly how we did this before. So this is going to be k final plus u final which is going to be 1/2 mv_{final}^{2} plus mgy_{final}.k+u So now what happens in the mechanical energy final term is because our y final is going to be 0, we no longer have any gravitational potential energy. We do not have any height. So what happens is we do have some kinetic energy, but no gravitational potential. So now all of our mechanical energy is now kinetic energy. To calculate this, we're going to have to figure out what this v final is, and I'm going to go ahead and just give you the shortcut. This really just comes down to a vertical motion equation. So we're just going to have to write out our 5 variables. We're going to have to pick out an equation. Just remember this equation right here. And we get a speed of 44.3. So what we can do is I can basically say this is 1/2 of 2 times this is going to be, 44.3^{2}. And what you get is the mechanical energy final once you plug it into your calculator and you round is going to be 1,960 joules. Okay. So we get the same exact number here. The mechanical energy at the top is the same as the mechanical energy at the bottom. So what happened? Basically what happened is that we had our mechanical energy initial which was all gravitational potential and it basically just became all kinetic energy at the bottom. But the numbers were the same. It remained the same. So when a system's mechanical energy gets transferred between potential and kinetic energies and there's no loss, we say that that energy is conserved. And that's exactly what the principle of conservation of mechanical energy is. It says that the meinitial is equal to me final. Now we usually are not going to know what the total mechanical energy is, our initial and final. So the way that we're always going to write this, we're always going to write this as kinetic initial plus potential initial equals kinetic final plus potential final. This is what's called the conservation of mechanical energy. It's going to be super helpful for us to solve problems that we haven't been able to solve before, and we can also solve some problems that we have seen before much quicker. Let me show you another example here. So now we're going to drop a ball from a 50 meter building, and we want to use conservation of energy to figure out the speed when it hits the ground. So whenever you have conservation of energy, you're going to draw a diagram. So let's go ahead and do this. We have a building like this. We got the ball and it's going to be falling. So very similar to our example before, we have an initial height of 50. It's going to fall down to the ground. Let me write that again. So it's going to be falling down to the ground. Our y final is going to be 0. So we want to use conservation of energy. We've got the diagram. Now we want to figure out we actually want to write that equation out. So we're always going to start with ki+ui equalskf+uf. So now what we want to do is we want to eliminate and expand out any of the terms. What do I mean by that? So we know from the previous example that if you drop something, the initial velocity is going to be 0. And what that means is that your initial kinetic energy is not going to be, is going to be 0. Right? So it's all just gravitational potential. You have some gravitational potential because you're at some height. And then finally what happens is that your kinetic energy final comes from the fact that you have some speed v final which is what we're actually trying to solve here. So we have some kinetic energy here. But because you're at the ground and your y final is equal to 0, you have no gravitational potential energy. So these are the only two terms that survive and we can expand them because we know that our Ug_{initial} is going to be mgy_{initial} and our kinetic energy final is going to be 1/2 mv_{final}^{2}.Ug This is an equation and we can do is we can say that the masses are going to cancel out. In fact, that's most of the time that's going to happen in these problems which is actually really important because we were told that this ball had an unknown mass. So we wouldn't have been able to solve this if we didn't know that. Alright? So now we can just figure out the speed. We can rearrange for this equation to solve for v final. What you're going to get is that v final is equal to the square roots. Once you move the one half to the other side, you're going to get 2g y initial. So remember I told you that this equation was going to be super important. I even highlighted it up there because these are 2 equations are actually going to be the same. Notice how we got this equation using, vertical motion and motion equations. Now we got the exact same equation from conservation of energy. So what I can do is I just have to plug in the numbers now. This is just going to be 2 times 9.8 times the initial height of 50. If you go ahead and work this out, what you're going to get is you're going to get 31.3 meters per second, and that is the speed at the bottom. Alright? So let me know if you guys have any questions. That's it for the conservation of energy equation. Let's go ahead and take a look at some work.

# Intro to Conservation of Energy - Online Tutor, Practice Problems & Exam Prep

### Conservation Of Mechanical Energy

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### Launching Up To A Height

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Hey, guys. So let's go ahead and check out this problem here. You're going to launch a 4 kilogram object directly up from the ground. So I go to the ground like this. You're going to take this 4 kilogram ball and you're going to throw it up with some initial speed \( v_0 = 40 \). And we want to figure out basically what happens when the ball finally comes to a stop here at some maximum height. So here, if you take the ground level to be 0, what we want to calculate is what is this \( y_{\text{max}} \) value here. Alright? So we've got some changing speeds and changing heights. We know we're going to use energy conservation. So we've already got the diagram and we're going to write our energy conservation equation. So we're going to write \( K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} \). Now we're going to eliminate and expand out each one of our terms here. Alright? We've got some initial kinetic energy. That's the initial speed, of 40 meters per second, so we've got that. But here, when we're at the ground level, if we take \( y = 0 \) to be the ground, right, where we're starting from, then that means that our gravitational potential energy is 0 here, so, therefore, there is no gravitational potential. Alright. So there's nothing there. What about \( K_{\text{final}} \)? So what happens is when this object gets up to its maximum height, the velocity is going to be 0. So here the velocity final equals 0. Therefore, there is no kinetic energy and there is going to be some potential energy because now we're at some height above the ground here. So let's go ahead and expand out our terms. What I've got here is \( \frac{1}{2} m v_{\text{initial}}^2 = mg y_{\text{max}} \). So I'm going to call this \( y_{\text{max}} \) here and I'm going to go ahead and solve for this. Well, one of the things we'll notice here is that the mass is going to cancel. Usually, that happens with these kinds of problems. And now we're just going to go ahead and figure out \( y_{\text{max}} \). So \( y_{\text{max}} \) or \( y_{\text{final}} \) is going to be \( \frac{1}{2} v_0^2 \div g \). So if you go ahead and plug in our numbers here, you're going to have \( \frac{1}{2} \times 40^2 \div 9.8 \), and you're going to get, 81.6 meters. And that's the answer. So it goes 81.6 meters high, that's actually really high, it's like 250 feet or something like that. So if you could actually throw this thing up with, with that speed, and of course, there was no air resistance, that's how far it would go. Alright. So that's it for this one, guys. Let me know if you have any questions.

### Throwing An Object Downwards

#### Video transcript

Hey, guys. Let's work this problem out together. In this problem, you're going to throw a 6-kilogram object down from an initial height of 20 meters. Let's go ahead and draw our diagram out. Remember that's step 1 for solving conservation of energy problems. So we have our height of 0, which is our ground level, and then we have at some point here at \( y = 20 \) meters, we have a ball at 6 kilograms, and we're going to spike it down. We're not going to drop it, right? We're not going to just release it. We're actually going to spike it down with some initial speed and that's actually what we want to calculate in the problem. What's that initial speed? The only other thing we know about this problem is that when the ball finally reaches the ground right before it hits, it has a final velocity of 30 or final speed of 30 meters per second. So what ends up happening is we can choose our upward direction to be positive, and therefore, this final velocity is going to be negative. We should also expect that when we calculate the initial velocity, that should also be a negative number. Okay. So let's check this out here. We're going to use energy conservation, so we're going to have to write out our big equation for this. This is going to be \( K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} \). So let's go through each one of our terms here. We have some kinetic energy because we have some initial speed. We also have some initial gravitational potential because we're at a height of 20. Right? So this is above our reference point \( y = 0 \) where there is no gravitational potential, so you have some stored energy here. So what about \( K_{\text{final}} \)? Well, this is going to come from the final velocity here which is we know is, is 30 meters per second or the final speed. And so we also, do we have any gravitational potential? Well, once we hit the floor, we actually have no gravitational potential because our height is 0. Alright? So let's go ahead and write out each of the terms here. I've got \( \frac{1}{2} m v_{\text{initial}}^2 + mgy_{\text{initial}} = \frac{1}{2} m v_{\text{final}}^2 \). So what we're looking for is \( v_{\text{initial}} \) and, we know that this height here \( y_{\text{initial}} \) is 20. Alright? So one thing we can do here is we can actually cancel out the masses because they appear in all the terms of the problem on the left and right side. What we can also do is, if I have fractions in some of the terms but not all of them, what I like to do is sort of multiply the equation by 2. It doesn't change anything. As long as you multiply everything by 2, nothing changes. But what you do do is sort of get rid of the fractions. So what this ends up being is you get \( v_{\text{initial}}^2 \) and then this is going to be plus \( 2g y_{\text{initial}} \) Initially, this was just \( mg y \), so you have to multiply it by 2 and it becomes \( 2g y \). So this is going to be \( v_{\text{final}}^2 \). This is \( v_{\text{final}}^2 \). Alright? So this is what we get. Now this hopefully should look a little familiar to you. This equation is really just equation number 2 sort of rewritten in a different way from back in our motion equations. So we can actually use conservation of energy to come back to the same equation that we saw when we looked at kinematics in motion. Alright? So let's go ahead and solve for this velocity here. What we're going to get is we're going to get the initial velocity squared equals we're going to move this over to the other side, and what we're going to get is, the \( v_{\text{final}}^2 - 2g y_{\text{initial}} \), and then we're just going to take the square root of this whole number here. So the answer I'm getting for \( v_{\text{initial}} \) is the square root, this is going to be \( 30^2 \). It actually doesn't matter if you plug it in as positive or negative, so you can plug this in as negative 30. The square will actually just make it and turn into a positive. Minus \( 2 \times 9.8 \times 20 \). So if you go ahead and plug this in, what you're going to get is you're actually going to get 2 answers for this. You're going to get positive or negative 22.5 meters per second. So what does that mean? It means that you could have thrown this thing either upwards or downwards. And in either case, you'll still have a final velocity of negative 30. Right? So what happens is we know that our initial velocity has to be negative because we're throwing it downwards. So, therefore, it's just going to be the negative number. So it's negative 22.5 meters per second. Alright. So that's it for this one, guys. Let's move on.

### Conservation Of Energy with Multiple Points

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Hey, guys. Let's work this problem out together here. So we're going to launch a ball directly upwards from the ground. Let me go ahead and start drawing that. So we've got the ground level like this, y equals 0. I've got a ball that I'm going to launch upwards with some initial speed. That's actually the first part of the problem. I want to calculate that launch speed. So this v equals something that I don't know. What I do know about this problem is that at some later time, the ball is still going upwards. I'm told that here, the ball has a speed of 20 meters per second, and the height is equal to 30. And because it's still moving upwards at some height, it can actually still be going even higher than this. So it's going to continue going upwards until it finally reaches its maximum height. Here, we know that the speed is going to be 0, and this is going to be y max. That's actually going to be part b of our problem. So part a, we're going to calculate this launch speed. Part b, we're going to calculate this y max. So this is our diagram here. We want to use energy conservation because we have changing heights and changing speeds. So what happens is we're going to write our energy conservation equation, but what is going to be our initial and our final? We actually have 3 different points here. So there's a couple of combinations for initial and final. I can go from here to here, or I can go from here to here. So what happens with these problems is that some problems will give you more than 2 points of information. Some points aren't just going to give you as simple as the initial and final. They'll give you some other information. So what happens is I'm going to label these as a, b, and c, similar to what we did with projectile motion. So a is when we're at the ground, b is when we know that we're, you know, 20 meters per second at 30 meters, and then c is going to be the maximum heights. So we're going to have to write an energy conservation equation, and the two points that we're going to pick, our initial and final, should be the given and the targets. The given should be the one that you know everything about, and the target variable should be, the target interval or the point of interest should be the thing that you actually are looking for. So for example, we're looking for the launch speed, which is v_{a}. So we want to pick A as one of our points of interest. And then between B and C, the one that makes more sense is B because we know everything about it. If we were to try to pick point C as our another point of interest, like from initial to final, we would actually sort of get stuck because we would have 2 unknowns in that equation, y_{max} and v_{a}.
So we actually wouldn't be able to solve it that way. Alright? So for part a here, I'm going to write my conservation of energy from point a to b. So I'm going to do the kinetic initial, which is k_{a}, plus the potential initial is equal to the kinetic final, plus the potential final. We're just using a's and b's because that's my initial and final here.
So let's go ahead and eliminate and expand all of our terms. We definitely have some kinetic energy. That's what we're looking for, because we're looking for the speed. We don't have any gravitational potential because y_{a} is equal to 0. This is our sort of floor level, our ground level. So we're going to just set, you know, the gravitational potential to be 0 there just so we can get rid of that term. And then there's definitely going to be some kinetic and potential when we get to point b because we have some speed and some height.
So let's go ahead and expand each of the terms here. We're going to have
v_{a}
2
/
2
, that's our target variable, equals
fencelongdiv
v_{b}
2
+
2
g
y_{b}
Alright. So what I'm going to do here is, as expected, our masses are going to cancel from, you know, all sides of the equation, and that actually is good for us because we didn't know what the mass of the ball was, but it turns out it doesn't matter. And we can also multiply this equation by 2 because we have a bunch of fractions in here. So we're going to get v_{a}^{2} equals v_{b}^{2} + 2 g y_{b}. So now we're just going to take the square root and start plugging in numbers. So v_{a} is going to be the square root. This is going to be 20^{2} plus 2 times 9.8 times the height at point b, which we know is 30. If you go ahead and work this out, you're going to get 31.4 meters per second. So that's the answer here at the sort of launch speed; our launch speed was actually equal to 31.4 meters per second, and then it is going to slow down when it gets to point b. It's going to only be traveling at 20 meters per second, and it continues on until it reaches point c, at which it stops. Alright. So that's the answer to part a.
Alright. So let's take a look at part b now, and now we want to calculate the maximum height. So that's going to be this guy over here. Now, again, we want to pick an interval in which we know something about one interval, and we're looking for y_{max}. And it turns out we can actually use either the interval from a to c or the interval from b to c. Notice how we have all this information now. So we know that y_{a} is equal to 0. So we know everything about both of these points of interest here. It actually doesn't matter which one we pick. So just to make things a little bit simpler, I'm going to choose from a to c just because we know that one of the terms is going to cancel out. So let's go ahead and write this. Right? We've got k_{a} + u_{a} equals, k_{c} + u_{c}. So we know there's no potential energy initially, and now what happens is at point c, there's going to be no kinetic energy final. And so what happens is we're going to get 1/2 mv_{a}^{2} equals, and this is going to be mgyc, or we can just call this actually y_{max} here as well. So that's our target variable. Our masses will cancel, and we actually know what this v_{a} is now. So we can go ahead and solve. Let's see. We're just going to get this y_{max} here is equal to this is going to be v_{a}, which is the 31.4 that we just figured out divided by 2 times 9.8. So I'm just doing one half of that velocity divided by the g once we move it over. And so what you end up getting here is a y_{max} of 50.3 meters. So that's the maximum height. Right? So with this thing goes 50.3 meters final, makes sense because here we were at 30 meters. We still had some speed, so we should expect this to be a higher number. Alright. So that's it for this one, guys. Let me know if you have any questions.

### Conservation Of Total Energy & Isolated Systems

#### Video transcript

Guys, now that we've covered conservation of mechanical energy, in this video, I want to cover a couple of conceptual points that you might need to know just in case you run into them on a problem or a test or something like that. So what I'm going to do in this video is introduce the first of 2 conceptual rules that explains when you have energy conservation. In this first video, we're going to talk about the conservation of total energy and what it means to have an isolated system. And< in the next couple of videos, we're going to talk about the second rule. So let's check this out. Basically, what this rule says is that the total energy E of a system is conserved if the system is isolated. So let me back up here because there are a couple of conceptual definitions I want to point out.

First is total energy. Remember that your total energy is just the sum of all your mechanical energy, kinetics and potentials and all that stuff, plus all of your non-mechanical energy, thermal U and basically everything else. So the total energy, all the different energy types have to remain the same number if your system is isolated. Right? So let's talk about this system here. A system is really just a collection of objects. So it could be as simple as one object, but most of the time it's going to be a couple of them that's arbitrarily chosen. So it could be chosen by you or the problem. So in our example down here, we're going to take a look at a box and a spring, and we're going to take a look at some forces and energies if the system is defined as only the box in part a and the box and the spring in part b. So sometimes they'll probably just pick for you.

Lastly, I want to talk about isolated. So what does it mean to be isolated? Well, I think of the word isolated as, like, you're off in a corner by yourself and nothing is bothering you. And that's kind of the idea here. A system is going to be isolated if there's no external forces that are doing work. So and only internal forces are going to be doing work. So what does external and internal mean? External just means outside of your system. Internal just means inside of your system. And so the rule says, if any net force is external, your system is not isolated. If all the forces are internal, then your system is going to be isolated. So let's check out our example here.

We have a spring that's pushing a box and it's going to accelerate. We're going to figure out the forces are internal, if the system is isolated, and if the total energy of the system is going to be conserved. So let's take a look here. In part a, we're only just going to consider the box only as our system. So what I like to do is just draw a little bubble around my box. So that's going to be my system. So are all the forces internal? That's the first part. Well, what are the forces that are actually acting on my system here? Well, if I have a box that's pushed up against the spring, then I have the spring force F_s that's actually pushing up against my system here. But notice how this spring force is actually coming from something that's outside of my system. It's coming from the spring, which I'm not considering as part of my system here. So this F_s here is actually an external force because it's coming from outside of my system. So the answer to this question is no. All my forces are not internal. I have one force that's external that's doing work. So what that means is that the system is not going to be isolated. Remember, if you have any net force that's external, the system is not going to be isolated. If this is no, then this is also no. So what does that mean for our energies? If we take a look here, what's our initial energy?

Our initial energy, if we just have the springs and kinetic energies, is going to be k_i+u_i. So the initial kinetic so initial energy is going to be k_i+u_i. Now what happens is the box itself doesn't have any potential energy. This potential energy belongs to the spring, not the box. So your initial energy for the box as your system is only just going to be the kinetic energy which is 20. Then what happens when the spring actually fully launches the box? Now it has a kinetic energy of 30 joules. So what happens is your E_f=k_f + u_f. Again, there's no potential energy of just the box. And your k_f is going to be 30 joules. So what happens is that you actually did not have energy conserved because your initial energy does not equal your final energy. And this is because your system was not isolated. There's an external force that's coming from outside of the system that is adding work or that's doing work and adding energy into your system here.

So let's talk now about the box and the spring. Now let's sort of expand our system and include the spring now. Now what happens? Well, if we take a look here, what we said is that there's going to be a spring force on the box. So there's going to be an F_s here, but what happens is because of action-reaction, the box also pushes back on the spring. So it's kind of weird to think about because we haven't really considered that before, but the spring pushes to the box and the box pushes back on the spring here. So what happens is our spring forces is actually going to be internal because it's within my bubble. There's nothing outside of the bubble that's acting on my system here. Right? So your spring force is going to be internal. So the answer to this question is yes. And so if all of your forces are internal, then your system is going to be isolated. So what happens to our energy now? Well, if you take a look at it now we're just looking at the box and the spring. So remember, E=k+u. So your energy initial is going to be k_i+u_i. So what happens is we have the initial energy of 20 plus the initial potential energy of 10 and your total energy is going to equal 30 joules. Right? So what about the energy final? Well, energy final, we're still just considering the box and the spring. So now what happens is you have k_f+u_f. What happens is your k is 30 and now your potential energy is 0. Basically, all of the elastic or stored energy that was inside the spring now just became kinetic energy of the box. But if you're considering both of these things as your system, then your final energy is 30 joules. So in this situation here, your energy was actually conserved. So because your system was isolated and you only had internal forces, your energy was actually conserved here. We have 30 initial and 30 final. So hopefully this kind of makes sense guys. Let me know if you guys have any questions on that.

### Systems & Conservative vs. Non-Conservative Forces

#### Video transcript

Hey, guys. So now that we understand the basics of conservation of mechanical energy, I want to go over some conceptual points in detail just in case you run across these in problems. So we're going to go over systems and what it means to be a conservative versus a nonconservative force. Let's check this out.

Conservation of energy often refers to a system, which is really just a collection of objects that is chosen. Sometimes it could be chosen by you. Most of the time, it is going to be chosen in your problems. It'll say what that system is. So I want to go over an example so I can show you really how this works and what it means.

Imagine that you have a spring pushing a box. Right? And we have the energies, the potential and the kinetic energies. We want to figure out in this problem whether the mechanical energy is conserved depending on how we choose our system. In part a, we're going to choose the box only, and in part b, we're going to choose the box and the spring. So let's check this out. Right? So we have an initial and final and if we're choosing the box only, what I like to do is I like to draw a little bubble around the object that we're considering as our system. So it's just going to be the box only. So I want to look inside this bubble and I want to figure out what are the energies inside of this bubble here.

So the mechanical energy inside is really just going to be if we are looking at the box only. It's just going to be the kinetic energy of the box, which is just 20 joules. Now if you look at the final, once the spring has released the box, the mechanical energy here is still just the kinetic energy of the box, but now it's equal to 30 joules. So what happens here is that these two answers are not equal to each other, which means that energy, mechanical energy, was not conserved. And it's basically just because you picked your bubble, you chose your bubble to be too small. You weren't including the fact that the spring is also doing some work or interacting with the box. So mechanical energy is not conserved here.

Now what happens if we include the box? So when I draw my little bubble to include the spring now. So I'm going to draw my little bubble to include the spring. And now when we look at our mechanical energies here, our mechanical energy inside is all the energies inside of this bubble. So it's going to be my initial kinetic plus potential. So really this is going to be 20 plus 10 and this equals 30 joules. Now when I took a look at the mechanical energy final, this is going to be k_{final} + u_{final}, and this is going to be 30 + 0, which equals 30 again. So here what I have is I have these two energies that actually do agree with each other, initial equals final. So what happens here is that energy was conserved because now I've included the spring. So it's conserved here.

Sometimes depending on how you choose your system might actually affect whether your mechanical energy is conserved or not. Alright? So I want to talk a little bit more about mechanical energy. We've already seen the mechanical energy in a system is conserved, but there's a specific rule where that happens. What you need to know is the mechanical energy is conserved if the only forces that are acting on an object are conservative. So mechanical energy is conserved if the forces are conservative. So I want to actually go ahead and talk about conservative versus nonconservative forces. But to do that, we're actually going to take a look at an example here. So for each of these situations that we have a through d, we're going to figure out if the mechanical energy is conserved or not, and we're going to describe any energy transfer. So let's take a look at the first one, a block falls without air resistance.

So we're actually going to take a look at this diagram here, which is kind of basically what, what what that looks like here. So your 2 conservative forces are going to be gravity and spring. And so what I just said is that mechanical energy is going to be conserved if the only forces that are doing work are these 2. Anytime you have these nonconservative forces like applied forces and friction, your mechanical energy will not be conserved. So what we say here is that the work done by nonconservative forces has to be 0 in order for the mechanical energy to be conserved.

Alright? So what's happening here? We have a block that's falling without air resistance. So as this block falls downwards, if there's no air resistance, it's being pulled down by gravity. And what happens is your gravitational potential is going down because you're losing height, but as a result, you are gaining speed. So what ends up happening is that the only force that's acting on this block here is gravity, and we said that the mechanical energy is going to be conserved. So what's happening? There's really just a transfer of energy. You're transferring gravitational potential to kinetic energy. So that's really what's going on here.

Now what ends up happening is that you could also reverse this process. Right? You could actually throw a block upwards, and what would happen is that your gravitational potential would go up and your kinetic would go down. So there's always this exchange of energy between gravitational potential and kinetic. Alright? So let's take a look at the second one. Now we have a moving block that hits a spring, and it deforms it and rebounds. That's actually basically this situation right here. And springs are also conservative forces. Here's what's going on. As the spring hits the as or as the block hits the spring, the spring compresses and it stores some energy. It stores some spring or elastic potential energy here. So that spring energy increases and the kinetic energy decreases because the box slows down. Then the reverse happens when the box shoots when the block sorry. When the spring shoots the block out, it releases its stored energy. So that's going to go down, but your kinetic energy is going to increase. So there's always this exchange of energy here between the elastic and the kinetic. And in general, that's what conservative forces do. In conservative forces, when you have conservative forces, the mechanical energy is going to be exchanged.

Now when you have nonconservative forces, the mechanical energy is going to be added or removed. Let's take a look at the second examples here. So sorry. Just to finish this off, a moving block is going to be conserved because the only force that's acting on it is the spring force, which is a conservative force, and the energy transfer is really just spring energy with kinetic energy. Alright? So the second so the third part is now we're going to push a block that's at rest and it's going to accelerate to the right. That's actually going to be this diagram right here. So you're pushing a block with some applied force, and then it's basically going to accelerate in this direction here. So if you take a look at our system, what's happening is that, basically, our kinetic energy is going to increase, and therefore, our mechanical energy is going to increase. There's no exchange of energy. It's not it's not gaining kinetic energy because it's losing some potential. We're actually doing some work on the box. We're giving it some energy. We're giving it some kinetic energy here.

Alright. So this is not going to be a conservative energy or a conservative system because there's energy actually being added to the system. And basically, that energy transfer is the work that is done by you that is now becoming kinetic energy of the box. Alright? So now the last one is a moving box that's slowly slowing down due to friction. So you're moving to the right and friction is going to act to the left. So this is going to be kinetic friction. What happens? Your speed is going to decrease. Therefore, your kinetic energy is going to decrease, but it's not an exchange of energy. What happens is friction is removing the energy from that system, so your total mechanical energy is going to go down. So the energy is not going to be conserved here because you have a nonconservative force. And, basically, what's happening is that this kinetic energy now is going into heat. That's what's dissipating this heat due to friction.

Alright? So one way I can kind of summarize conservative versus nonconservative forces, one way I like to think about it, is that conservative forces are reversible. What this means is that whatever you do, right, whatever action you do, you can always sort of hit the undo button, and you can gain any lost energy back. What I mean by that is that here we have gravitational potential that becomes kinetic. But if you reverse the action, right, if you actually throw a box up, now you have gravitational potential increasing and kinetic that's decreasing. One analogy I like to use is it's kind of like money in a bank. Right? You can always put money in and take money out. And in some banks, you can do that without having to pay a fee. That's like your conservative forces. And then your nonconservative forces are where you take money out and you actually have to pay a fee each time. Right? You're losing energy as you're sort of withdrawing and putting that energy back in.

Let me know if you guys have any questions. That's it for this one.

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