 ## Physics

Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

10. Conservation of Energy

# Intro to Conservation of Energy

1
concept

## Conservation Of Mechanical Energy 6m
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2
example

## Launching Up To A Height 2m
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Hey guys, So let's go and check out this problem here, you're gonna launch a four kg object directly up from the ground so that the ground like this, you're gonna take this four kg ball and you're gonna throw it up with some initial speed. V not equals 40. And we want to figure out basically what happens when the ball finally comes to a stop here at some maximum heights. So here, if we take the ground level to be zero, what we want to calculate is what is this? Y max value here. All right. 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 gonna write our energy conservation equation. So we're gonna write K initial plus you initial equals K final plus you final. Now we're going to eliminate and expand out each one of our terms here. All right. So, we've got some initial kinetic energy. That's the initial speed of 40 m per second. So, we've got that. But here, when we're at the ground level, if we take y equals zero to be the grounds, right? We're starting from that means that our gravitational potential energy is zero here. So, therefore there is no gravitational potential. All right. So, there's nothing there. What about K final? So, what happens is when this object gets up to its maximum height, the velocity is going to be zero. So here the velocity final equals zero. 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 our terms. What I've got here is one half M V initial squared equals And then the gravitational potential energy is going to be M G H. Or rather MG Y max. So, I'm gonna call this Y max here. I'm gonna go ahead and solve for this. Well, one of things we notice here is that the mass is going to cancel. Usually that happens to these kinds of problems. And now we're just gonna go ahead and figure out why final? So sorry, why max? So why max or why final is going to be one half of V not squared divided by G. So you go ahead and plug in our numbers here are going to have one half of 40, squared divided by 9.8. And you're gonna get um 81.6 m. And that's the answer. So it goes 81.6 m high. That's actually really high. It's like 250 ft or something like that. So, if you can actually could throw this thing up with with that that speed. And of course there was no air resistance. That's how far it would go. All right, so that's if this one guys let me know if you have any questions.
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example

## Throwing An Object Downwards 4m
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Hey guys, let's work this problem out. Together in this problem, you're going to throw a six kg object down from an initial height of 20 m. Let's go ahead and draw a diagram out. Remember that. Step one for solving conservation of energy problems. So we have our height of zero which is just our ground level. And then we have at some point here at y equals 20 m. We have a ball that six kg and we're going to spike it down. We're not gonna drop it right, we're not gonna just release it. We're actually gonna 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 m per second. So what ends up happening is we can choose our upward direction to be positive and therefore this final velocity is gonna 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 gonna use energy conservation. So we're gonna have to write our big equation for this. This is gonna be K. Initial plus you initial equals K. Final plus you 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 equals zero. Where there is no gravitational potential. So you have some stored energy here. So what about K final? Well, this is going to come from the final velocity here, which is we know it is 30 m per second of the final speed. And so we also um do we have any gravitational potential? Well, once we hit the floor, we actually have no gravitational potential because our height zero. All right. So let's go ahead right on each of the terms here, I've got one half M V initial squared plus mg Y initial equals uh And this is gonna be one half K mv final squared. So we're looking for is the initial and um we know that this heights here, Y initial is 20. All right. So one thing we can do here is we actually cancel out the masses because they appear in all the terms of the problem on the left and right side. We can also do is one thing I like to do is um 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 two. It doesn't change anything as long as you multiply everything by two, nothing changes. But what you do do is sort of get rid of the fractions, so it doesn't end up being is you get the initial squared and then this is gonna be plus two G. Y. Initial. Right, initially this was just M G Y. You have to multiply it by two and it becomes to G. Y. So this is going to be now uh the final square does this is the final squared. Alright, so this is what we get now. This hopefully should look a little familiar to you. This equation is really just the equation number two 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 um cinematics in motion. All right, 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 gonna move this over to the other side. And what we're gonna get is um The the final squared -2 G. Y. initial. And then we're just going to take the square roots. So we're just gonna take the square root of this whole number here. So, what you end up getting for the initial is the square roots. This is going to be 30 squared. 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 -2 times 9, 8 times the initial height of 20. So if you go ahead and plug this in, what you're gonna get is you're actually gonna get two answers for this, You're going to get positive or negative 22. m 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 gonna be a negative number. So it's negative 22.5 m/s. All right, so that's what this one guys, let's move on.
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example

## Conservation Of Energy with Multiple Points 5m
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5
concept

## Conservation Of Total Energy & Isolated Systems 5m
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