10. Conservation of Energy
Intro to Conservation of Energy
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Hey, guys. So let's go ahead and check out this problem here. You're gonna launch a 4 kilogram object directly up from the ground. So I go to the ground like this. You're gonna take this 4 kilogram ball and you're gonna throw it up with some initial speed v naught equals 40. And we wanna figure out basically what happens when the ball finally comes to a stop here at some maximum heights. So here, if you take the ground level to be 0, what we wanna calculate is what is this y max value here. Alright? So we've got some changing speeds and changing heights. We know we're gonna 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 u initial equals k final plus u final. Now we're gonna 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 equals 0 to be the grounds, 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 no nothing there. What about k final? So what happens is when this object gets up to its maximum height, the velocity is gonna be 0. So here the velocity final equals 0. Therefore, there is no kinetic energy and there is gonna be some, potential energy because now we're at some height above the ground here. So So let's go ahead and expand out our terms. What I've got here is 1 half m v initial squared equals, and then the gravitational potential energy is gonna be mgh, or rather mgymax. So I'm gonna call this ymax here and I'm gonna go ahead and solve for this. Well, one of the things we'll notice here is that the mass is gonna cancel. Usually, that happens with these kinds of problems. And now we're just gonna go ahead and figure out y final. So oh, sorry. Y max. So y max or y final is gonna be 1 half of v naught squared divided by g. So if you go ahead and plug in our numbers here, you're gonna have 1 half of 40 squared divided by 9.8, and you're gonna 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 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. Alright. So that's it for this one, guys. Let me know if you have any questions.
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