10. Conservation of Energy

Springs & Elastic Potential Energy

# Anderson Video - Energy- Springs

Professor Anderson

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>> Hello class. Let's talk about a slightly different conservation of energy problem. Let's talk about a spring problem. We didn't really talk a lot about springs yet. But what we did say was that springs have potential energy in them when they are compressed. And so, let's do the following problem. Let's say we have an object that is sliding along a frictionless surface, okay with some initial speed v sub i. And then it's going to hit a spring that is attached to the wall. Okay, this is the before picture. And we all know what's going to happen, right? When it hits that spring, it's going to compress it slightly. Spring gets compressed, block is stuck right there at a final speed of zero, when the spring has compressed the maximum amount. Now, one thing that we need to know is how far did that spring compress. It compressed a distance x, x is always measured from equilibrium to maximum compression. And so, that is the distance x that we're worried about. And it looks like we have all the other terms that we need. Okay this is the before picture, this is the after picture. And we're going to say that this surface is frictionless. Okay let's write down conservation of energy and see if we can make some sense of it. So, energy in this initial case, there is energy in the moving block if it's mass m, then it has energy one half m v i squared. There's no energy anywhere else in the system. In the final picture, the block has come to rest, there's no kinetic energy in the block anymore, but the spring has compressed. It compressed a distance x. k is the spring constant for that particular spring. Depends on the material, depends on how tightly its wound, depends on how thick the material is. It is dependent on the particular spring that you're using. But x is how far it gets compressed. And let's say now that we're trying to solve this thing for x. Alright, these are equal. So, I get one half k x squared equals one half m v i squared. And now I can quickly solve this thing for x, the halves go away. I can write this as x is equal to m over k v i squared. And I'm going to take the square root of that whole thing. If I take the square root of v i square, I can in fact bring it out of the square root. And so, we can write the whole thing like that. How far does the spring compress? It is square root of m over k times v i. And now when you look at this solution, you should make sur that it makes sense to you. If that mass is heavier, that spring is going to get compressed more. m is there in the numerator, so that will increase x as m increases. If the spring is really strong, then k is a big number. A big spring constant means a strong spring. And so, it will compress less. And that makes sense because it's in the numerator, I mean in the denominator, right there. If k goes up, then x of course goes down. And then finally, if I throw in this mass at a faster initial speed, it's going to compress that spring even more. Okay, so all those limits seem to indicate that we must have the right answer. And if you look at the units, you can evaluate the units, you have to be careful because k has some funky units on it. Okay, but once you plug those in, you will get units of meters when you're all done. So, conservation of energy again, but now we're dealing with potential energy in this spring, as opposed to potential energy due to gravity. Alright, questions about that? Everybody okay with that one? Okay. Good.

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