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University of Michigan - Ann Arbor

Numerade Educator

University of Washington

Simon Fraser University

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Welcome to our fourth example video considering spring potential energy in this video, we're going to consider the problem of a spring hanging from the ceiling. Now, normally speaking, spring is going to have some standard length Call little L. And when we attach a mass to it, it's going to have an extended length. Okay? And we're going to call the distance of this extended length capital l here, Um, if I were to draw a free body diagram for this and we're going to start with the forces argument here, but it will work our way back into an energy argument. When I draw the free body diagram here, I know I have a force mg down and then I have a force due to the spring pulling it back up, and that's gonna be a k times the length L because that's how far it's been pulled down. Now, when I then say well, F net is equal to K l minus m g, which equals zero eso I could actually solve for Al here. I could say that l is equal to M g over K. Meanwhile, when I take a look at, let's say if I pushed it up a little bit. Okay, A distance. Delta y. What would be the force here? Well, I still have mg down, but unless I have compressed it by an amount mg over K, then I'm still having a force up due to the spring. In this case, the elastic force would be equal to que times l minus delta. Why? Because we're no longer compressed. The air extended the entire distance. L it's been reduced by an amount Delta y. Now, when I goto ask what's the Net force here? What I'm going to get is Well, I've got f elastic minus m g, which is K times L minus delta Y minus M g and then notice up here. K times l minus M G is equal to zero. That's the condition for equilibrium. And here, if I distribute K through, I'll have a k l minus Delta y minus M g or, in other words, a k l minus m g minus delta y times K. So this is zero. Which means that our net force in this situation is simply negative. K Delta y. Well, that's just Hook's law all over again. Which means when we go to look at this vertical spring with the mass on the end and we're letting it oscillate back and forth. All we have to worry about is Hook's law. So the energy in the spring is gonna look like the U Total. Actually, we country as just one half K X squared or y squared, I should say for why is the distance from the extended length L So we're now considering our new equilibrium to be at the point where we've extended it a distance. L and we're oscillating around that position. And if we do that, then we can completely eliminate gravity from are concerned in our forces. Which means that because of the relationship between potential energy and force, I can also eliminate it from my concerns for potential energy. So if I were to say, What's my potential energy and kinetic energy interaction gonna look like? Well, if I pull it down a distance, why, from my new equilibrium, my potential energy be will will be one half que y squared. My kinetic energy would be zero and then I release it and it passes through my new equilibrium. Why equals zero, which remember is the extended length L. And I'll have my maximum kinetic energy here. And then when it goes up a distance, why all once again have one half K y squared in my potential energy and I'll have zero jewels in my kinetic energy. So we don't even have to worry about a gravitational term because it actually gets eliminated from the force since the forces simply negative K Delta y. And if you remember U is equal to or rather Delta U is equal to negative the integral of F. D Y. This is the only term we have to worry about. It's a neat little trick. Just remember that you have to let it extend by the natural distance it wants to, which will be L or MG over K.

University of North Carolina at Chapel Hill

Energy Conservation

Moment, Impulse, and Collisions

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Gravitation