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Kathleen Tatem
(II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 $g$'s. Calculate the force on a 65-kg person accelerating at this rate.What distance is traveled if brought to rest at this rate from 95 km/h?
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Muhammed Shafi
(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?
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welcome to the section on spring potential energy or elastic potential energy. As you might guess, this is going to be associated with the elastic or spring force. So we know that that force is equal to negative K X. So if we plug that in, then we know that work is equal to FDX. Therefore, work is he gonna be equal to integral negative K x d x, And when we calculate this will come up with work is equal to negative one half. Okay, times X squared, have an X final squared and then we'll have a one half. Okay, X initial squared. Okay, So work is negative. One half K Delta X squared. We're to plug out the one hat, pull out the one half in the K, and we could say then that you will be equal to one half k x squared and dealt to you will be equal to one half k X final squared minus one half k X initial squared. Okay, so this is going to be our potential energy expression for elastic or spring forces. I'm going to denote that by writing a little e l for elastic here as the subscript. You'll find that different books use different sub scripts. So in this video we're going to consider how do we use spring potential energy as our potential energy instead of gravitational potential energy, It's gonna work much the same as away as it has with gravitational, where we know that we have a total mechanical energy that is equal to kinetic plus potential and will use this interaction in order to solve problems. Except now we have two possibilities. We have gravitational potential energy, and we have elastic potential energy. So energy will be split could be split between all three of these. For example, in a system where you have a spring hailing hanging from the ceiling and you attach a mass to it and then it oscillates up and down. So if it does this, then you have changes in gravitational potential energy. You have changes in spring potential energy, and you have changes in kinetic energy. So this is a lot of variable t keep track of, and we don't have very many equations, which means we're going to need to kind of established markers and really be clever with where we pick y equals zero notice that by default with F equals negative K x, we choose X equals zero to the be the point at which the spring is neither compressed nor stretched. So that's X equals zero. If we compress it, we get a negative X. If we pull it out, we get a positive X. We need to make sure we keep track of that. So this has already been dictated for us. What? X equals zero. As though notice here. As I stated in a previous video on Springs, that when you turn it upside down with a mass on the end, you will get some stretch. Okay, you'll have looking at the free body diagram again. You have mg down and you have Okay, why up? So that means why is going to be equal to M G over K. And that will be the amount of stretch you get when thes two are equal to each other. Because we'll have k y minus m g equals zero at equilibrium. Okay, so it stretched a little bit. So what you do is you simply pick a new Y equals zero. You'd pick here to be y equals zero for the f equals K Y equation, as opposed to the zero for the spring. Being at the original length, it's gonna be at this stretched length instead, Um, s Oh, there's a few little tricks like that that we're going to take a look at in the coming videos and figure out how it is that energy is stored in a spring and how we can get it out of the spring again.
Energy Conservation
Moment, Impulse, and Collisions
Rotation of Rigid Bodies
Dynamics of Rotational Motion
Gravitation
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