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02:55

Keshav S.

(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?

0:00

Suman Saurav T.

(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?

00:56

Donald A.

I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?

01:24

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welcome to our first example video. Looking at restoring and dissipated forces in this video, we're going to continue to look at Hook's Law. F equals negative K X and some of the things we need to be aware with it. OK, eso say we have a spring that is oriented horizontally with a mass that comes up to it and we are we're to compress the spring amount will say we compress it one centimeter and it has a K equal to 100 kilograms per second squared. Okay, So given this, then how much force is it pushing back with? We would say, Well, f is going to be equal to negative 100 kilograms per second. Squared, multiplied by one centimeter is 0.1 meters. Remember this is in the negative. I had direction. So in the eye hat, we have a negative times a negative 0.1 times 100. That's gonna bring it back to decimal places. So all the way back to one, and so we'll end up with force is equal to one Newton. So at that point where we have compressed it by one centimeter, it is pushing out with a force of one Newton. And if it's being held stable, it means we must be pushing in with the fourth of one Newton so, And that would mean that it would stay there with one centimeter compression for as long as we continue to Maine, that one Newton force. If, on the other hand, we had pulled it out by one Newton, we would be the force on the right here pulling it out, and it would be the force on the left trying to pull it back in each again with the force of one Newton. Now, if we were to turn a spring vertically, hang it from a ceiling, for example, and then tie a mass to it. When we draw the free body diagram for the mass year, we're going to have MG down, and then we'll have a spring force up, which will mean that we need a spring force minus M. G equals zero Newton's when it is at an equilibrium point where it's not moving anymore. Now for spring, many students will make the mistake of saying okay for spring is equal to negative K X, and they blindly plug in negative k x here and make a sign air. We need to make sure that we remember we've already assigned directions by putting in negative mg and drawing our picture, which means that we can simply plug in K X minus m g equals. See your own Newton's, which means that this spring will be extended by an amount M g over K, as we found over here so it will hang down mg over K farther. That means it will be stretched that amount and then it will just sit there as long as we don't give it any sort of disturbance. If we carefully hook on the mass and then slowly lower it now, having done that, then if we were to press, compress or pull down this system with the mass already on it, say we were to add a secondary mass, then the spring would behave just like it had before it, except it would have a new equilibrium. Point noticed that when we were working with the horizontal position, the equilibrium point was simply where it is neither stretched or compressed. When we're working with a vertical spring, it's wherever it happens to be stretched to based on the amount of gravity pulling on it, that is to say, the amount of weight that we attached to it. So we attach a weight mg. It pulls it out on amount M g. K and down an amount M g k. And then it will sit there, and that becomes our new X equals zero point. And if we compress this way, we'll have ah, Delta X up. If we pull it down this way, will have a negative Delta X down The most effective way, though, to deal with these vertical springs isn't to try and maintain the X equals zero you had before. It's to calculate where your new X equals zero is and then operate with your zero there. Remember, as I've said before, with position that zero is determined by you, don't try to feel like you need to stick to some sort of external choice. I'm telling you right here that this is the easiest choice, and so I definitely recommend that you do it. You'll notice I've used X here simply because we're using a spring but is in the Y direction. If you're more comfortable writing out F equals negative K y. In this case, by all means, go ahead and do it

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