su•per•struc•ture
*
(so̅o̅
′
pər struk′chər),
*
n.

Rieker Y0580, Bottes Classiques femme Noir schwarz/schwarz / 00

per•struc

- Civil Engineering, Architecturethe part of a building or construction entirely above its foundation or basement.
- Civil Engineeringany structure built on something else.
- the overlying framework or features of an organization, institution, or system, built or superimposed on a more fundamental base.
- Nautical, Naval Termsany construction built above the main deck of a vessel as an upward continuation of the sides. Cf. deckhouse.
- Civil Engineeringthe part of a bridge that rests on the piers and abutments.
- anything based on or rising from some foundation or basis:a complex ideological superstructure based on two hypotheses.

Rieker Y0580, Bottes Classiques femme Noir schwarz/schwarz / 00

per•struc

- super- + structure 1635–45

Collins Concise English Dictionary © HarperCollins Publishers::

superstructure
/ˈsuːpəˌstrʌktʃə/n

- the part of a building above its foundation
- any structure or concept erected on something else
- any structure above the main deck of a ship with sides flush with the sides of the hull
- the part of a bridge supported by the piers and abutments

Introduction aux notions d'énergie cinétique et d'énergie potentielle. Par Sal Khan.

Transcription de la vidéo

Welcome back. In the last video, I showed you or hopefully, I did show you that if I apply a force of F to a stationary, an initially stationary object with mass m, and I apply that force for distance d, that that force times distance, the force times the distance that I'm pushing the object is equal to 1/2 mv squared, where m is the mass of the object, and v is the velocity of the object after pushing it for a distance of d. And we defined in that last video, we just said this is work. Force times distance by definition, is work. And 1/2 mv squared, I said this is called kinetic energy. And so, by definition, kinetic energy is the amount of work-- and I mean this is the definition right here. It's the amount of work you need to put into an object or apply to an object to get it from rest to its current velocity. So its velocity over here. So let's just say I looked at an object here with mass m and it was moving with the velocity v. I would say well, this has a kinetic energy of 1/2 mv squared. And if the numbers are confusing you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram object and it's moving at 7 meters per second. So I would say the kinetic energy of this object is going to be 5-- 1/2 times the mass times 5 times 7 squared, times velocity squared. It's times 49. So let's see. 1/2 times 49, that's a little under 25. So it'll be approximately 125 Newton meters, which is approximately-- and Newton meter is just a joule-- 125 joules. So this is if we actually put numbers in. And so when we immediately know this, even if we didn't know what happened, how did this object get to this speed? Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force. So let's say the force that I apply is equal to mass times the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already moving because you would actually have to apply a force slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if it was slowing down upwards, then the force being applied would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? And so, let's see. This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it doesn't have any kinetic energy. But it now has a lot of potential to do work. And what do I mean by potential to do work? Well after I move an object up 100 meters into the air, what's its potential to do work? Well, I could just let go of it and have no outside force other than gravity. The gravity will still be there. And because of gravity, the object will come down and be at a very, very fast velocity when it lands. And maybe I could apply this to some machine or something, and this thing could do work. And if that's a little confusing, let me give you an example. It all works together with our-- So let's say I have an object that is-- oh, I don't know-- a 1 kilogram object and we're on earth. And let's say that is 10 meters above the ground. So we know that its potential energy is equal to mass times gravitational acceleration times height. So mass is 1. Let's just say gravitational acceleration is 10 meters per second squared. Times 10 meters per second squared. Times 10 meters, which is the height. So it's approximately equal to 100 Newton meters, which is the same thing is 100 joules. Fair enough. And what do we know about this? We know that it would take about 100-- or exactly-- 100 joules of work to get this object from the ground to this point up here. Now what we can do now is use our traditional kinematics formulas to figure out, well, if I just let this object go, how fast will it be when it hits the ground? And we could do that, but what I'll show you is even a faster way. And this is where all of the work and energy really becomes useful. We have something called the law of conservation of energy. It's that energy cannot be created or destroyed, it just gets transferred from one form to another. And there's some minor caveats to that. But for our purposes, we'll just stick with that. So in the situation where I just take the object and I let go up here, up here it has a ton of potential energy. And by the time it's down here, it has no potential energy because the height becomes 0, right? So here, potential energy is equal to 100 and here, potential energy is equal to 0. And so the natural question is-- I just told you the law of conservation of energy, but if you look at this example, all the potential energy just disappeared. And it looks like I'm running out of time, but what I'll show you in the next video is that that potential energy gets converted into another type of energy. And I think you might be able to guess what type that is because this object is going to be moving really fast right before it hits the ground. I'll see you in the next video.

FRANCE 3
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Mis à jour le 25/06/2017 13:25

publié le 25/06/2017 13:25

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