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### Course: AP®︎/College Physics 1>Unit 5

Lesson 5: Spring potential energy and Hooke's law

# Intro to springs and Hooke's law

Discover the phenomena of springs and Hooke's Law. Explore how force applied to a spring results in compression or elongation, and how this relationship is linear. Uncover the concept of restorative force and how it counteracts applied force, keeping our spring in equilibrium. Created by Sal Khan.

## Want to join the conversation?

• I have a simple question: On a microscopic level, what force(s) makes the spring want to restore itself to its original position?
• The equilibrium position of the string minimizes the potential energy of the metal atoms from which the spring is constructed. I don't know how much inorganic chemistry you know, but what it basically comes down to is that chemical bonds have a sort of "best distance of separation." Atoms consist of subatomic constituents that carry electric charges (nuclei are positively charged due to protons and the space around the exterior of the atom has a negative charge due to the electrons). Bring two nuclei too close together and their positive like charges will repel--this is the case with spring compression. Separate the bound atoms by too great a distance and the "electron glue" that holds them together will create tension that draws the atoms back together--this is stretching the spring. Keep in mind that this is a very very simplified explanation of what really happens, to truly understand the nature of this elastic energy you must study the wave-nature of subatomic particles and quantum mechanics.
• What does the constant K of the spring determine ? Does it vary from spring to spring depending on the material it's made of ? If not , then what ?
• K is a constant that represents the elasticity of a spring (and therefore stiffness). It varies between spring to spring, depending on what it is made of, the shape of the spring, and the width of the wire.
• What are elastomers ?
• Elastomers are polymers that are designed to be highly elastic (they can stretch far without breaking). For example, yoga pants, running tights, nylons, and the rubber bands are all composed of elastomers.
The concept was first discovered with the sap of the rubber tree, which was distilled into highly flexible material. Today, manufacturing companies can artificially design/construct their own materials.
• As Sal explained in the video that the restoring for is ' F = -kx", he also take the restoring force to be equal to the force applied by Newton's 3rd Law. But how is it necessary to be the same. If k has big value, we will have to apply large force for very small displacement in that case restoring force will be lesser because of very small value of 'x'.
So how can Sal take it to be equal to the force applied?
• Sal is right. If k has a big value, you will have to apply a huge force to cause a small displacement. But that also means (k having a big value) that for a small displacement the restoring force will be huge.
In the end everything is balanced and Newton's 3rd law still holds :D
• well what does x stand for?
• x is the displacement of the spring's end from its equilibrium point - how much the spring is stretched (or, in the other direction, compressed)
• I have a query regarding springs in a series combination.
Say, there's a thread attached to the ceiling, connecting three springs vertically and a block of mass 'M' connected at the bottom.
It is said that the force is same for all springs in series (=Mg in this case). How is this possible if we have mass-full springs?
Or is it applicable solely for 'mass-less' springs?
• great point.

yes, usually we have 'mass-less' springs... ie their mass is very very small compared to the mass being suspended by the springs.

OR

we may be given the 'mass per unit length' of each spring and therefore able to figure out the extension due to the mass of the springs, but this is probably an unlikely situaiton (though not impossible in some courses)

ok??
• k is always positive so it cant be -2 at
F=-kx <->-2 = -kx <-> -2=-k <-> k=2; is the correct equation
• Almost missed the last part, the last five to ten seconds there's an assumption to the problem that's changed, very hard to capture that change in a matter of five seconds.
• I saw F= k*x^2/2 what is that different from Hooke's law?
• I believe you mean U = k*x^2/2. This is equivalent to Hooke's law. (U is potential energy stored by spring device)
• Are the units for "K" kg/s^2 ? In that case does the number have this units for an experimental reason or are the units in there to make the result in newtons.
• F = kx. What MUST the units of k be if x is in meters and F has to be in Newtons?
That's how constants work.

## Video transcript

Let's learn a little bit about springs. So let's say I have a spring. Let me draw the ground so that we know what's going on with the spring. So let me see, this is the floor. That's the floor, and I have a spring. It's along the floor. I'll use a thicker one, just to show it's a spring. Let's say the spring looks something like this. Whoops, I'm still using the line tool. So the spring looks like this. This is my spring, my amazingly drawn spring. Let's say at this end it's attached to a wall. That's a wall. And so this is a spring when I don't have any force acting on it, this is just the natural state of the spring. And we could call this, where it just naturally rests, this tip of the spring. And let's say that when I were to apply a force of 5 Newtons into the spring, it looks something like this. Redraw everything. So when I apply a force of 5 Newtons-- I'll draw the wall in magenta now. When I apply a force of 5 Newtons, the spring looks like this. It compresses, right? We're all familiar with this. We sit on a bed every day or a sofa. So let's say it compresses to here. If this was the normal resting-- so this is where the spring was when I applied no force, but when I applied 5 Newtons in that direction, let's say that this distance right here is 10 meters. And so a typical question that you'll see, and we'll explain how to do it, is a spring compresses or elongates when you apply a certain force by some distance. How much will it compress when you apply a different force? So my question is how much will it compress when I apply a 10-Newton force? So your intuition that it'll compress more is correct, but is it linear to how much I compress it? Is it a square of how much I compress it? How does it relate? I think you probably could guess. It's actually worth an experiment. Or you could just keep watching the video. So let's say I apply a 10-Newton force. What will the spring look like? Well, it'll be more compressed. Drop my force to 10 Newtons. And if this was the natural place where the spring would rest, what is this distance? Well, it turns out that it is linear. What do I mean by linear? Well, it means that the more the force-- it's equally proportional to how much the spring will compress. And it actually works the other way. If you applied 5 Newtons in this direction, to the right, you would have gone 10 meters in this direction. So it goes whether you're elongating the spring or compressing the spring within some reasonable tolerance. We've all had this experience. If you compress something too much or you stretch it too much, it doesn't really go back to where it was before. But within some reasonable tolerance, it's proportional. So what does that mean? That means that the restoring force of the spring is minus some number, times the displacement of the spring. So what does this mean? So in this example right here, what was the displacement of the spring? Well, if we take positive x to the right and negative x to the left, the displacement of the spring was what? The displacement, in this example right here, x is equal to minus 10, right? Because I went 10 to the left. And so it says that the restorative force is going to be equal to minus K times how much it's distorted times minus 10. So the minuses cancel out, so it equals 10K. What's the restorative force in this example? Well, you might say, it's 5 Newtons, just because that's the only force I've drawn here, and you would be to some degree correct. And actually, since we're doing positive and negative, and this 5 Newton is to the left, so to the negative x-direction, actually, I should call this minus 5 Newtons and I should call this minus 10 Newtons, because obviously, these are vectors and we're going to the left. I picked the convention that to the left means negative. So what's the restorative force? Well, in this example-- and we assume that K is a positive number for our purposes. In this example, the restorative force is a positive number. So what is the restorative force? So that's actually the force, the counteracting force, of the spring. That's what this formula gives us. So if this spring is stationary when I apply this 5-Newton force, that means that there must be another equal and opposite force that's positive 5 Newtons, right? If there weren't, the spring would keep compressing. And if the force was more than 5 Newtons, the spring would go back this way. So the fact that I know that when I apply a 5-Newton force to the left, or a negative 5-Newton force, the spring is no longer moving, it means that there must be-- or no longer accelerating, actually, it means that there must be an equal and opposite force to the right, and that's the restorative force. Another way to think about it is if I were to let-- well, I won't go in there now. So in this case, the restorative force is 5 Newtons, so we can solve for K. We could say 5 is equal to 10K. Divide both sides by 10. You get K is equal to 1/2. So now we can use that information to figure out what is the displacement when I apply a negative 10-Newton force? When I push the spring in with 10 Newtons in the leftward direction? So first of all, what's the restorative force here? Well, if the spring is no longer accelerating in either direction, or the tip of the spring is no longer accelerating in either direction, we know that the restorative force must be counterbalancing this force that I'm compressing with, right? The force that the spring wants to expand back with is 10 Newtons, positive 10 Newtons, right? And we know the spring constant, this K for this spring, for this material, whatever it might be, is 1/2. So we know the restorative force is equal to 1/2 times the distance, right? And the formula is minus K, right? And then, what is the restorative force in this example? Well I said it's 10 Newtons, so we know that 10 Newtons is equal to minus 1/2x. And so what is x? Well, multiply both sides by minus 1/2, and you get minus 20. I'm sorry, multiply both sides by minus 2, you get minus 20 is equal to x. So x goes to the left 20 units. So that's all that it's telling us. And this law is called Hooke's Law, and it's named after-- I'll read it-- a physicist in the 17th century, a British physicist. And he figured out that the amount of force necessary to keep a spring compressed is proportional to how much you've compressed it. And that's all that this formula says. And that negative number, remember, this formula gives us the restorative force. So it says that the force is always in the opposite direction of how much you displace it. So, for example, if you were to displace this spring in this direction, if you were to apply a force and x were a positive and you were to go in that direction, the force-- no wait, sorry. This is where the spring rests. If you were to apply some force and take the spring out to here, this negative number tells us that the spring will essentially try to pull back with the restorative force in the other direction. Let's do one more problem and I think this will be clear to you. So let's say I have a spring, and all of these problems kind of go along. So let's say when I apply a force of 2 Newtons, so this is what I apply when I apply a force of 2 Newtons. Well, let's say it this way. Let's say when I stretch the spring. Let's say this is the spring, and when I apply a force of 2 Newtons to the right, the spring gets stretched 1 meter. So first of all, let's figure out what K is. So if the spring is stretched by 1 meter, out here, its restorative force will be 2 Newtons back this way, right? So its restorative force, this 2 Newtons, will equal minus K times how much I displaced it. Well I, displaced it by 1 meter, so then we multiply both sides by negative 1, and we get K is equal to minus 2. So then we can use Hooke's Law to note the equation for this-- to figure out the restorative force for this particular spring, and it would be minus 2x. And then I said, well, how much force would I have to apply to distort the spring by 2 meters? Well, it's 2 times 2, it would be 4. 4 Newtons to displace it by 2 meters, and, of course, the restorative force will then be in the opposite direction, and that's where we get the negative number. Anyway, I've run out of time. I'll see you in the next video.