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### Course: Physics library > Unit 5

Lesson 2: Springs 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?(26 votes)
- 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.(21 votes)

- 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 ?(9 votes)
- 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.(26 votes)

- What are elastomers ?

Please help. Thank you.(14 votes)- 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.(16 votes)

- 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?(7 votes)- 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(9 votes)

- well what does x stand for?(6 votes)
- 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)(7 votes)

- 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?(4 votes)- 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??(6 votes)

- k is always positive so it cant be -2 at9:38

F=-kx <->-2 = -kx <-> -2=-k <-> k=2; is the correct equation(6 votes) - 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.(4 votes)
- I saw F= k*x^2/2 what is that different from Hooke's law?(2 votes)
- I believe you mean U = k*x^2/2. This is equivalent to Hooke's law. (U is potential energy stored by spring device)(4 votes)

- 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.(2 votes)
- 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.(2 votes)

## 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.