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## Class 11 Physics (India)

### Unit 10: Lesson 5

Spring potential energy and Hooke's law- Intro to springs and Hooke's law
- Potential energy stored in a spring
- Spring potential energy example (mistake in math)
- Calculating spring force
- Calculating elastic potential energy
- Spring potential energy and Hooke's law review
- What is Hooke's Law?
- What is elastic potential energy?

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# Potential energy stored in a spring

Learn about the force required to compress a spring, and the work done in the process, and how this relates to Hooke's Law, which defines the restorative force of a spring. Using a graph, see how force increases proportionally with displacement, and how one can use the area under the graph to calculate the work done to compress the spring. Finally, relate this work to the potential energy stored in the spring. Created by Sal Khan.

## Want to join the conversation?

- Spring constant k will vary from spring to spring, correct?(19 votes)
- Yes, the word 'constant' might throw some people off at times. The k constant is only constant for that spring, so a k of -1/2 may only apply for one spring, but not others depending on the force needed to compress the spring a certain distance. Hope this helps!(20 votes)

- if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it). So where does the other half go?(18 votes)
- You're analysis is a bit off here. This is because the force with which you pull the spring is not 4N the entire time.

What do I mean by this? Imagine you were only stretching the spring by small incremental amounts, say increments of 10 cm. The first 10 cm would correspond to an applied force of 0.2 N. This force will steadily increase until you get to your 2 m and 4 N.

Because the applied force is not constant, you cannot use the simple W=F*d. You must instead use this idea of chopping the stretch up into little bits and add up the work done during each of these little bits. If you are familiar with calculus, you should recognize this as a simple definite integration problem. W = ∫(kx) dx from x=0 to d. With this method, you will find that all of your work goes into spring energy.(26 votes)

- why is work work area under the line? if work = f*d and if f= kx and d = x then shouldn't work=kx^2 why is it just the triangle and not the square?(11 votes)
- the formula we've learnt here is assuming F_initial to the spring is 0, not the same as F_final which you may be given in the problem description

if you want to use squre to get W, you need a constant F. thus F_initial = F_final = kx, which can't be 0 at starting of spring's motion(1 vote)

- At7:13sal says thw work is going to be the area below the slope ... a triangle.... but why not the whole rectangle ?(4 votes)
- We are looking for the area under the force curve. We only have a rectangle-like graph when the force is constant. Unfortunately, the force changes with a spring. Basically, we would only have a rectangle graph if our force was constant!

Say you're pushing a box 5 meters. If you apply a CONSTANT force of 13N, then the graph of work (which would be a graph of force on the y axis and displacement on the x axis) would be a rectangle; the vertical component (the force) would be all the way up to 13N (again, on the y axis), while the x component would just be all the way up to 5m.

Why is it not the whole rectangle in the case of a spring? In the case of a spring, the force that one must exert to compress a spring 1m is LESS than the force needed to compress it 2m or 3m, etc. The force needed CHANGES; this is why we are given an EQUATION for the force: F = kx, yes? If the F = a constant, we would, indeed, have a rectangle. When we are given an equation for the force, that means the force is changing and we cannot straight up do W=Fd because F is changing. Because F is changing, there will be different y values for different x's. To make it a little easier, we can revert to integration (which is a b word), because when we integrate a function (say, a function for a force), then we are able to account for the differing Y.

I'm sorry if this was too long of a response, I have had so much trouble with this topic and I still do... Hope this helped! :D(11 votes)

- why is the restorative force -kx, negative(4 votes)
- Because it is in the opposite direction of the displacement, x.(6 votes)

- How would you calculate the equation if you were putting force on the spring from both directions?(2 votes)
- You are always putting force on the spring from both directions. If you weren't, it would move away from you as you tried to push on it.(6 votes)

- So, if the work done is equal to the area under the graph, couldn't the equation just be force times extension divided by 2? Why use a more complex version of the equation, or is it used when the force value is not known?(6 votes)
- Let's say that the graph were a curved shape and to find the area under the curves, we would have to use calculus of course !(1 vote)

- What happens to the potential energy of a bubble whenit rises up in water? Also explain y it is so(4 votes)
- Hello Shunethra,

I like this question but I'm going to turn it back to you.

Assume you are in a swimming pool. Would it take work to "push" a balloon to the bottom?

This is very hard to do. As a kid I remember trying to move a basketball to the bottom of the pool. I couldn't do it. The force on the ball is equal to the weight of the water it displaces. As you go deeper the weight of this water increases. Perhaps you have seen ho air bubbles in crease in size as they rise in the water column...

Regards,

APD(2 votes)

- i dont understand how to find the force constant k of a spring. can you give me some tips on how to start a problem like that.(2 votes)
- So you have F=kx, say you had a 2m spring

F = the force applied, and say I'm pulling the string with a force of 2N

and that stretched the spring 0.2m, so x would be 0.2m

2=k0.2

divide both sides by 0.2

k = 10

now you can apply that to find the force required to stretch the spring any distance, or how much force you need to stretch the string a certain distance(3 votes)

- going off f=-kx, the greater the displacement, the greater the force.

if the spring is being compressed at a constant velocity, the force should not change. but the above equation implies that force will increase as displacement increases. Could someone explain??(2 votes)- The force from a spring is not proportional to the rate of compression. With an ideal spring the more you compress it the more force it will increase. If you compress a spring by X takes half the force of compressing it by 2X.(3 votes)

## Video transcript

Welcome back. So we have this green spring
here, and let's see, there's a wall here. This connected to the wall. And let's say that this is where
the spring is naturally. So if I were not to push on the
spring, it would stretch all the way out here. But in this situation, I pushed
on the spring, so it has a displacement
of x to the left. And we'll just worry about
magnitude, so we won't worry too much about direction. So what I want to do is think
a little bit-- well, first I want to graph how much force
I've applied at different points as I compress
this spring. And then I want to use that
graph to maybe figure out how much work we did in compressing
the spring. So let's look at-- I know I'm
compressing to the left. Maybe I should compress to the
right, so that you can-- well, we're just worrying about the
magnitude of the x-axis. Let's draw a little
graph here. That's my y-axis, x-axis. So this axis is how much I've
compressed it, x, and then this axis, the y-axis, is how
much force I have to apply. So when the spring was initially
all the way out here, to compress it a little
bit, how much force do I have to apply? Well, this was its natural
state, right? And we know from-- well, Hooke's
Law told us that the restorative force-- I'll write
a little r down here-- is equal to negative K, where K is
the spring constant, times the displacement, right? That's the restorative force,
so that's the force that the spring applies to whoever's
pushing on it. The force to compress it is just
the same thing, but it's going in the same direction
as the x. If I'm moving the spring, if I'm
compressing the spring to the left, then the force I'm
applying is also to the left. So I'll call that the force
of compression. The force of compression
is going to be equal to K times x. And when the spring is
compressed and not accelerating in either
direction, the force of compression is going
to be equal to the restorative force. So what I want to do here is
plot the force of compression with respect to x. And I should have drawn it the
other way, but I think you understand that x is increasing
to the left in my example, right? This is where x is equal
to 0 right here. And say, this might be x is
equal to 10 because we've compressed it by 10 meters. So let's see how much
force we've applied. So when x is 0, which is right
here, how much force do we need to apply to compress
the spring? Well, if we give zero force, the
spring won't move, but if we just give a little, little
bit of force, if we just give infinitesimal, super-small
amount of force, we'll compress the spring just
a little bit, right? Because at that point, the force
of compression is going to be pretty much zero. So when the spring is barely
compressed, we're going to apply a little, little bit of
force, so almost at zero. To displace the spring zero,
we apply zero force. To displace the spring a little
bit, we have to apply a little bit more force. To displace soon. the spring 1
meter, so if this is say, 1 meter, how much force
will we have to apply to keep it there? So let's say if this is
1 meter, the force of compression is going to
be K times 1, so it's just going to be K. And realize, you didn't apply
zero and then apply K force. You keep applying a little
bit more force. Every time you compress the
spring a little bit, it takes a little bit more force to
compress it a little bit more. So to compress it 1 meters,
you need to apply K. And to get it there, you have to
keep increasing the amount of force you apply. At 2 meters, you would've been
up to 2K, et cetera. I think you see a
line is forming. Let me draw that line. The line looks something
like that. And so this is how much force
you need to apply as a function of the displacement of
the spring from its natural rest state, right? And here I have positive x going
to the right, but in this case, positive
x is to the left. I'm just measuring its
actual displacement. I'm not worried too much about
direction right now. So I just want you to think
a little bit about what's happening here. You just have to slowly keep
on-- you could apply a very large force initially. If you apply a very large force
initially, the spring will actually accelerate much
faster, because you're applying a much larger force
than its restorative force, and so it might accelerate and
then it'll spring back, and actually, we'll do a little
example of that. But really, just to displace the
spring a certain distance, you have to just gradually
increase the force, just so that you offset the
restorative force. Hopefully, that makes sense,
and you understand that the force just increases
proportionally as a function of the distance, and
that's just because this is a linear equation. And what's the slope of this? Well, slope is rise
over run, right? So if I run 1, this is
1, what's my rise? It's K. So the slope of this
graph is K. So using this graph, let's
figure out how much work we need to do to compress
this spring. I don't know, let's
say this is x0. So x is where it's the
general variable. X0 is a particular
value for x. That could be 10 or whatever. Let's see how much
work we need. So what's the definition
of work? Work is equal to the force
in the direction of your displacement times the
displacement, right? So let's see how much
we've displaced. So when we go from zero
to here, we've displaced this much. And what was the force
of the displacement? Well, the force was gradually
increasing the entire time, so the force is going to be be
roughly about that big. I'm approximating. And I'll show you that you
actually have to approximate. So the force is kind of that
square right there. And then to displace the next
little distance-- that's not bright enough-- my force is
going to increase a little bit, right? So this is the force, this
is the distance. So if you you see, the work I'm
doing is actually going to be the area under the
curve, each of these rectangles, right? Because the height of the
rectangle is the force I'm applying and the width is
the distance, right? So the work is just going to
be the sum of all of these rectangles. And the rectangles I drew are
just kind of approximations, because they don't get
right under the line. You have to keep making the
rectangle smaller, smaller, smaller, and smaller, and just
sum up more and more and more rectangles, right? And actually I'm touching on
integral calculus right now. But if you don't know
integral calculus, don't worry about it. But the bottom line is the work
we're doing-- hopefully I showed you-- is just going to
be the area under this line. So the work I'm doing to
displace the spring x meters is the area from here to here. And what's that area? Well, this is a triangle, so we
just need to know the base, the height, and multiply
it times 1/2, right? That's just the area
of a triangle. So what's the base? So this is just x0. What's the height? Well, we know the slope is K, so
this height is going to be x0 times K. So this point right here
is the point x0, and then x0 times K. And so what's the area under the
curve, which is the total work I did to compress
the spring x0 meters? Well, it's the base, x0, times
the height, x0, times K. And then, of course, multiply by
1/2, because we're dealing with a triangle, right? So that equals 1/2K
x0 squared. And for those of you who know
calculus, that, of course, is the same thing as the
integral of Kx dx. And that should make sense. Each of these are little dx's. But I don't want to go too
much into calculus now. It'll confuse people. So that's the total work
necessary to compress the spring by distance of x0. Or if we set a distance
of x, you can just get rid of this 0 here. And why is that useful? Because the work necessary to
compress the spring that much is also how much potential
energy there is stored in the spring. So if I told you that I had a
spring and its spring constant is 10, and I compressed it 5
meters, so x is equal to 5 meters, at the time that it's
compressed, how much potential energy is in that spring? We can just say the potential
energy is equal to 1/2K times x squared equals 1/2. K is 10 times 25, and
that equals 125. And, of course, work and
potential energy are measured in joules. So this is really what you
just have to memorize. Or hopefully you don't
memorize it. Hopefully, you understand where
I got it, and that's why I spent 10 minutes doing it. But this is how much work is
necessary to compress the spring to that point and how
much potential energy is stored once it is compressed
to that point, or actually stretched that much. We've been compressing,
but you can also stretch the spring. If you know that, then we can
start doing some problems with potential energy in springs,
which I will do in the next video. See