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Welcome back.

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So we have this green spring
here, and let's see,

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there's a wall here.

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This connected to the wall.

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And let's say that this is where
the spring is naturally.

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So if I were not to push on the
spring, it would stretch

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all the way out here.

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But in this situation, I pushed
on the spring, so it

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has a displacement
of x to the left.

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And we'll just worry about
magnitude, so we won't worry

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too much about direction.

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So what I want to do is think
a little bit-- well, first I

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want to graph how much force
I've applied at different

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points as I compress
this spring.

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And then I want to use that
graph to maybe figure out how

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much work we did in compressing
the spring.

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So let's look at-- I know I'm
compressing to the left.

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Maybe I should compress to the
right, so that you can-- well,

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we're just worrying about the
magnitude of the x-axis.

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Let's draw a little
graph here.

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That's my y-axis, x-axis.

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So this axis is how much I've
compressed it, x, and then

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this axis, the y-axis, is how
much force I have to apply.

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So when the spring was initially
all the way out

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here, to compress it a little
bit, how much force

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do I have to apply?

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Well, this was its natural
state, right?

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And we know from-- well, Hooke's
Law told us that the

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restorative force-- I'll write
a little r down here-- is

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equal to negative K, where K is
the spring constant, times

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the displacement, right?

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That's the restorative force,
so that's the force that the

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spring applies to whoever's
pushing on it.

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The force to compress it is just
the same thing, but it's

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going in the same direction
as the x.

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If I'm moving the spring, if I'm
compressing the spring to

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the left, then the force I'm
applying is also to the left.

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So I'll call that the force
of compression.

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The force of compression
is going to be

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equal to K times x.

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And when the spring is
compressed and not

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accelerating in either
direction, the force of

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compression is going
to be equal to

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the restorative force.

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So what I want to do here is
plot the force of compression

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with respect to x.

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And I should have drawn it the
other way, but I think you

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understand that x is increasing
to the left in my

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example, right?

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This is where x is equal
to 0 right here.

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And say, this might be x is
equal to 10 because we've

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compressed it by 10 meters.

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So let's see how much
force we've applied.

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So when x is 0, which is right
here, how much force do we

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need to apply to compress
the spring?

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Well, if we give zero force, the
spring won't move, but if

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we just give a little, little
bit of force, if we just give

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infinitesimal, super-small
amount of force, we'll

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compress the spring just
a little bit, right?

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Because at that point, the force
of compression is going

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to be pretty much zero.

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So when the spring is barely
compressed, we're going to

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apply a little, little bit of
force, so almost at zero.

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To displace the spring zero,
we apply zero force.

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To displace the spring a little
bit, we have to apply a

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little bit more force.

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To displace soon. the spring 1
meter, so if this is say, 1

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meter, how much force
will we have to

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apply to keep it there?

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So let's say if this is
1 meter, the force of

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compression is going to
be K times 1, so it's

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just going to be K.

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And realize, you didn't apply
zero and then apply K force.

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You keep applying a little
bit more force.

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Every time you compress the
spring a little bit, it takes

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a little bit more force to
compress it a little bit more.

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So to compress it 1 meters,
you need to apply K.

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And to get it there, you have to
keep increasing the amount

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of force you apply.

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At 2 meters, you would've been
up to 2K, et cetera.

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I think you see a
line is forming.

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Let me draw that line.

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The line looks something
like that.

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And so this is how much force
you need to apply as a

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function of the displacement of
the spring from its natural

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rest state, right?

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And here I have positive x going
to the right, but in

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this case, positive
x is to the left.

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I'm just measuring its
actual displacement.

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I'm not worried too much about
direction right now.

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So I just want you to think
a little bit about what's

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happening here.

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You just have to slowly keep
on-- you could apply a very

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large force initially.

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If you apply a very large force
initially, the spring

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will actually accelerate much
faster, because you're

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applying a much larger force
than its restorative force,

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and so it might accelerate and
then it'll spring back, and

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actually, we'll do a little
example of that.

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But really, just to displace the
spring a certain distance,

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you have to just gradually
increase the force, just so

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that you offset the
restorative force.

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Hopefully, that makes sense,
and you understand that the

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force just increases
proportionally as a function

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of the distance, and
that's just because

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this is a linear equation.

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And what's the slope of this?

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Well, slope is rise
over run, right?

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So if I run 1, this is
1, what's my rise?

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It's K.

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So the slope of this
graph is K.

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So using this graph, let's
figure out how much work we

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need to do to compress
this spring.

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I don't know, let's
say this is x0.

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So x is where it's the
general variable.

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X0 is a particular
value for x.

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That could be 10 or whatever.

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Let's see how much
work we need.

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So what's the definition
of work?

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Work is equal to the force
in the direction of your

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displacement times the
displacement, right?

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So let's see how much
we've displaced.

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So when we go from zero
to here, we've

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displaced this much.

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And what was the force
of the displacement?

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Well, the force was gradually
increasing the entire time, so

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the force is going to be be
roughly about that big.

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I'm approximating.

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And I'll show you that you
actually have to approximate.

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So the force is kind of that
square right there.

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And then to displace the next
little distance-- that's not

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bright enough-- my force is
going to increase a little

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bit, right?

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So this is the force, this
is the distance.

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So if you you see, the work I'm
doing is actually going to

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be the area under the
curve, each of

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these rectangles, right?

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Because the height of the
rectangle is the force I'm

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applying and the width is
the distance, right?

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So the work is just going to
be the sum of all of these

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rectangles.

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And the rectangles I drew are
just kind of approximations,

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because they don't get
right under the line.

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You have to keep making the
rectangle smaller, smaller,

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smaller, and smaller, and just
sum up more and more and more

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rectangles, right?

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And actually I'm touching on
integral calculus right now.

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But if you don't know
integral calculus,

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don't worry about it.

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But the bottom line is the work
we're doing-- hopefully I

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showed you-- is just going to
be the area under this line.

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So the work I'm doing to
displace the spring x meters

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is the area from here to here.

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And what's that area?

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Well, this is a triangle, so we
just need to know the base,

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the height, and multiply
it times 1/2, right?

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That's just the area
of a triangle.

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So what's the base?

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So this is just x0.

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What's the height?

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Well, we know the slope is K, so
this height is going to be

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x0 times K.

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So this point right here
is the point x0, and

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then x0 times K.

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And so what's the area under the
curve, which is the total

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work I did to compress
the spring x0 meters?

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Well, it's the base, x0, times
the height, x0, times K.

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And then, of course, multiply by
1/2, because we're dealing

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with a triangle, right?

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So that equals 1/2K
x0 squared.

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And for those of you who know
calculus, that, of course, is

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the same thing as the
integral of Kx dx.

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And that should make sense.

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Each of these are little dx's.

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But I don't want to go too
much into calculus now.

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It'll confuse people.

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So that's the total work
necessary to compress the

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spring by distance of x0.

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Or if we set a distance
of x, you can just get

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rid of this 0 here.

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And why is that useful?

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Because the work necessary to
compress the spring that much

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is also how much potential
energy there is

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stored in the spring.

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So if I told you that I had a
spring and its spring constant

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is 10, and I compressed it 5
meters, so x is equal to 5

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meters, at the time that it's
compressed, how much potential

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energy is in that spring?

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We can just say the potential
energy is equal to 1/2K times

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x squared equals 1/2.

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K is 10 times 25, and
that equals 125.

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And, of course, work and
potential energy

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are measured in joules.

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So this is really what you
just have to memorize.

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Or hopefully you don't
memorize it.

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Hopefully, you understand where
I got it, and that's why

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I spent 10 minutes doing it.

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But this is how much work is
necessary to compress the

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spring to that point and how
much potential energy is

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stored once it is compressed
to that point, or actually

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stretched that much.

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We've been compressing,
but you can

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also stretch the spring.

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If you know that, then we can
start doing some problems with

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potential energy in springs,
which I will

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do in the next video.

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See

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