Video Transcript
In this video, we’re talking about
elastic potential energy. This sort of potential energy gets
my vote for being the most fun kind, because it’s behind things like trampolines,
rubber bands, and bows and arrows. Of all the objects we associate
with elastic potential energy though, the one that comes up over and over, probably
more than any other, is the spring. In fact, springs show up when we
talk about this kind of potential energy so often that sometimes elastic potential
energy is called spring potential energy. So if we hear that phrase, it means
the same thing as elastic potential energy. But we’ll keep the term elastic
because it’s a bit more general.
Now to get a better understanding
of this term, let’s consider it word by word, starting with this last word
“energy.” We can think of energy as the
ability of something to do work, that is, to exert a force on some object over some
distance. So that’s energy. And one type of energy we know is
potential energy. This word potential means stored up
or held in reserve. And then finally, this word
elastic. This term describes an object that
if we change its shape, then the object returns quickly to its original shape
afterward. So, for example, considering our
spring here, which is an elastic object. If we were to push on the end of
the spring and compress it in some distance. Then the moment we would release
our hand, the spring would spring back to its original or natural length. In that way, it returns to its
original shape after it’s been deformed. So this spring is an elastic
object.
So when we talk about elastic
potential energy, we’re talking about the ability to do work. Which is stored up in some object
due to or because of some deformation of that object and its tendency to elastically
return to its original shape. So in the case of the spring, we
would need to either compress it or extend it beyond its natural length in order to
give it elastic potential energy.
Now, in order to get a deeper
understanding of this topic, it’s helpful to talk not about energy, but about
force. Say that we do this, say we take
our hand. And we start to push to the left on
the spring. We exert a force on it to compress
it. Let’s say that as we press on the
spring, we compress it at distance 𝑥 from its natural length or its equilibrium
length. Now, there’s a helpful law, which
tells us just how much force we have to exert on the spring in order to do this. That law is known as Hooke’s
law. And here’s what it tells us.
It says that the force we need to
apply to an elastic object, like a spring, is directly proportional to the
compression or extension of that spring. In other words, if we needed to
apply a force 𝐹 to compress a spring in amount 𝑥, then if you wanted to double our
compression. Say we wanted to compress it by two
𝑥 that distance, we would need to apply twice as much force. So the force, Hooke’s law says,
varies with the compression or the extension. It could go the other way too. And in particular, that force is
proportional to the change in the length of the spring from its natural length.
Now there’s another equivalent
mathematical way of writing this expression. We could also write it like
this. We could say that the force we
apply is equal to some constant of proportionality, typically called 𝑘, multiplied
by displacement. These two relationships are saying
the same thing. Often though, when we see Hooke’s
law written out, we see it written this way. 𝐹 is equal to 𝑘 times 𝑥. And when we’re working with springs
such as we are over here. Then this constant of
proportionality 𝑘 has a special name. It’s called the spring
constant.
The spring constant 𝑘 of a spring
is unique to that spring. If we look at the units of 𝑘,
those units are newtons per meter. If we want to stretch or compress a
given spring by a distance of one meter, then the spring constant 𝑘 tells us how
much force in newtons we would need to do that. So this spring we have here as well
as any spring has some spring constant. And that tells us just how much
force would be necessary in order to extend or compress the spring.
Now, one last thing about Hooke’s
law. This expression of it, 𝐹 is equal
to 𝑘𝑥. We can see that the force that we
need to apply to a spring in order to stretch or compress it depends on how much the
spring has been stretched or compressed. For example, over here, if we
wanted to compress our spring even farther than the distance we have so far. We would need to press that much
harder. We would need to apply more
force. This is a way of saying that the
force in Hooke’s law depends on 𝑥, the displacement. And we can write that
mathematically this way. We can say that 𝐹 is a function of
𝑥. That is, 𝐹 depends on 𝑥.
Okay, now that we’ve talked about
force, let’s remember what we said about energy earlier. We said that energy is the capacity
of something to do work. Work, which we usually symbolize
using a capital 𝑊, is equal to the force applied to an object multiplied by that
object’s displacement. Now, we bring this up because we’re
about to do a little experiment on our spring, which will help us understand elastic
potential energy. Say that we set up a
two-dimensional plot, where on the horizontal axis, we plot out distance. That’ll be the distance that our
spring travels from its equilibrium length. And then, on the vertical axis, we
plot out the force being applied to the spring in order to displace it this given
distance.
So here we go. Starting out, we put our hand to
the end of the spring. And we start to push in. We start to compress it. After we apply some amount of
force, the spring compresses in some distance. And let’s say we plot that force
and distance point on our graph. Let’s say that point is right
here. So, in other words, we’re applying
this much force to the spring. And that’s causing it to compress
by this distance. Then, let’s say we double the force
we’re applying to the spring and therefore compress it even further, twice that
original distance. If we plot that point in the graph,
it will go right there.
At this point, if we were to hold
steady and continue to apply the same force, the spring wouldn’t keep
compressing. But rather we would just hold it in
place. Remember that the force we need to
apply to compress or extend a spring is proportional to that compression or
extension. So if at any moment we keep the
force applied the same, that will mean that the spring extension or compression
stays the same too. In other words, if we want more
compression, we need to push with greater force. So we do. We push harder and we get more
compression. And that data point looks like
this.
Say then, as part of this
experiment that we keep going. That we keep pressing harder and
harder and compressing the spring farther and farther. By this point, we’ve compressed our
spring some overall distance from its equilibrium position. We could call this distance 𝑥 and
mark it in on our graph so that there is 𝑥. And then, if we look back over at
Hooke’s law, we see that the force needed in order to displace our spring at
distance 𝑥 is equal to the spring constant 𝑘 times 𝑥. So we can mark that value in on our
vertical axis. Now, what we have is a plot of
force versus displacement of our spring from equilibrium. And according to our equation for
work, we know that if we multiply these two values together, force times
distance. We’ll get the work done on the
spring. Note that each time we do, we’re
using the specific force needed to stretch the spring a specific distance. That these values change as the
spring extends. And we need to take that into
account when we calculate work.
The question is, what does that
look like graphically. To figure that out, let’s do
this. Let’s draw a line of best fit
through these five data points. Since these data points all lie
along the same line, this line moves exactly through them. And it also passes through the
origin. From a graphical perspective, if we
were to multiply the value on the vertical axis, force, by the value on the
horizontal axis, distance. That product would be equal to the
area under this line of best fit that we just drew in. In other words, it would be equal
to the area of this right triangle, an area we can call capital 𝐴. We’re interested in solving for
this area because if we can, our equation for work tells us that we’ve solved for
the work done on the spring.
So what is that area under that
curve? What is the area of the
triangle? We can recall that a triangle’s
area, we’ll call it 𝐴 sub 𝑡, is equal to one-half its base multiplied by its
height. So that means that 𝐴 is equal to
one-half the base of this triangle, which we know is 𝑥, multiplied by its height,
which we know is 𝑘 times 𝑥. If we multiply these terms
together, we find that this area is equal to one-half the spring constant 𝑘 times
𝑥 squared. And as we said, this value is equal
to the total work done by our hand on the spring to compress it. And here’s where things get
interesting. Because the work done by our hand
on the spring is also equal to the elastic potential energy we give the spring. That is, this spring’s elastic
potential energy, now that we’ve compressed it, is equal to one-half the spring
constant multiplied by its displacement from equilibrium squared.
Now, we developed this equation for
the elastic potential energy based on a specific spring and a specific experiment
that we did. But notice that the equation is
quite general. We haven’t specified what the
spring constant is. We’ve left that as 𝑘. And also this displacement from
equilibrium 𝑥 in our case was a compression. But nothing about this equation
constrains it to only being a compression. It could also be an extension of a
spring or other elastic object. What we found then is a general
equation for elastic potential energy. So long as we know the displacement
of an object from equilibrium and we know its spring constant, we can solve for this
energy.
In just a minute, we’ll get some
practice with this equation. But before we do, it’s important to
realize what this 𝑥 is and what it is not. We’ve been saying that this 𝑥 is a
displacement from equilibrium. That means that if we have some
spring, which is sitting at its natural length, say that natural length is 𝐿. Then 𝐿 is not equal to 𝑥 in this
equation. That’s because, as it is, this
spring is neither extended nor compressed. It’s at the length that it
naturally wants to be at. But if we do change the length of
the spring, say we extended some distance beyond that equilibrium length. And that extension is 𝑥. Then that’s the distance that we’ll
use in this equation for elastic potential energy. That’s just something to watch out
for as we move on using this equation. Alright. All that being said, let’s try out
an example exercise involving elastic potential energy.
A spring has 50 joules of energy
stored in it when it is extended by 2.5 meters. What is the spring’s constant?
Okay. In this example, we have a
spring. Let’s say that this is our
spring. And we can imagine that it starts
out at its natural or its equilibrium length. In other words, in this position,
the spring isn’t being stretched or compressed at all. Now, by the way, if a spring is
completely unstretched or uncompressed, that means that no energy is stored in
it. We can say there are zero joules of
energy stored in the spring when it’s at its natural length. But then the spring is stretched
out. And we’re told it’s extended a
distance of 2.5 meters. And that, under these conditions,
the spring then has 50 joules of energy stored in it. So as a result of this change, the
spring has gained internal energy. And we know that it’s elastic
potential energy because it resulted from a stretching of the spring. This stretching was resisted by the
spring, according to a value known as the spring’s constant, symbolized lowercase
𝑘. It’s that value that we want to
solve for. And we can do it by recalling an
equation for elastic potential energy in terms of spring constant.
The elastic potential energy of an
object, sometimes also called its spring potential energy, is equal to one-half 𝑘,
the spring constant, multiplied by the displacement of that object from equilibrium
squared. And it’s important to know that
this distance 𝑥 is not equal to the natural length of a spring or other elastic
object. Instead, it’s equal to the
displacement of that spring from equilibrium, whether by compression or, in our
case, extension. So understanding this equation,
it’s not the elastic potential energy we want to solve for. But it’s the spring constant
𝑘. To solve for that, we can rearrange
this equation. We can multiply both sides by two
divided by 𝑥 squared. Now on the right-hand side, the
factor of one-half cancels with the two. And 𝑥 squared divided by 𝑥
squared is equal to one. We find then that two times the
elastic potential energy of an object divided by its displacement from equilibrium
squared is equal to its spring constant.
In our case, to solve for this
particular spring constant applying to our particular spring. We’ve been told that the elastic
potential energy stored in the spring is 50 joules. And that this much energy is stored
in the spring when it’s displaced from equilibrium by 2.5 meters. So to solve for 𝑘, we only need to
calculate the right-hand side of this equation. When we compute it, we find a
result of 16 newtons per meter. In other words, for this particular
spring, if we wanted to compress or extend it by a distance of one meter, we would
need to apply 16 newtons of force. That’s the spring constant of the
spring.
Let’s look now at another example
involving elastic potential energy.
A spring with a constant of 80
newtons per meter is extended by 1.5 meters. How much energy is stored in the
extended spring?
In this example, we start out with
this spring. Let’s say this is it right
here. And we’re told that our spring has
a spring constant. We’ll symbolize it with a lowercase
𝑘 of 80 newtons per meter. This means that if we want to
extend or compress the spring by a distance of one meter, we would need to apply a
force of 80 newtons. But at the moment, our spring isn’t
extended or compressed. We can say that it’s at its natural
or its equilibrium length. But we’re told that it doesn’t stay
that way. Instead, the spring is extended
from this original length. And that extension, we can call it
𝑥, is given as 1.5 meters. Based on all this, we want to know
how much energy is stored in the extended spring.
Now, the reason there’s energy at
all is because the spring wants to return to its natural length. It has a native capacity to do
that. In doing so, this currently
extended spring is capable of doing work. And that amount of work it can do
is equal to how much energy is stored in it while it’s extended. When we consider what type of
energy this is then, we know that it’s potential energy. The energy isn’t manifest now. But it will be if the spring is
released from its extended length. And we also know the energy is
elastic potential energy. That’s because it has its source in
the compression or, in our case, extension of an elastic body away from
equilibrium.
To solve for this amount of energy
then, we can recall a mathematical relationship describing elastic potential
energy. An object’s elastic potential
energy is equal to one-half its spring constant multiplied by its displacement from
equilibrium squared. As we solve for the elastic
potential energy stored in our spring. The great thing is that we’re given
𝑘, the spring constant, as well as the spring’s extension. So the elastic potential energy
stored in this extended spring is equal to one-half the spring constant, 80 newtons
per meter, multiplied by the displacement from equilibrium, 1.5 meters squared. When we calculate the right-hand
side of this expression, we find a result of 90 newton meters. But at this point, we can recall
that one newton, the base unit of force, multiplied by a meter, the base unit of
distance, is equal to a joule, the base unit of energy. So our final answer is 90
joules. That’s how much energy is stored in
this extended spring.
Let’s take a moment now to
summarize what we’ve learned in this lesson about elastic potential energy. We saw, first of all, that elastic
potential energy is stored-up ability to do work due to the deformation. That is, the extension or
compression of a body that will return to its natural shape. That’s a mouthful. But it basically means the energy
due to an elastic object, like a spring or a rubber band being either stretched out
or compressed. We also learned this. Hooke’s law says that the force on
an elastic object is equal to the displacement of that object from equilibrium
multiplied by something called the spring constant. And that that spring constant,
which is unique to any spring or other elastic object, is a measure of how much
force is required to stretch or compress the object a certain distance.
It was analyzing Hooke’s law and
combining it with the notion of work that led us to an equation for elastic
potential energy. That it’s equal to one-half the
spring constant of the object multiplied by the object’s displacement from
equilibrium squared. And lastly, we noted that 𝑥, in
that equation for elastic potential energy, is not the natural length of an elastic
object. But rather its displacement from
that length, whether through stretching or through compression.
