### Video Transcript

In this video, we’re going to learn
about elastic potential energy. We’ll learn what this type of
energy is, how it relates to the work done on an object, and we’ll see how to
combine elastic potential energy with other energy types.

To start out, imagine that you have
made a tree fort in your backyard. And as the fort is one of your
favorite places to spend time, you find yourself climbing up and down the steps to
get there frequently. After a while though, you get tired
of all this climbing and you have an idea. You purchase what is essentially a
human-sized elastic slingshot and string it up between two small trees near the tree
fort. In order to figure out just how far
back you’ll need to stand in the human slingshot so that it will shoot you up into
the tree fort, we’ll want to know a little bit about elastic potential energy.

The first thing we can say about
elastic potential energy is that whenever a spring is stretched or compressed, it
possesses potential energy, that is the ability to do work. And when we say spring, that can
include not just literal springs, but springy materials or substances that rebound
after they’re deformed.

If we were to take a spring resting
at its equilibrium length and then stretch that spring out at distance 𝑥 displaced
from its natural length, we know that the stretched spring would exert a restoring
force, we can call 𝐹 sub 𝑠, whose magnitude is equal to the spring constant 𝑘
times that displacement from equilibrium 𝑥.

We recall a basic equation for work
done, which is that it’s equal to force times displacement. But this equation for work assumes
a force which does not vary with displacement, while our spring force does. It depends on the displacement
𝑥. For the work done then by our
spring force, that work will be equal to the integral from zero to the maximum
displacement 𝑥 of that force times 𝑑𝑥.

Plugging in for the magnitude of
that force, we now have an expression for the work done by a stretched or compressed
spring. Integrating this expression from
zero to 𝑥, we find a result of one-half 𝑘 times 𝑥 squared. This is equal to the work done by a
spring and it’s also equal to the spring potential energy, or elastic potential
energy.

One way to understand elastic
potential energy is to think of it like gravitational potential energy. We know that if we take a mass and
elevate it in a gravitational field some amount of height, then that mass now has
potential energy and, therefore, the ability to do work. Likewise, if we put a mass on the
end of a spring and then compress that spring, that mass likewise has potential
energy. In this case, due to the elastic
energy of the spring. And just like gravitational
potential energy, elastic potential energy fits within energy conservation
schemes. Let’s get some practice with
elastic potential energy through a couple of examples.

The spring of a spring gun has a
force constant 𝑘 equals 12 newtons per centimeter. When the gun is aimed vertically
upward, a 15-gram projectile is shot to a height of 5.0 meters above the end of the
expanded spring, as shown in the accompanying diagram. How much was the spring compressed
initially?

In the diagram, we’ve called this
distance 𝑑. And we’re also told that the spring
constant of the spring is 12 newtons per centimeter and the mass of the shot being
fired is 15 grams. Looking at the diagram of this
scenario, we see that the initial and final conditions are connected through energy
conservation.

If we call the state where the
spring is compressed the initial state and the point where the ball is at its
highest point in flight the final state, then we can write that the initial sum of
kinetic plus potential energy is equal to the final amount of kinetic plus potential
energy. Both at the start and at the
finish, the ball isn’t in motion, so its kinetic energy in both those cases is
zero.

When we consider the initial state
of the ball, we know that its potential energy is due to the elastic energy
compressed in the spring. We can call this PE sub 𝑠, for
spring potential energy. On the other hand, the final
potential energy of our system is entirely gravitational as the ball is motionless a
known height above the expanded spring. So, we can write that spring
potential energy equals gravitational potential energy. And we recall that spring potential
energy is equal to one-half the spring constant times the displacement from
equilibrium squared and that gravitational potential energy is equal to 𝑚 times 𝑔
times ℎ.

For our scenario, we write the
spring potential energy is one-half 𝑘 times 𝑑 squared, where 𝑑 is the distance we
want to solve for. And on the potential energy side,
we let the height ℎ be equal to 5.0 meters, the height of the ball when it’s above
the fully extended spring and not in motion.

Rearranging this expression to
solve for 𝑑, we find it’s equal to the square root of two times 𝑚 times 𝑔 times ℎ
all divided by 𝑘. 𝑔, the acceleration due to
gravity, we’ll treat as exactly 9.8 meters per second squared. When we plug in for these values,
we’re careful to convert our mass into a value in units of kilograms and our spring
constant 𝑘 into a value in units of newtons per meter.

Calculating this result, we find 𝑑
is 3.5 centimeters. That’s the amount the spring must
be compressed in order to fire the ball up to this given height ℎ.

Now, let’s look at another elastic
potential energy example.

A string has a force constant of
53.0 newtons per meter. An object, initially at rest, with
a mass of 0.960 kilograms is suspended from it. The object descends, stretching the
spring, oscillates, and then comes to rest. How much is the spring stretched
when the object has come to rest after oscillating? Calculate the decrease in the
gravitational potential energy of the object between its position at the point at
which it is attached to the unextended spring and its position at the point at which
it comes to rest after oscillating. Calculate the energy stored in the
spring by its extension.

In this three-part exercise, we
first wanna solve for the displacement of the spring after the mass has been hung on
it and comes to rest. Next, we want to solve for the
change in gravitational potential energy of the mass between the two points from
which it is hung on the unextended spring and the point at which it comes to
rest. And finally, we wanna calculate the
energy stored in the spring by its extension. We’ll call this PE sub 𝑠.

In this scenario, we start off with
a vertically oriented spring at its natural, or equilibrium, length. The spring, we’re told, has spring
constant 𝑘 of 53.0 newtons per meter. Then, we take a mass, we’ve called
it 𝑚, of value 0.960 kilograms and attach it to the free end of the spring. In response to this, the spring
stretches, oscillates, and then comes to rest at a displacement we’ve called Δ𝑥
from its original position.

To solve for Δ𝑥, let’s start by
considering the forces that are acting on the mass 𝑚. When the mass has come to rest
after oscillating on the end of the spring, we know that the gravitational force
acting on the mass is equal in magnitude to the spring force on the mass. So, we can write 𝑚 times 𝑔 is
equal to the spring constant 𝑘 times Δ𝑥. Rearranging this expression, we see
that Δ𝑥 is equal to 𝑚 times 𝑔 over 𝑘, where 𝑔 we’ll treat as exactly 9.8 meters
per second squared.

Plugging in for these three values,
when we calculate Δ𝑥, we find that, to three significant figures, it’s 17.8
centimeters. That’s the amount the spring is
stretched because of the mass 𝑚. Now, let’s move on to solving for
the change in gravitational potential energy of the mass 𝑚 over this
displacement.

Recalling that, in general,
gravitational potential energy is equal to an object’s mass times the acceleration
due to gravity times its height relative to some reference level. Since we’re calculating the change
in gravitational potential energy of the mass, we can write that it’s equal to 𝑚
times 𝑔 times Δ𝑥. No negative sign needed.

When we plug in for these three
values, being careful to write our displacement Δ𝑥 in units of meters, we find that
ΔPE sub 𝑔 is 1.67 joules. That’s the change in the mass’
gravitational potential energy. Finally, we want to solve for the
potential energy stored up in the spring as result of this extension.

We recall that in general that
potential energy is equal to half a spring‘s constant multiplied by its displacement
from equilibrium squared. So, we can write that PE sub 𝑠 is
equal to one-half 𝑘 times Δ𝑥 squared, where 𝑘 the spring constant is known and
Δ𝑥 we solved for earlier. Substituting in for these values
using a distance in meters for our displacement Δ𝑥, we find that PE sub 𝑠 is 0.835
joules. That’s how much potential energy
the spring has gained by virtue of being extended.

We may wonder why the spring
potential energy is not equal to the change in gravitational potential energy. Recall that, before coming to rest,
our mass 𝑚 oscillated up and down on the spring. During that up-and-down motion, it
lost energy through friction with the air as well as through the spring. That helps account for the
difference between these two values.

Let’s summarize what we’ve learnt
so far about elastic potential energy. We’ve seen that a stretched or
compressed spring has potential energy. We’ve said that that potential
energy, we could call it PE sub 𝑠, is equal to one-half the spring constant 𝑘
multiplied by its displacement from its equilibrium, or natural length, 𝑥
squared.

We’ve also seen that the potential
energy stored in a spring, PE sub 𝑠, is equal to the work done on the spring. And we’ve seen that elastic, or
spring, potential energy in an isolated system is part of the overall energy
conservation picture, including kinetic energy and gravitational potential
energy. In other words, it’s an energy term
just like other energy terms.