KAHIBARO
Table of Contents
Stored energy in a stretched or compressed spring
Elastic potential energy is the energy stored when an object is deformed and can return to its original shape. The most common example in beginner physics is a spring. If you stretch a spring or compress it, you do work on it, and that work is stored as elastic potential energy.
A spring resists deformation. For an ideal spring, the restoring force follows Hooke's law,
$$
F = -kx
$$
where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. The minus sign means the force points back toward equilibrium.
Elastic potential energy belongs to the spring or elastic object itself. It depends on how far the object is stretched or compressed from its natural length.
For an ideal spring, the elastic potential energy is
$$
U = \frac{1}{2}kx^2
$$
where $x$ is measured from the equilibrium position.
Why the formula has one half
The force from a spring is not constant. When the spring is barely stretched, the force is small. As the stretch increases, the force increases linearly. Because of this, the work needed to stretch the spring from $x=0$ to some final displacement $x$ is not simply $Fx$ with a single constant force.
The force magnitude changes from $0$ to $kx$. The average force during this process is therefore
$$
F_{\text{avg}} = \frac{0 + kx}{2} = \frac{kx}{2}
$$
So the work done to stretch or compress the spring is
$$
W = F_{\text{avg}} x = \frac{kx}{2}x = \frac{1}{2}kx^2
$$
That work becomes stored elastic potential energy.
Meaning of the variables
The formula
$$
U = \frac{1}{2}kx^2
$$
uses quantities with the following meanings.
| Symbol | Meaning | SI unit |
|---|---|---|
| $U$ | elastic potential energy | J |
| $k$ | spring constant, stiffness of spring | N/m |
| $x$ | stretch or compression from equilibrium | m |
A larger value of $k$ means a stiffer spring. A larger value of $x$ means more deformation. Since the displacement is squared, doubling the stretch makes the stored energy four times larger.
Elastic potential energy depends on $x^2$, so it is always zero or positive for an ideal spring.
Stretching and compressing by the same distance store the same energy:
$$
U\left(+x\right) = U\left(-x\right)
$$
Equilibrium position
The displacement $x$ must always be measured from the equilibrium position, not from some arbitrary point. The equilibrium position is where the spring is neither stretched nor compressed.
If $x=0$, then
$$
U = 0
$$
This is the minimum elastic potential energy for the ideal spring model.
Force and energy comparison
Force and potential energy are related, but they are not the same thing. The spring force tells how strongly the spring pushes or pulls at a particular displacement. The elastic potential energy tells how much energy is stored at that displacement.
| Displacement $x$ | Force magnitude $ | F | = k | x | $ | Elastic potential energy $U = \frac{1}{2}kx^2$ |
|---|---|---|---|---|---|---|
| small | small | small | ||||
| doubled | doubled | quadrupled | ||||
| zero | zero | zero |
This difference is important. Force grows linearly with $x$, but energy grows with $x^2$.
Graph of elastic potential energy
The graph of elastic potential energy versus displacement is a parabola opening upward. It has its minimum at the equilibrium position.
The graph shows that the stored energy increases whether the spring is stretched to the right or compressed to the left.
Work done on the spring
If an external agent slowly stretches the spring from $x_1$ to $x_2$, the change in elastic potential energy is
$$
\Delta U = \frac{1}{2}k x_2^2 - \frac{1}{2}k x_1^2
$$
This is equal to the work done on the spring, provided no energy is lost.
Change in elastic potential energy:
$$
\Delta U = \frac{1}{2}k x_f^2 - \frac{1}{2}k x_i^2
$$
If the spring goes from equilibrium to displacement $x$, then
$$
U = \frac{1}{2}kx^2
$$
Example
Suppose a spring has spring constant
$$
k = 200\ \text{N/m}
$$
and it is compressed by
$$
x = 0.10\ \text{m}
$$
Then the elastic potential energy is
$$
U = \frac{1}{2}(200)(0.10)^2
$$
$$
U = 100 \times 0.01 = 1.0\ \text{J}
$$
So the compressed spring stores $1.0\ \text{J}$ of energy.
If the same spring were compressed by $0.20\ \text{m}$, then
$$
U = \frac{1}{2}(200)(0.20)^2 = 4.0\ \text{J}
$$
The compression doubled, but the energy became four times as large.
Physical interpretation
Elastic potential energy is a way to track how energy is stored in deformable systems. In simple mechanics problems, a spring can store energy and later release it, causing motion. When the spring returns toward equilibrium, the stored energy decreases and can be transformed into kinetic energy or other forms of energy.
For absolute beginners, the key idea is simple. If you deform an elastic object, you may be storing energy in it. For an ideal spring, that stored energy is given by a very useful formula.
Key result for ideal springs:
$$
U_{\text{elastic}} = \frac{1}{2}kx^2
$$
Measure $x$ from the equilibrium position, and remember that the energy is the same for stretching and compression by the same amount.