Elastic Potential Energy

Elastic potential energy is the energy stored in an object that has been stretched, compressed or deformed elastically.

$E_e = \frac{1}{2}kx^2$

where $k$ is the spring constant (N/m) and $x$ is the extension or compression (m).

Relationship to the force-extension graph

The elastic potential energy equals the area under the force-extension graph (within the proportional region):

$E_e = \frac{1}{2}Fx = \frac{1}{2}kx^2$

Energy conversions with springs

When a stretched spring is released:

• Elastic PE converts to kinetic energy (if attached to a mass)
• Maximum KE occurs when the spring returns to its natural length ($x = 0$)

In a spring-mass system oscillating vertically, energy continuously converts between elastic PE, gravitational PE, and kinetic energy.

Worked example: A spring with $k = 40$ N/m is stretched by 0.15 m. How much elastic PE is stored?

$E_e = \frac{1}{2}(40)(0.15^2) = 0.45 \text{ J}$

Common errors and how to correct them

• Using $E = kx^2$ without the $\frac{1}{2}$ factor.
• Using this equation beyond the limit of proportionality (it only applies within the Hooke’s law region).
• Confusing extension ($x$) with total length.

How to apply this in an exam

Identify the spring constant and extension. Substitute into $E = \frac{1}{2}kx^2$. If energy converts to KE, set $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ and solve.