When a material is stretched or compressed but stays within its proportional limit, the force applied is directly proportional to the extension. This relationship is described by Hooke’s Law:
\$F = kx\$
Think of a rubber band: pull it gently and it stretches, but if you pull too hard it snaps. The part where it stretches and returns to its original length is the elastic region.
- Proportional Limit: The maximum point on the force–extension graph where Hooke’s Law still holds. The graph is a straight line up to this point.
- Elastic Limit: The highest stress a material can withstand and still return to its original shape. It usually coincides with the proportional limit for many metals.
Beyond the elastic limit, the material undergoes plastic deformation – it keeps its new shape even after the force is removed. Imagine bending a paperclip: once you bend it past a certain point, it stays bent.
The energy stored in a stretched or compressed spring (or any elastic material) is the area under the force–extension curve up to the point of interest. For a linear (Hookean) spring this area is a triangle:
\$EPE = \frac{1}{2} k x^2 = \int_{0}^{x} F \, dx\$
So, if you know the spring constant \$k\$ and the extension \$x\$, you can calculate the elastic potential energy directly.
Suppose a spring has a spring constant \$k = 200 \, \text{N/m}\$ and is stretched by \$x = 0.05 \, \text{m}\$ (5 cm). The elastic potential energy stored is:
| Step | Calculation | Result |
|---|---|---|
| 1. Hooke’s Law (force) | \$F = kx = 200 \times 0.05\$ | \$F = 10 \, \text{N}\$ |
| 2. Elastic Potential Energy | \$EPE = \frac{1}{2} k x^2 = \frac{1}{2} \times 200 \times (0.05)^2\$ | \$EPE = 0.25 \, \text{J}\$ |
So the spring stores \$0.25\$ joules of elastic potential energy when stretched 5 cm.
Remember: stretch gently, and the material will give back. Pull too hard, and it may stay stretched. 🚀