Forces
Hooke's Law and springs
Elastic and inelastic deformation
Elastic deformation: the object returns to its original shape and size when the force is removed.
Inelastic (plastic) deformation: the object does not fully return to its original shape — the deformation is permanent.
Inelastic (plastic) deformation: the object does not fully return to its original shape — the deformation is permanent.
A spring or wire shows elastic behaviour up to a certain force. Beyond the elastic limit, the deformation becomes inelastic (permanent). The object is said to have been permanently deformed.
Hooke's Law
Within the limit of proportionality, the extension of a spring is directly proportional to the applied force.
\[ F = ke \]
A larger spring constant means a stiffer spring (more force needed per unit extension).
| Symbol | Quantity | Unit |
|---|---|---|
| \( F \) | Force applied | newton, N |
| \( k \) | Spring constant (spring stiffness) | N/m |
| \( e \) | Extension (increase in length from natural length) | metre, m |
Extension \( e \) is the increase in length: \( e = \text{new length} - \text{natural length} \). Do not use the total length of the spring as the extension.
Force–extension graph
Reading the graph:
- O → A: straight line through the origin — Hooke's Law obeyed. Gradient = spring constant \( k \).
- A: limit of proportionality — beyond here, \( F \) and \( e \) are no longer proportional.
- A → B: still elastic (spring returns to original shape) but not proportional.
- Beyond B: elastic limit exceeded — permanent (inelastic) deformation. Dashed line = spring does not return to origin when force removed.
Calculations using F = ke
Example 1 — find spring constant
A spring has a natural length of 12 cm. When a force of 6.0 N is applied, it stretches to 18 cm. Find the spring constant.
Extension: \( e = 18 - 12 = 6\,\text{cm} = 0.06\,\text{m} \)
\( k = \dfrac{F}{e} = \dfrac{6.0}{0.06} = 100\,\text{N/m} \)
Example 2 — find extension
A spring with \( k = 250\,\text{N/m} \) has a force of 15 N applied. Find the extension.
\( e = \dfrac{F}{k} = \dfrac{15}{250} = 0.060\,\text{m} = 6.0\,\text{cm} \)
Investigating Hooke's Law
Apparatus: spring, clamp stand, ruler, set of 100 g masses, pointer.
Method:
Method:
- Measure and record the natural length of the spring with no masses.
- Add masses one at a time (e.g. 100 g = 1.0 N). Record the new length after each addition.
- Calculate extension = new length − natural length for each force.
- Plot a graph of force (y-axis) against extension (x-axis).
- The gradient of the straight-line section = spring constant \( k \).
If the question asks you to find \( k \) from a graph, calculate the gradient of the straight-line section only. Use two widely spaced points on the line (not data points) and show the working as \( k = \Delta F / \Delta e \).