5.1 Hooke's law — Learn

Prerequisites: force in newtons, and reading a graph.

Objectives: state Hooke's law; use $F = kx$; explain what the limit of proportionality is and what the graph shows on either side of it.

By the end of this section, you can use Hooke's law in either direction and describe what a spring does when you stretch it too far.

Force and extension

Hold a spring and pull. The harder you pull, the further it stretches. For a metal spring, over the range where it behaves, the two are related in the simplest possible way: twice the force gives twice the extension.

That statement is Hooke's law.

Hooke's law. The extension of a spring is directly proportional to the force applied to it, provided the limit of proportionality is not exceeded.

• $F$ is the applied force, in newtons (N);
• $x$ is the extension, in metres (m) — how much longer the spring is than its natural length, not its total length;
• $k$ is the spring constant, in newtons per metre (N/m).

The spring constant measures stiffness. A large $k$ means a stiff spring: it takes a large force to stretch it a little.

Example 1. A spring has a spring constant of 40 N/m. Calculate the force needed to stretch it by 0.15 m.

Example 2. A force of 12 N stretches a spring by 0.04 m. Calculate its spring constant.

Rearranging $F = kx$ for $k$:

Example 3. A spring of spring constant 250 N/m is pulled with a force of 8.0 N. Calculate its extension.

Rearranging for $x$:

That is 32 mm.

Watch out. $x$ is the extension, not the length. If a spring is 0.20 m long naturally and 0.26 m long when loaded, the extension is $0.26 - 0.20 = 0.06$ m, not 0.26 m. Using the total length makes the answer several times too large, and it is the most common mistake in this section.

The limit of proportionality

Hooke's law does not hold forever. Stretch a spring far enough and the relationship breaks down.

The limit of proportionality is the point beyond which the extension is no longer proportional to the force. Plotting force on the vertical axis against extension on the horizontal axis shows it clearly.

• Below the limit, the points lie on a straight line through the origin. The gradient of that line is the spring constant, because $F = kx$ is a straight line through the origin with gradient $k$.
• At the limit, the line begins to bend.
• Beyond the limit, the graph curves upwards: each extra newton produces a larger extension than the one before. The spring has been stretched past what its material can recover from.

Stretch it further still and you reach the elastic limit. Past that point the spring does not return to its natural length when the force is removed. It has been permanently deformed.

Predict, then test. Before you move the slider, write down what you expect the force to be at an extension of 0.20 m. Then move it there and check. Now push the extension past the limit of proportionality and describe what happens to the relationship between force and extension. Explain why the graph curves the way it does once it is past that point.

Progress check

Back to the chapter problem. You hang weights on a spring and find that each extra newton stretches it the same amount — until it does not.

Write down, in one sentence, which part of the graph shows the spring behaving as Hooke's law says. Then write a second sentence describing what the spring is doing in the part where the line bends. Keep both: Section 5.4 uses a graph like yours to find the spring constant, and you will need to know which part of it to use.

You should now be able to

• state Hooke's law and say what each symbol in $F = kx$ means, with its unit;
• find a force, a spring constant or an extension from the other two;
• tell an extension from a length, and say why the difference matters;
• describe a force-extension graph below and beyond the limit of proportionality, and say what the gradient of the straight part means.