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A.2.9—Hooke’s law restoring force

Syllabus
First assessment 2025
Objective
—
Level
HL

Apply Hooke’s Law to Elastic Restoring Force

Restoring force

For an ideal elastic element within its proportional range,

F⃗H=−kx⃗\vec F_H=-k\vec x

The minus sign means the force acts opposite the displacement from equilibrium.

Use extension correctly

For a spring, xx is extension or compression relative to its natural length. In a vertical equilibrium, the spring tension can balance weight, but the extension is not the total spring length.

Respect the model boundary

Hooke’s law is a linear approximation. Beyond the limit of proportionality, the force–extension graph is no longer linear and the same kk cannot be used.

A.2.9 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

The evidence asks for the spring constant from natural length, loaded length and mass, requiring the extension and the equilibrium force balance.

Command terms

Calculate / Identify

What earns marks

Use the spring’s extension \(\Delta x=l-l_0\), not its total length, and apply the stated equilibrium or Hooke relationship. Rearrange symbolically before substituting and give \(k\) in N m⁻¹.

Watch for

Using the loaded length instead of the extension when calculating \(k\).

Representative question

Question 1

[Maximum number: 1]

A spring of negligible mass and length l0l_{0} hangs from a fixed point. When a mass m is attached to the free end of the spring, the length of the spring increases to l. The tension in the spring is equal to kΔxk \Delta x, where k is a constant and Δx\Delta x is the extension of the spring. What is k ?

A

mgl0\frac{m g}{l_{0}}

B

mgl\frac{m g}{l}

C

mgl−l0\frac{m g}{l-l_{0}}

D

mgl0−l\frac{m g}{l_{0}-l}