A.3.11—Elastic potential energy

Syllabus
First assessment 2025
Objective
Level
SL

Calculate Elastic Potential Energy

Elastic store

For a spring within its linear range,

Ep,elastic=12k(Δx)2E_{p,elastic}=\frac12k(\Delta x)^2

where Δx\Delta x is extension or compression from the natural length.

Area under the graph

The elastic potential energy equals the work done in stretching or compressing the spring. On a force–extension graph it is the area under the graph.

Worked example from local Question Bank row 31357

A spring with k=100Nm1k=100\,\mathrm{N\,m^{-1}} is compressed by 0.10m0.10\,\mathrm{m}.

Ep,elastic=12(100)(0.10)2=0.50JE_{p,elastic}=\frac12(100)(0.10)^2=0.50\,\mathrm{J}

This is the energy available for transfer when the ideal spring is released.

Common trap

Do not use the total spring length as Δx\Delta x, and remember that doubling extension quadruples the stored energy in the ideal model.

A.3.11 Exam Analysis

Assessment in practice

1–3 marks
How it is assessed

The evidence asks for spring constant from work and compression, and for maximum elastic potential energy in a spring system.

Command terms

Calculate

What earns marks

Use Eh=1/2k(Δx)² with extension or compression from the unstretched length. If a graph or work value is given, connect the area or work to the spring constant and report N m⁻¹ or J as requested.

Watch for

Using Δx rather than (Δx)² or confusing spring constant with elastic energy.

Representative question

Question 1

[Maximum number: 1]

0.25 J\quad 0.25 \mathrm{~J} of work is done to compress a spring by a distance of 0.10 m from its unstretched length. What is the spring constant?

A

2.5Nm12.5 \mathrm{Nm}^{-1}

B

5.0Nm15.0 \mathrm{Nm}^{-1}

C

25Nm125 \mathrm{Nm}^{-1}

D

50Nm150 \mathrm{Nm}^{-1}