Not given in the exam, recall it.
What it is for
Elastic potential energy is used in Force and Motion II (SPM Physics Form 5). Keep the SI units in every line of working. Force and Motion II
Symbols and SI units
| Symbol | Meaning | SI unit |
|---|---|---|
| E | elastic potential energy | J |
| F | force | N |
| x | extension | m |
| k | spring (force) constant | N m⁻¹ |
Rearrangements
k = 2E / x²x = √(2E / k)F = 2E / x
Worked example
Common trap
Understanding elastic potential energy
E is the elastic potential energy in joules (J), F is the stretching force in newtons (N), x is the extension in metres (m), and k is the spring (force) constant in newtons per metre (N m⁻¹). When a spring is stretched or compressed within its elastic limit, work is done against the restoring force and stored as elastic potential energy.
Because the force grows linearly with extension (F = kx), the average force is ½F, giving E = ½Fx; substituting F = kx gives E = ½kx².
- To find the spring constant, rearrange to k = 2E / x².
- To find the extension for a stored energy, use x = √(2E / k).
- The force follows from F = 2E / x.
Keep x in metres so the energy is expressed in joules.
A worked example, step by step
A spring of constant k = 150 N m⁻¹ is compressed by x = 8.0 cm. Find the elastic potential energy stored.
- Convert the extension to metres: x = 8.0 cm = 0.080 m.
- Write the formula: E = ½kx².
- Substitute with units: E = ½ × 150 N m⁻¹ × (0.080 m)² = ½ × 150 N m⁻¹ × 0.0064 m².
- Work the square first, then multiply: 150 × 0.0064 = 0.96, and half of that is 0.48.
The answer is E = 0.48 J, quoted to two significant figures. The unit resolves correctly: N m⁻¹ × m² = N m = J, confirming joules.
Common mistakes and how this is tested
The most common slip is forgetting to square the extension in ½kx², square x before multiplying, and remember x is the extension, not the spring's total length. The formula is valid only within the elastic limit; beyond it the spring no longer obeys F = kx.
- Leaving x in centimetres gives an answer 10⁴ times too big.
- Dropping the factor ½ halves the error.
- Confusing E = ½kx² with the area under a force–extension graph, when they are in fact the same triangle area (½ × base × height).
Questions phrase this as calculate the elastic potential energy stored, determine the spring constant when energy and extension are given, or explain why energy is stored when a spring is stretched. Show the substitution line to earn method marks.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English), Sijil Pelajaran Malaysia: Format Pentaksiran mulai 2021, Fizik (4531) (Bahagian Pembangunan Kurikulum (BPK), KPM)