Hooke’s law says that force is proportional to extension, F = kx. The stored energy in the stretched spring is ½kx², which is half of force times extension, not the whole.
This lesson is part of SPM Physics force and motion II. It leads into interpreting force-extension graphs.
What is the units-first routine?
Convert before you substitute.
- Write the extension x = new length − original length.
- Convert x to metres by dividing centimetres by 100.
- Substitute into F = kx or Ep = ½kx².
- State the unit: N m⁻¹ for k, J for energy.
Skipping step 2 is the main reason k comes out 100 times too small, because 5.0 is used instead of 0.050. The units and significant figure checker can check a conversion.
Worked example 1: spring constant and energy
A spring extends by 5.0 cm when an 8.0 N force is applied. Find k and the elastic energy stored.
Extension x = 5.0 cm = 0.050 m. The spring constant k = F ÷ x = 8.0 ÷ 0.050 = 160 N m⁻¹. The energy stored is Ep = ½kx² = ½ × 160 × 0.050² = ½ × 160 × 0.0025 = 0.20 J.
Check with ½Fx: ½ × 8.0 × 0.050 = 0.20 J, which matches.
Worked example 2: a launched ball
A spring with k = 250 N m⁻¹ is compressed by 4.0 cm and launches a 0.10 kg ball along a smooth horizontal surface. Find the launch speed, ignoring losses.
Compression x = 0.040 m. Force at maximum compression = 250 × 0.040 = 10 N. Energy stored = ½ × 250 × 0.040² = ½ × 250 × 0.0016 = 0.20 J.
All the stored energy becomes kinetic energy: ½mv² = 0.20, so v² = 2 × 0.20 ÷ 0.10 = 4.0 and v = 2.0 m s⁻¹. The words “ignoring losses” are the assumption behind this answer, so write them in your working.
The mistake to avoid
The common mistake is to use Ep = Fx. That gives twice the real energy, because the force is not constant.
| Formula | Result for example 1 | Correct? |
|---|---|---|
| Ep = Fx | 8.0 × 0.050 = 0.40 J | No, double |
| Ep = ½Fx | 0.20 J | Yes |
| Ep = ½kx² | 0.20 J | Yes |
The average force over the stretch is half the final force, which gives the factor ½. A graph of force against extension makes this visible as a triangle.
Check yourself
A spring has k = 400 N m⁻¹. Find the force needed to extend it by 3.0 cm, and the energy stored.
Answer
x = 0.030 m. Force = kx = 400 × 0.030 = 12 N. Energy = ½kx² = ½ × 400 × 0.030² = ½ × 400 × 0.0009 = 0.18 J.
Check with ½Fx = ½ × 12 × 0.030 = 0.18 J.
What to study next
Read the same ideas from data in interpreting force-extension graphs. Test the calculations in the force and motion II practice set, and revisit balance with applying equilibrium conditions.
If you would like a teacher to go through your unit conversions with you, see online one-to-one Physics tuition.