Elastic Potential Energy — Archery Bow (Worked Example)

Given

• k = 400 N/m

• x = 0.50 m

Find

The elastic potential energy PE.

Principle

Stretching a spring stores energy as the area under the F–x graph — a triangle: PE = ½kx².

Formula

Solution

1. Write the formula: $PE = \frac{1}{2}kx^2$

2. Substitute: $PE = \frac{1}{2}(400)(0.50)^2$

3. Calculate: $PE = 200 \times 0.25 = 50\ \mathrm{J}$

Unit check

N/m × m² = (kg·m/s²)/m × m² = kg·m²/s² = J — energy, correct.

What this means physically

50 J stored in the drawn bow — released in a fraction of a second, most of it becomes the arrow's kinetic energy. That is why arrows leave the bow near 90 m/s.