Elastic Potential Energy — Bungee Cord (Worked Example)

Medium

Problem

A bungee cord acts like a spring with k = 100 N/m. At the lowest point of a jump it has stretched 15 m beyond its natural length. How much elastic potential energy is stored?

Given

k = 100 N/m
x = 15 m

Find

The elastic potential energy PE.

Principle

The cord is a giant spring: PE = ½kx² with a huge x, so the energy is large despite the soft spring constant.

Formula

PE = \frac{1}{2}kx^2

PE = ½kx²

Solution

Write the formula: $PE = \frac{1}{2}kx^2$
Substitute: $PE = \frac{1}{2}(100)(15)^2$
Calculate: $PE = 50 \times 225 = 11250\ \mathrm{J}$

1.125 × 10⁴ J (11.25 kJ)

Unit check

N/m × m² = J — correct.

What this means physically

11.25 kJ — roughly the energy of a 70 kg jumper falling 16 m, which checks out: the cord absorbed the entire fall. Cord designers tune k so this stretch stays within safe deceleration limits.

Common mistake

Using the total cord length instead of the stretch x. Only the extension beyond natural length stores energy.