Elastic Potential Energy — Mousetrap (Worked Example)

Easy

Problem

The spring of a mousetrap has k = 60 N/m and flexes 0.050 m when set. How much elastic potential energy does the set trap store?

Given

k = 60 N/m
x = 0.050 m

Find

The elastic potential energy PE.

Principle

Small spring, small displacement — but the x² law still applies exactly.

Formula

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

PE = ½kx²

Solution

Write the formula: $PE = \frac{1}{2}kx^2$
Substitute: $PE = \frac{1}{2}(60)(0.050)^2$
Calculate: $PE = 30 \times 0.0025 = 0.075\ \mathrm{J}$

0.075 J

Unit check

N/m × m² = J — correct.

What this means physically

Only 0.075 J — tiny, but released in milliseconds it snaps the bar shut fast enough to catch a mouse. Power (energy ÷ time) is what makes traps lethal, not total energy.

Common mistake

Expecting a large number because traps feel violent. Small x dominates: energy scales with x², so 5 cm of flex stores very little despite the snap.