KinexiaLEARN PHYSICS

Hooke's Law — Trampoline (Worked Example)

Medium
Problem

A trampoline mat behaves like a spring with k = 5000 N/m. When a jumper lands, the mat deflects 0.35 m. What upward force does it exert at maximum deflection?

Given

  • k = 5000 N/m

  • x = 0.35 m

Find

The spring force F.

Principle

Maximum deflection gives maximum restoring force via F = kx — this peak force is what launches the jumper back up.

Formula

Solution

  1. Write the law: F=kxF = kx

  2. Substitute: F=(5000)(0.35)F = (5000)(0.35)

  3. Multiply: F=1750 NF = 1750\ \mathrm{N}

1750 N upward

Unit check

N/m × m = N — correct.

What this means physically

1750 N is about 2.5 times the jumper's weight — that extra upward net force is what accelerates them skyward. Trampolines feel fun because the force builds smoothly instead of hitting all at once.

Common mistake

Assuming the force is constant during the bounce. F = kx varies with x — 1750 N is only the peak at maximum stretch.