Skip to main content
Pearson+ LogoPearson+ Logo
All Calculators & ConvertersAll calculators

Elastic Potential Energy Calculator

Calculate elastic potential energy stored in a spring using E = ½kx². Switch modes to solve for energy, spring constant, stretch/compression, or combine two springs. Includes quick picks, step-by-step, and a mini spring + energy visual.

Background

When a spring is stretched or compressed by a distance x, it stores energy. For an ideal (linear) spring, Hooke’s law is F = kx and the stored energy is E = ½kx². (Same formula for compression or stretch — just use the magnitude of x.)

Enter values

Tip: You can type scientific notation like 2.5e3.

x in meters (m)
  • x in meters (m)
  • x in centimeters (cm)
  • x in millimeters (mm)
k in N/m
  • k in N/m
  • k in N/cm

Internally the calculator converts to SI: x in meters and k in N/m.

Optional: add a rated stretch so the visual shows a max marker (x_max) and an energy limit notch (E_max).

E = ½kx². Example: k=200 N/m and x=0.10 m → E=1.0 J.

Options:

Chips fill values and solve immediately.

Result:

No results yet — enter values and click Calculate.

How to use this calculator

  • Choose a mode: solve for E, k, x, or keq.
  • Select units for x (m/cm/mm) and k (N/m or N/cm).
  • Optional: enter x_max to see limit warnings and an E_max notch on the energy bar.
  • Click Calculate to see outputs, steps, and the visual.

How this calculator works

  • Hooke’s law: F = kx
  • Elastic potential energy: E = ½kx²
  • Solve for k: k = 2E/x²
  • Solve for x: x = √(2E/k)
  • Combine springs: Parallel k_eq = k₁ + k₂, Series 1/k_eq = 1/k₁ + 1/k₂

Formula & Equation Used

Elastic energy: E = ½kx²

Hooke’s law: F = kx

Rearranged: k = 2E/x², x = √(2E/k)

Energy limit from x_max: E_max = ½k x_max²

Example Problems & Step-by-Step Solutions

Example 1 — Find energy from k and x

A spring has k = 200 N/m and is stretched x = 0.10 m. Find E.

  1. E = ½kx²
  2. E = 0.5 × 200 × (0.10)²
  3. E = 100 × 0.01 = 1.0 J

Example 2 — Solve for k

A spring stores E = 2.0 J when stretched x = 0.20 m. Find k.

  1. k = 2E/x²
  2. k = (2×2.0) / (0.20)² = 4 / 0.04
  3. k = 100 N/m

Example 3 — Solve for x

A spring with k = 400 N/m stores E = 5.0 J. Find x.

  1. x = √(2E/k)
  2. x = √(10/400) = √(0.025)
  3. x ≈ 0.158 m (about 15.8 cm)

Frequently Asked Questions

Q: Does compression use the same formula as stretching?

Yes. Use the magnitude of displacement: E = ½k|x|².

Q: Why is there a ½ in the energy formula?

Because the force increases from 0 to kx as you stretch the spring. The average force is ½kx, so work is (½kx)×x.

Q: What does a larger k mean physically?

A larger k is a stiffer spring. For the same x, it stores more energy.

Q: When does this calculator stop being accurate?

If the spring isn’t linear (it doesn’t follow F=kx), then E=½kx² may not match reality.