Spring Force Calculator
Spring force and stored energy from stiffness and stretch, both systems.
What it does
Springs answer stretch with proportional force, which is why scales and suspensions read true. A 200 N/m spring pulled 10 cm answers with 20 N and banks 1 joule for release. The tool above prices that pair in metric or imperial units.
The metric default multiplies straight through: 200 times 0.10 m is 20 N, and half times 200 times 0.01 m2 is 1 J. Imperial mode mirrors it at 10 lb/in over 2 in for 20 lbf with 20 in-lb stored. Both readings assume the elastic range where the line stays straight.
Textbooks anchor both formulas. OpenStax college physics states Hooke force as F = -kx with energy half k x squared, working a toy gun spring at 50 N/m and 0.150 m to 0.563 J, which this tool reproduces digit for digit. LibreTexts frames the same law as deformation proportional to applied force.
Elastic range is the quiet condition behind every number. Small pulls spring back fully, while yanks past the limit sag, crack, or snap. Car suspensions hide damping beside the spring for exactly that reason: steel alone would bounce every bump down the road.
How to use
- Pick metric (N/m, cm) or imperial (lb/in, in).
- Type the spring stiffness k.
- Type the stretch from rest length.
- Read force first for the pull or push.
- Read stored energy for launch and recoil math.
Formula
- k
- Stiffness in N/m or lb/in.
- x
- Stretch from rest length in matching units.
- E
- Elastic energy in joules or inch pounds.
Linear range only; force direction opposes the stretch. Series and parallel spring banks combine with different rules outside this page.
Worked example
Example: 200 N/m at 10 cm
Metric defaults: k 200 N/m, stretch 10 cm.
- Meters: 10 cm = 0.10 m.
- Force: 200 x 0.10 = 20 N.
- Energy: 0.5 x 200 x 0.01 = 1 J.
- Check: average force 10 N over 0.10 m is 1 J.
- Tool shows 20 N, 1 J, 10 cm.
Reference table
| Stretch | Force | Energy |
|---|---|---|
| 2 cm | 2 N | 0.02 J |
| 5 cm | 5 N | 0.125 J |
| 10 cm | 10 N | 0.5 J |
| 15 cm | 15 N | 1.125 J |
Limits
- Linear elastic range only.
- Spring mass and damping ignored.
- Static holds; oscillation needs more math.
- Temperature shifts real stiffness slightly.
Common mistakes
- Typing cm where meters belong, off by 100x.
- Forgetting the half in stored energy.
- Using force where stiffness belongs.
- Stretching past yield and trusting the line.
- Mixing N/m with inch stretches.
Try it yourself
Quick checks. Answers sit tucked under each question.
50 N/m at 15 cm?
50 x 0.15 = 7.5 N, energy 0.563 J.
100 N/m at 5 cm?
100 x 0.05 = 5 N, energy 0.125 J.
10 lb/in at 3 in?
10 x 3 = 30 lbf, energy 0.5 x 10 x 9 = 45 in-lb.
Questions
What is 200 N/m at 10 cm?
20 N of force with 1 J stored.
Why half k x squared?
Force climbs from 0 to kx, so average force is half.
What does k mean?
Stiffness: newtons per meter of stretch.
Stretch or compress?
Both follow F = kx inside the elastic range.
Why did my spring sag?
It passed yield; Hooke lines stop applying.
N/m or lb/in?
Same law: switch systems to read either.