Elastic Potential Energy Calculator

elastic potential energy = ½ × spring constant × displacement² spring force = spring constant × displacement

k

Spring constant (stiffness) in newtons per metre — the field also takes kN/m, N/mm and lbf/in. It is the force needed to move the spring one unit, so a stiff spring has a large k

x

Displacement in centimetres — how far the spring is stretched or compressed from its resting length, measured in the field's default unit. The sign does not matter: the energy is the same in tension and in compression

E

Elastic potential energy in joules, the work done in deforming the spring; the third output row gives the same energy in foot-pounds

F

Spring force in newtons, from Hooke's law, F = kx — the force at the fully stretched position, not the average over the pull

Use this page when a spring, a rubber band, a bow, a trampoline or a suspension is involved and you need to know how much energy it holds or how hard it pulls. The two questions that come up most are sizing a spring for a given energy — a launcher, a counterbalance, a door mechanism — and working out what happens if it lets go. Two habits make the answers more useful. First, check the unit on the displacement before trusting the number: the field defaults to centimetres, and a figure meant as 10 mm entered as 10 cm is a hundred times the energy. Second, remember that the energy scales with the displacement squared while the force is only linear, so the last part of a compression holds most of the energy — a spring compressed 90 percent of the way is still holding only 81 percent, and the last tenth adds the remaining fifth.

1. A 1 lbf/in spring pulled out 2 inches

   1. 1 lbf/in = 175.127 N/m (the stiffness field takes N/m, so the pound-per-inch spring goes in as this)
   2. Displacement: 2 in = 5.08 cm
   3. Spring force: 175.127 × 0.0508 = 8.90 N, which is the 2 lbf you would feel on the end
   4. Displacement squared: 0.0508 × 0.0508 = 0.002581
   5. Energy: ½ × 175.127 × 0.002581 = 0.226 J
   6. In the other unit: 0.226 ÷ 1.3558179 = 0.167 ft·lb

   This is the reference table's row for 2 in, and the numbers are deliberately small — a light spring pulled a couple of inches holds a fifth of a joule, about the energy of a dropped apple. The interesting part is the ratio between the two outputs: the force is 2 lbf and the energy is 0.167 ft·lb, so the spring has been moved 2/12 of a foot against a force that averaged 1 lbf. Energy is force times distance, and the force here was not constant — it went from zero to 2 — which is exactly where the half in ½kx² comes from.

2. A 180 lbf/in suspension spring compressed 3 inches

   1. 180 lbf/in = 31,522.830 N/m
   2. Displacement: 3 in = 7.62 cm
   3. Spring force: 31,522.830 × 0.0762 = 2,402.04 N, or about 540 lbf
   4. Displacement squared: 0.0762 × 0.0762 = 0.005806
   5. Energy: ½ × 31,522.830 × 0.005806 = 91.518 J
   6. In the other unit: 67.5 ft·lb

   A real car spring, and the number to take away is the force: 540 lbf on that corner, which for a 3,300 lb car is about a sixth of the vehicle's weight at this compression. Note also that 91.5 J is less than the kinetic energy a thrown ball carries — springs are not energetic because they are stiff, they are energetic because they move a long way. A suspension spring only travels three inches; a bow travels two feet, and that is where its energy comes from.

3. The same spring at 6 inches — twice as far

   1. Displacement: 6 in = 15.24 cm, twice the example above
   2. Spring force: 31,522.830 × 0.1524 = 4,804.08 N, exactly twice — Hooke's law is linear
   3. Displacement squared: 0.1524 × 0.1524 = 0.023226, which is four times the 0.005806 above
   4. Energy: ½ × 31,522.830 × 0.023226 = 366.071 J, exactly four times the 91.518 J above
   5. In the other unit: 270 ft·lb

   Run this one against the example above and the whole page is in the two ratios. Twice the compression gives twice the force and four times the energy. That is the reason a spring that feels manageable in its first inch can be genuinely dangerous by its fourth, and the reason a bow is rated by the energy it stores rather than by the peak draw weight — the draw weight is the linear part, the energy is the square.

This is Hooke's law, which only holds while the spring is behaving elastically. Push a real spring past its yield point and the force stops being proportional to the displacement; push it to the end of its travel and the coils touch, at which point the stiffness changes completely and the ½kx² above overstates the energy. There is no damping, no hysteresis and no mass: a real spring loses a little energy to internal friction on every cycle, and a spring with mass has kinetic energy of its own while it moves. The page does not model a spring that is pre-loaded — a valve spring, a spring behind a bolt — where the energy stored is the difference between two states rather than the total against a zero-displacement reference, so the number it gives is the energy from the resting length. It assumes the whole spring moves by the same amount: a tapered spring, a series stack of two different springs, or anything where the coils do not share the displacement equally is not described by a single k. It is also static: it says nothing about the oscillation that follows when the spring is released, which needs the mass as well and belongs to a different page. Finally, it does not care whether the stored energy will be released usefully — the same 366 J can launch a projectile or simply be lost as heat in a damper — and it assumes the displacement is measured along the spring's axis, not at an angle.

What is the formula for elastic potential energy?

E = ½kx², where k is the spring constant in newtons per metre and x is the displacement in metres. The half is there because the force grows from zero to kx as the spring moves, so the average force over the whole displacement is only half the final value. A 1 lbf/in spring pulled out 2 in stores 0.226 J, and its pages of arithmetic are above if you want to check that by hand.

Why is the displacement squared and not just multiplied?

Because the work done on the spring is the force times the distance, and the force is not constant — it is zero at the start and kx at the end, so the average is ½kx and the work is ½kx times x. The practical consequence is that twice the displacement is four times the energy but only twice the spring force, which is why the last part of a compression is the hard part and why a crossbow is rated by stored energy rather than by draw weight.

What is the spring constant and how do I find it?

It is the force needed to move the spring by one unit of length, so it is measured in N/m — or in N/mm and lbf/in, both of which the field accepts. Measure it by hanging a known weight on the spring, noting how far it stretches, and dividing: a spring that stretches 4 cm under 8 N has k = 8 ÷ 0.04 = 200 N/m. A car suspension spring is around 30 N/mm, which is 30,000 N/m, and a light laboratory spring is nearer 100 N/m.

Is the energy the same in compression as in tension?

Yes, as long as the spring is symmetric, which most coil springs are. The formula uses x², so a 5 cm compression and a 5 cm extension both store ½ × k × 0.0025 J. Real springs are not perfectly symmetric — a coil spring can buckle when compressed far enough, and a spring with a helper coil is much stiffer in one direction than the other — so the equality holds over the working range, not to the ends of the travel.

How much energy does a spring hold when it is fully compressed?

Take the displacement at full compression and apply ½kx². A 30 N/mm suspension spring compressed 10 cm holds 150 J; the same spring at 20 cm holds 600 J, four times as much for twice the travel. Do check where the travel actually ends, though — once the coils touch, the spring stops behaving like a spring and the formula no longer describes what is happening.

How do I convert a spring rate from lbf/in to N/m?

Multiply by 175.127 — that is one pound-force (4.448222 N) divided by one inch (0.0254 m). So a 180 lbf/in spring is 31,522.83 N/m, and a 300 lbf/in spring is 52,538.05 N/m. Both factors are exact by definition, so the conversion adds no error; if your figure is out by a factor of about 9.8 you have probably used pounds mass instead of pounds force somewhere.

• Hooke's Law: Stress and Strain Revisited (College Physics 2e, §16.1) — F = kx and the elastic potential energy stored in a deformed spring — OpenStax
• Elasticity: Stress and Strain (College Physics 2e, §5.3) — the elastic limit and the point past which the linear relation between force and displacement stops holding — OpenStax
• NIST Guide to the SI — the newton per metre as the coherent unit of stiffness, and the exact factors for pounds (0.45359237 kg) and inches (0.0254 m) that the lbf/in conversion is built from — National Institute of Standards and Technology