Hooke's Law and Elastic Potential Energy Calculator

Energy stored and energy released. ½kx² is the area under the force–extension line, so it goes as the square of the extension — and the page takes the stored energy back out again as the launch speed and the height a given mass reaches, then tells you when the extension has gone past the limit of proportionality and the formula no longer applies.

Last updated 11 Nov 2026

Energy stored in a spring, and what it launches

½kx² in, launch speed out

Spring constantIn newtons per metre — the force needed per metre of extension, also called the stiffness or the force constant. A value quoted in N/cm or N/mm must be converted: 2.5 N/cm is 250 N/m, and 2.5 N/mm is 2500. This is the single commonest unit error on the topic and it moves the stored energy by a factor of ten or a hundred without looking wrong.

Extension or compressionIn metres, and measured from the spring’s natural length — not its total length. A spring 30 cm long stretched to 42 cm has an extension of 0.12 m, which is the default here. Compression works identically and the energy is positive either way, because the formula squares it. Centimetres typed here instead of metres make the energy ten thousand times too large.

Mass the spring launchesIn kilograms. Used only for the release half of the page: the stored energy is converted into the kinetic energy of this mass, giving a launch speed, and then into height. 50 g is the default because that is the order of a toy projectile or a laboratory trolley weight. The mass has no effect at all on the energy stored, which is a property of the spring and the extension alone.

Extension at the limit of proportionalityIn metres: the extension beyond which the spring stops obeying Hooke’s law, from the question or from a measured force–extension graph. Past it the graph curves, the force is no longer kx, and the area under it is no longer ½kx² — so both answers on this page become wrong rather than approximate. If the question does not give you one, set it generously high and read the page as unconditional.

1.800JExample

A 250 N/m spring extended 0.12 m, launching a 50 g mass, proportional to 0.20 m

½kx² is a triangle, and the square is the whole story

F = kx · E_p = ½kx² = ½Fx · ½kx² = ½mv² ⇒ v = x√(k/m)

k: the spring constant, in newtons per metre. A property of the spring, not of how far you pull it: it is the gradient of the force–extension line. Quoted in N/cm or N/mm as often as in N/m, and converting it is where most of the wrong answers on this topic begin.
x: the extension from the natural length, in metres. Compression is the same formula and gives the same positive energy, because x is squared. Measured from the unloaded length, which is not the same as the length the spring is sitting at when there is already a load on it.
½kx²: the area under the force–extension graph: a right triangle with base x and height kx, so ½ × base × height. That is the whole derivation, and Cambridge 9702 asks for it that way at outcome 6.2.3 — from the area under the graph. The ½ is not a fudge; it is there because the force started at nothing and grew linearly, so the average force over the stretch was half the final one.
the square: double the extension and the energy goes up four times; triple it and nine. The force only doubles. This is the single most testable consequence of the formula and the one most often answered as “twice”: the force is linear in x and the energy is quadratic, and the page prints both so the difference is visible.
v = x√(k/m): the release, from equating ½kx² with ½mv². Useful as a shape rather than as a formula to memorise: the launch speed is proportional to the extension, and to the square root of the stiffness-to-mass ratio. Quadruple the mass and the speed halves. It assumes every stored joule reaches the projectile, which no real spring manages.

Worked example

A 250 N/m spring extended 0.12 m, launching a 50 g mass, proportional to 0.20 m
THE FORCE FIRST, AS A CHECK ON THE UNITS. F = kx = 250 × 0.12 = 30.0 N. If you typed the extension as 12 rather than 0.12 this comes out as 3000 N, which is three hundred kilograms hanging on a laboratory spring, and that is the moment to notice. Hooke's law wants metres.
THE ENERGY STORED. E_p = ½kx² = 0.5 × 250 × 0.12² = 0.5 × 250 × 0.0144 = 1.800 J. The arithmetic you should do beside it is the triangle: the force went from 0 to 30 N, so the average was 15 N, over 0.12 m, which is 15 × 0.12 = 1.8 J. Same answer, and it is where the ½ comes from.
WHAT THE SQUARE DOES. At 0.24 m — twice the extension — the force doubles to 60 N and the energy becomes 0.5 × 250 × 0.0576 = 7.200 J, which is four times, not twice. At 0.06 m it is 0.450 J, a quarter. Any question that changes an extension and asks about the energy is testing this and nothing else.
NOW RELEASE IT. All 1.800 J into a 0.050 kg mass: ½mv² = 1.8, so v = √(2 × 1.8 / 0.05) = √72 = 8.485 m/s. Fired straight up, that mass rises h = E/mg = 1.8 / (0.05 × 9.806 65) = 3.671 m — using standard gravity, which is 9.806 65 m/s² exactly by definition. At the 9.8 a question usually prints, the height is 3.673 m, so the choice of g shows up in the third decimal and nowhere higher.
AND THE LIMIT, WHICH IS WHY THAT LAST FIELD EXISTS. 0.12 m is 60 per cent of the way to the 0.20 m limit of proportionality, so F = kx and ½kx² both hold. At the limit itself the spring holds 50 N and stores 5.000 J. Ask this page for 0.25 m and it will still return a number — 7.8 J — and that number is wrong rather than approximate, because past the limit the graph is not a straight line and its area is not ½kx².

Force is linear, energy is quadratic: a 250 N/m spring

The third and fourth columns are the point. The force column doubles from 0.06 m to 0.12 m; the energy column quadruples. The last row is marked “claimed” deliberately: at an extension past the limit of proportionality the straight-line graph has stopped applying, so F = kx and ½kx² are no longer descriptions of this spring and the figures in that row are what the formula says rather than what the spring does. For most metals the real graph bends *below* the line, so the true force and the true stored energy are smaller than printed. The launch-speed column assumes every stored joule reaches the mass, which is the standard school idealisation; a real spring keeps a little in its own moving coils and loses a little to internal friction, so measured launch speeds come out a few per cent low. Note also that this page is the school treatment of a spring. Designing a real compression spring — wire diameter, coil count, solid height, buckling, allowable stress — is an engineering calculation and lives in the mechanical engineering section of this site rather than here.
Extension	Force F = kx	Energy ½kx²	Energy as a multiple of the 0.12 m value	Launch speed for 50 g
0.00 m	0 N	0.000 J	0	0.000 m/s
0.03 m	7.5 N	0.113 J	0.0625	2.121 m/s
0.06 m	15 N	0.450 J	0.25	4.243 m/s
0.09 m	22.5 N	1.013 J	0.5625	6.364 m/s
0.12 m	30 N	1.800 J	1	8.485 m/s
0.18 m	45 N	4.050 J	2.25	12.728 m/s
0.20 m — limit of proportionality	50 N	5.000 J	2.7778	14.142 m/s
0.24 m — past the limit	60 N claimed	7.200 J claimed	4	16.971 m/s claimed

A triangle under a straight line, and the two limits that end it

Hooke’s law says the force is proportional to the extension, F = kx, and everything else on this page is a consequence of that one straight line. If you plot force against extension you get a line through the origin with gradient k. The work done in stretching the spring is the area under that line, and the area under a straight line from the origin is a triangle: ½ × base × height = ½ × x × kx = ½kx². That is the derivation in full, and Cambridge 9702 asks for it in exactly that form at outcome 6.2.3 — the elastic potential energy is to be determined “from the area under the force–extension graph”. The ½ is not a correction factor: the force started at zero and grew linearly, so the average force over the stretch was half the final force, and energy is average force times distance.

The square is what the questions are about. The force is linear in the extension and the energy is quadratic, so doubling the extension doubles the force and quadruples the energy. Pull a spring to three times its original extension and you have nine times the energy in it. This asymmetry is the whole reason the topic is examined: it is counter-intuitive, it is easy to state, and the wrong answer — twice — is the one that comes to mind first. It is also why the last millimetres of a catapult or a bow matter so much more than the first.

Then take it back out, which is the half of the topic that gets less attention. A spring holding 1.8 J is a spring that will deliver 1.8 J, and the question is what to. Equate ½kx² with ½mv² and the launch speed is x√(k/m): proportional to the extension, and inversely proportional to the square root of the mass, so four times the mass leaves half the speed. Convert instead to height and h = ½kx²/mg. Both of those assume every stored joule arrives at the projectile, which no real spring achieves — some stays in the moving coils, some is lost to internal friction in the metal, and a vertical launch gives more away to the air. Measured launch speeds therefore come out a few per cent below the ideal, which is a result worth expecting rather than being surprised by in a laboratory.

Two limits end the formula and they are not the same limit. The limit of proportionality is where the force–extension graph stops being straight: beyond it F is no longer kx, the area is no longer ½kx², and both of this page’s answers become wrong rather than imprecise. The elastic limit is a little further along and is where the spring stops returning to its natural length when released — beyond it the deformation is partly permanent and some of the energy you put in has gone into rearranging the material and is not coming back. Exam questions distinguish these two routinely, and the reason to keep them apart is that one is about the shape of the graph and the other is about whether the process is reversible at all. If a question asks for the energy stored in a spring stretched past the limit, the only honest method is the area under the measured curve.

What this page is not. It is the school treatment: one spring, one constant, metres and newtons. Designing a real spring — choosing a wire diameter, counting coils, checking it against solid height, buckling and the allowable stress of the wire — is a mechanical engineering calculation with a quite different set of inputs, and this site keeps those in its mechanical engineering section. If you came here with a wire diameter and a coil count rather than a spring constant, that is the section you want.

Frequently asked questions

Why is there a ½ in ½kx²?

Because the force was not constant while you were stretching the spring — it started at zero and grew to kx. Energy is average force times distance, and the average of a quantity growing linearly from 0 to kx is kx/2, so the energy is (kx/2) × x = ½kx². Graphically it is the area under the force–extension line, which is a triangle rather than a rectangle. The practical consequence is that multiplying the final force by the extension gives an answer exactly twice too large, which is the second commonest mistake on this topic.

If I double the extension, does the stored energy double?

No — it quadruples. The energy depends on x², so doubling x multiplies the energy by four, tripling it by nine, and halving it leaves a quarter. The force is the quantity that doubles. On the default spring here, 0.12 m stores 1.8 J and 0.24 m stores 7.2 J while the force goes from 30 N to 60 N. Keeping those two straight — force linear, energy quadratic — is most of what this topic examines.

Does the formula work for compression as well as extension?

Yes, identically, as long as the spring is within its proportional range in compression too. x is squared, so the energy comes out positive whichever way you deform it, and the stored energy in a spring compressed 0.12 m is the same 1.8 J as in one stretched 0.12 m. The practical caveats differ: a compression spring can buckle sideways if it is long and thin, and it eventually reaches its solid height where the coils touch and the force rises almost vertically. Neither of those is in F = kx, and both are design calculations rather than school physics.

What is the difference between the limit of proportionality and the elastic limit?

The limit of proportionality is where the force–extension graph stops being a straight line, so F = kx and ½kx² stop being true. The elastic limit is slightly further on and is where the material stops returning to its original shape when the load is removed. Between the two, the spring is still elastic — it will come back — but the force is no longer proportional to the extension, so you need the graph rather than the formula. Past the elastic limit the deformation is partly permanent, and part of the work you did has gone into rearranging the material instead of into recoverable energy. Cambridge 9702 attaches the phrase “within its limit of proportionality” to the energy formula at outcome 6.2.4 for exactly this reason.

Why is my measured launch speed lower than the calculated one?

Because the calculation assumes every stored joule reaches the projectile, and three things take a share. The spring’s own coils have mass and have to be accelerated, so some of the kinetic energy ends up in the spring rather than in the load — this is the biggest effect for a light projectile on a heavy spring. Internal friction in the metal, called hysteresis, means the force–extension curve on release sits slightly below the curve on loading, and the area between them is lost as heat. And anything launched through air loses to drag immediately. A measured speed a few per cent below the ideal is the expected result, not a mistake; a measured speed well above it means the extension or the spring constant is wrong.

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References

Cambridge International AS & A Level Physics 9702, syllabus for use in 2025, 2026 and 2027, Version 1 (cambridgeinternational.org, read 11 November 2026). Quoted by outcome number so that the difference from the Indian syllabus is checkable rather than asserted. Outcome 6.1.3 asks the candidate to “recall and use Hooke’s law”. The energy comes in two steps that are worth noticing separately: 6.2.2 is that “the area under the force–extension graph represents the work done”, 6.2.3 asks for the elastic potential energy to be determined from that area, and only then does 6.2.4 give the formula — E_P = ½Fx = ½kx² “for a material deformed within its limit of proportionality”. The qualifier is part of the outcome, which is why the limit is an input on this page rather than a footnote. 6.2.1 defines elastic and plastic deformation and the elastic limit as separate terms.
PHYSICS (042), Class XI-XII, Central Board of Secondary Education senior secondary curriculum document for the session 2025-26 (cbseacademic.nic.in, read 11 November 2026). Cited for what the syllabus asks for, never as any kind of endorsement. Unit IV, Chapter 5, lists “potential energy of a spring” within the work-energy unit rather than in a separate materials chapter, and the conservative-force material sits beside it — which is the natural home for the point that a spring gives back what you put in and friction does not.
Fundamental Physical Constants — Complete Listing, 2022 CODATA adjustment, NIST Standard Reference Database 121 (read 11 November 2026 from the machine-readable ASCII listing). Three lines are used across this batch and all three are quoted as printed: the Newtonian constant of gravitation is 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻² with a standard uncertainty of 0.000 15 × 10⁻¹¹ — a relative uncertainty of 2.2 × 10⁻⁵, which makes G the worst-known constant on this page by four orders of magnitude; the standard acceleration of gravity is 9.806 65 m s⁻² (exact), exact because it is a defined conventional value and not a measurement of anywhere; and the standard atmosphere is 101 325 Pa (exact). The 2022 adjustment is the current one and it did not move G: the value and its uncertainty are unchanged from 2018.