Elastic potential energy in a stretched wire

Pull back a catapult and you feel it fight you harder the further you stretch it — and when you let go, that fight turns into flight. The same thing happens, on a stiffer and less visible scale, inside a steel wire, a bungee cord, or the string of a guitar the instant you tune it. Stretching any elastic body means doing work against its internal restoring forces, and that work does not vanish — it is stored, silently, as elastic potential energy, ready to be paid back the moment the load is removed. This article answers a very concrete question: exactly how much energy is stored, and where does the formula come from? The answer turns out to need nothing beyond a force-extension graph and the area of a triangle, because a wire obeying Hooke's law behaves exactly like a spring — the restoring force grows in direct proportion to the extension, from zero up to its final value. That simple fact is enough to derive one of the most quoted results in this chapter: energy stored per unit volume, $u = \tfrac12\sigma\varepsilon = \tfrac12 Y\varepsilon^2$ , sitting right alongside its twin $u = \sigma^2/(2Y)$ . Along the way we will stretch a steel wire and a rubber cord side by side, see why the rubber — despite carrying far less stress — can store far more energy, and connect that geometric picture to the very same area-under-the-graph idea you already used for work done by a variable force. By the end you should be able to derive the formula from scratch, choose the right variant of it in a numerical, and explain precisely why it stops working once a wire is stretched past its elastic limit.

Key takeaways

Stretching a wire that obeys Hooke's law does work against an internal restoring force that grows linearly with extension — from $0$ up to the final value $F$ .
That work equals the area of the triangle under the force-extension graph, $W = \tfrac12 F\,\Delta L$ , and it is stored as elastic potential energy, fully recoverable within the elastic limit.
Energy density (energy per unit volume) is $u = \tfrac12\sigma\varepsilon$ , which becomes $u = \tfrac12 Y\varepsilon^2$ or $u = \sigma^2/(2Y)$ once Hooke's law $\sigma = Y\varepsilon$ is substituted.
Total energy stored in a wire of volume $V = AL$ is $U = u\,V = \tfrac12 F\,\Delta L$ — always check whether a question wants density (J/m³) or total energy (J).
This energy equals the area under the linear part of any stress-strain or force-extension graph, and is released as sound, heat, or kinetic energy if the wire suddenly relaxes.
The factor of $\tfrac12$ exists only because the force builds up gradually; it disappears if you (wrongly) use the final force for the whole extension.
Key idea: because a Hookean wire's restoring force rises linearly from zero, the work done stretching it is half the final force times the extension, giving stored energy density $u = \tfrac12\sigma\varepsilon = \tfrac12 Y\varepsilon^2 = \sigma^2/(2Y)$ , valid only up to the elastic limit.

The core insight

A wire stretched within its elastic limit stores energy $u = \tfrac12\sigma\varepsilon = \tfrac12Y\varepsilon^2$ per unit volume, exactly the triangular area under its linear force-extension graph, because the restoring force it works against grows from zero to its final value as the extension builds up.

Where it fits in the syllabus

This is the last quantitative result of the CBSE Class 11 unit Mechanical Properties of Solids, arriving right after stress, strain, Hooke's law, Young's modulus, and Poisson's ratio. It sits squarely inside the prescribed 2025-26 scope — the board explicitly lists "elastic potential energy per unit volume $u = \tfrac12\sigma\varepsilon = \tfrac12Y\varepsilon^2$ " as examinable — so every formula and worked number here is board-safe, with no out-of-syllabus material used. What follows extends the Pass-1 treatment with a fuller derivation, an SVG force-extension graph, a second worked example at competitive-exam pace, and a compact exam card, while keeping every original number intact.

The physical picture

Picture a wire of natural length $L$ and cross-sectional area $A$ , clamped at the top and pulled from the bottom by a slowly increasing force. At every instant during the pull, the applied force is only just enough to balance the wire's internal restoring force at that extension — otherwise the wire would accelerate rather than stretch quasi-statically. Because Hooke's law holds, that restoring force grows in direct proportion to how far the wire has already stretched, exactly like a spring being drawn out.

F(x) = k x, \qquad 0 \le x \le \Delta L

where: $F(x)$ is the force needed to hold the wire at extension $x$ , $k$ is an effective stiffness (fixed by the wire's geometry and Young's modulus), and $\Delta L$ is the final, total extension. Because $F$ rises in a straight line from $0$ to its final value $F = k\,\Delta L$ , a plot of force against extension is a straight line through the origin — and the work done stretching the wire, being the sum of $F\,dx$ over every tiny step, is simply the area under that line.

Figure

A force versus extension graph for a wire showing a straight diagonal line from the origin to a final point, with the triangular area beneath the line shaded to represent the work done, labelled delta L on the horizontal axis and F on the vertical axis

Since the graph is a triangle of base $\Delta L$ and height $F$ , the work done — and hence the elastic potential energy stored, because none of it is lost within the elastic limit — is

U = W = \frac{1}{2}F\,\Delta L

where: $F$ is the final applied force, and $\Delta L$ is the total extension produced. To turn this into a property of the material rather than of one particular wire, divide by the volume $V = AL$ and rewrite $F/A$ as stress $\sigma$ and $\Delta L/L$ as strain $\varepsilon$ :

u = \frac{U}{V} = \frac{1}{2}\sigma\varepsilon = \frac{1}{2}Y\varepsilon^{2} = \frac{\sigma^{2}}{2Y}

where: $u$ is elastic potential energy density (J/m³, the same dimensions as stress, since energy per volume is force times length divided by volume), $\sigma = F/A$ is longitudinal stress, $\varepsilon = \Delta L/L$ is longitudinal strain, and $Y$ is Young's modulus, related to $\sigma$ and $\varepsilon$ by Hooke's law $\sigma = Y\varepsilon$ . All three forms are algebraically the same quantity; you pick whichever variables a problem hands you — stress and strain together, strain alone, or stress alone.

Derivation AElastic potential energy density of a stretched wire

The goal is to derive the energy stored per unit volume of a wire that has been stretched by an external force obeying Hooke's law throughout, starting only from the definition of work done by a variable force and the geometry of a straight-line force-extension graph, and ending at the three interchangeable forms $u = \tfrac12\sigma\varepsilon = \tfrac12Y\varepsilon^2 = \sigma^2/(2Y)$ that the board expects a student to both derive and apply.

Step 1. Set up the force law. As the wire is stretched quasi-statically from zero extension to a final extension $\Delta L$ , Hooke's law guarantees the restoring (and hence the applied) force at any intermediate extension $x$ is $F(x) = kx$ , where $k$ is a constant fixed by the wire's dimensions and Young's modulus. This is exactly the spring-like behaviour that makes the whole calculation possible: without linearity, the force-extension curve would not be a straight line and the triangle argument below would fail.

Step 2. Compute the work done as an area. Work done against a variable force is $W = \int_0^{\Delta L} F(x)\,dx$ , and because $F(x)$ traces a straight line from the origin $(0,0)$ to the final point $(\Delta L, F)$ , this integral is exactly the area of the right triangle beneath that line — base $\Delta L$ , height $F$ . So $W = \tfrac12 \times \text{base} \times \text{height} = \tfrac12 F\,\Delta L$ . Physically, this factor of one-half says the average force during the stretch is half the final force, since the force built up linearly from zero.

Step 3. Identify this work as stored energy. Within the elastic limit no energy is dissipated as heat or permanent deformation — every joule of work done on the wire is recoverable — so the elastic potential energy equals the work done: $U = \tfrac12 F\,\Delta L$ .

Step 4. Convert to a per-volume quantity. The wire's volume is $V = AL$ (area times natural length), so dividing gives $u = U/V = \dfrac{\tfrac12 F\,\Delta L}{AL} = \tfrac12\left(\dfrac{F}{A}\right)\left(\dfrac{\Delta L}{L}\right) = \tfrac12\sigma\varepsilon$ , since $F/A$ is stress and $\Delta L/L$ is strain by definition.

Step 5. Substitute Hooke's law to get the alternate forms. Since $\sigma = Y\varepsilon$ , replacing $\sigma$ gives $u = \tfrac12(Y\varepsilon)\varepsilon = \tfrac12 Y\varepsilon^2$ ; replacing $\varepsilon = \sigma/Y$ instead gives $u = \tfrac12\sigma(\sigma/Y) = \sigma^2/(2Y)$ . All three expressions are exactly equal for a wire obeying Hooke's law, and the total energy in the whole wire is simply $U = u\,V$ .

Worked examples

Board level - steel wire stretched by a known force

A steel wire of length $2.0\ \text{m}$ , cross-sectional area $1.0\ \text{mm}^2$ , and Young's modulus $Y = 2.0\times10^{11}\ \text{Pa}$ is stretched by a force of $F = 100\ \text{N}$ . Find (a) the extension, (b) the elastic potential energy stored in the wire.

Step 1. Convert the area to SI units before doing anything else: $A = 1.0\ \text{mm}^2 = 1.0\times10^{-6}\ \text{m}^2$ . Skipping this step is the single most common arithmetic slip in this kind of problem.

Step 2. Find the extension from Young's modulus, $Y = FL/(A\,\Delta L)$ , rearranged to $\Delta L = FL/(AY)$ : $\Delta L = \dfrac{(100)(2.0)}{(1.0\times10^{-6})(2.0\times10^{11})} = \dfrac{200}{2.0\times10^{5}} = 1.0\times10^{-3}\ \text{m}$ , that is, $1.0\ \text{mm}$ .

Step 3. Find the elastic potential energy using $U = \tfrac12 F\,\Delta L$ , since we already have both the final force and the total extension directly: $U = \tfrac12(100)(1.0\times10^{-3}) = 0.050\ \text{J}$ .

Step 4. Sanity check with the density route: $\sigma = F/A = 100/(1.0\times10^{-6}) = 1.0\times10^{8}\ \text{Pa}$ , $\varepsilon = \Delta L/L = 5.0\times10^{-4}$ , so $u = \tfrac12\sigma\varepsilon = \tfrac12(1.0\times10^{8})(5.0\times10^{-4}) = 2.5\times10^{4}\ \text{J/m}^3$ . Volume $V = AL = 2.0\times10^{-6}\ \text{m}^3$ , giving $U = uV = (2.5\times10^{4})(2.0\times10^{-6}) = 0.050\ \text{J}$ — the same answer, confirming both routes agree.

Answer: extension $\Delta L = 1.0\ \text{mm}$ ; elastic potential energy $U = 0.050\ \text{J}$ , the energy that would be released as sound, heat, or kinetic energy if the wire were suddenly allowed to relax.

Competitive level - rubber cord at large strain, then compared with the steel wire

A rubber cord has volume $V = 4.0\times10^{-5}\ \text{m}^3$ and, at a certain extension, carries a longitudinal stress of $\sigma = 3.0\times10^{5}\ \text{Pa}$ and a longitudinal strain of $\varepsilon = 0.60$ . Find (a) the elastic potential energy density, (b) the total elastic potential energy stored, and (c) how many times more energy the rubber stores than the steel wire of Worked Example 1, despite its much lower stress.

Step 1. Energy density directly from $u = \tfrac12\sigma\varepsilon$ , since stress and strain at the same state are both given: $u = \tfrac12(3.0\times10^{5})(0.60) = 9.0\times10^{4}\ \text{J/m}^3$ .

Step 2. Total energy from $U = u \times V$ : $U = (9.0\times10^{4})(4.0\times10^{-5}) = 3.6\ \text{J}$ .

Step 3. Compare with the steel wire's $U = 0.050\ \text{J}$ from Worked Example 1: the ratio is $3.6/0.050 = 72$ . The rubber cord stores $72$ times more total energy than the steel wire, even though its stress ( $3.0\times10^{5}\ \text{Pa}$ ) is more than a hundred times smaller than the steel wire's ( $1.0\times10^{8}\ \text{Pa}$ ).

Answer: $u = 9.0\times10^{4}\ \text{J/m}^3$ , $U = 3.6\ \text{J}$ , about $72\times$ the steel wire's stored energy, because energy density scales with the product of stress and strain, and rubber's enormous strain ( $0.60$ , versus steel's $5.0\times10^{-4}$ ) far outweighs its modest stress — exactly why elastomers, not stiff metals, are chosen for catapults, bungee cords, and shock-absorbing mounts. Timing note: this is a two-line plug-and-chug once you recognise $u = \tfrac12\sigma\varepsilon$ needs no unit conversion when $\sigma$ is already in pascals and $\varepsilon$ is dimensionless — aim to finish parts (a) and (b) in under thirty seconds each. The only place competitive papers add a twist is asking for a ratio or ranking between two materials, as in part (c); keep both energies in the same units (joules) before dividing, and do not accidentally compare a density to a total.

Common traps

Forgetting the factor of $\tfrac12$ : because the restoring force grows linearly from zero (Hooke's law), the work done is the average force times the extension, never the final force times the extension.
Confusing energy per unit volume ( $u$ , in J/m³) with total energy ( $U$ , in J): always check whether a problem gives, or asks for, the volume-specific or the total quantity, and multiply or divide by volume $V = AL$ as needed.
Applying $u = \tfrac12\sigma\varepsilon$ beyond the elastic limit: the derivation assumes Hooke's law holds throughout the stretch, so the formula is valid only within the linear, fully recoverable part of the stress-strain curve — beyond the elastic limit, work is also spent on permanent deformation and is not stored as recoverable elastic energy.
Mixing stress and strain from two different loading states, such as stress read off at one extension and strain at another, instead of using both values at the same, final state of extension.
Treating $u = \tfrac12Y\varepsilon^2$ and $u = \sigma^2/(2Y)$ as different results rather than the same formula written in different variables — pick whichever the given data supports directly, rather than converting unnecessarily.

Common trapThis energy is only recoverable within the elastic limit, and the one-half is not optional

Two errors dominate lost marks here. First, students often write $U = F\,\Delta L$ (dropping the half) by analogy with work done by a constant force — but the force on a Hookean wire is not constant, it ramps up from zero, so only the average force (half the final value) times the extension gives the correct work and correct stored energy. Second, students apply $u = \tfrac12\sigma\varepsilon$ to a wire stretched beyond its elastic limit, where the stress-strain graph curves and is no longer described by $\sigma = Y\varepsilon$ ; some of the work done there goes into permanent (plastic) deformation and is never returned when the load is removed, so treating all of it as recoverable elastic energy overstates the energy available for spring-back.

You should be able to…

Derive $u = \tfrac12\sigma\varepsilon$ starting from the work done stretching a wire that obeys Hooke's law, using the triangular area under a force-extension graph.
Write and freely switch between the equivalent forms $u = \tfrac12\sigma\varepsilon$ , $u = \tfrac12Y\varepsilon^2$ , and $u = \sigma^2/(2Y)$ , choosing whichever suits the data given in a problem.
Calculate total elastic potential energy $U$ either from energy density and volume ( $U = uV$ ) or directly from force and extension ( $U = \tfrac12F\,\Delta L$ ), and get the same answer both ways.
Relate elastic potential energy to the area under the linear part of a stress-strain or force-extension graph.
Explain why the stored energy is only fully recoverable within the elastic limit, and why the one-half factor arises specifically because the restoring force builds up linearly from zero.
Compare the elastic energy stored in two different stretched materials, correctly separating the roles of stress, strain, and volume.

Ask the tutor: Want to see how this elastic energy formula maps exactly onto the spring potential energy $\tfrac12kx^2$ you already know from mechanics — ask the AI tutor to walk through the analogy term by term.

Why is there a factor of one half in the energy formula? A spring's energy formula also has one half, is that a coincidence?

Not a coincidence at all — it is the same physics. Both a Hookean wire and an ideal spring obey a restoring force that grows linearly with displacement or extension ( $F = kx$ for a spring, and the same shape of law for a wire under Hooke's law). Whenever a force ramps up in a straight line from zero, the work done against it equals the average force (which is half the final force) multiplied by the total displacement. That is exactly where every one-half in this chapter's energy formulas comes from — it is a geometric fact about triangles, not a special rule for wires or springs separately.

If I double the extension of a wire while staying inside the elastic limit, does the stored energy also double?

No — it quadruples. Since $u = \tfrac12Y\varepsilon^2$ , energy density depends on the square of strain, not on strain directly. Doubling the extension doubles the strain $\varepsilon$ , and squaring that doubled strain gives four times the energy density. This is the same reason doubling a spring's compression quadruples its stored energy $\tfrac12kx^2$ — it is a direct analogue, and worth remembering because many exam questions test exactly this square-law dependence rather than the plain formula.

Once I stretch a wire and let go, where does the stored elastic potential energy actually go?

Within the elastic limit, essentially all of it converts back into other forms of energy the instant the wire is released — kinetic energy of the wire snapping back, and eventually heat and sound as the vibrations die away (think of a plucked guitar string, or a rubber band snapping against your skin). None of it is permanently lost to the material itself, because no permanent deformation occurred. This is precisely why the formula only applies below the elastic limit: beyond that point, some energy is spent rearranging the internal structure of the material permanently, and that fraction is not returned when the load is removed.

How is this different from the work-energy theorem I already know from mechanics?

It is the same underlying idea — work done equals energy transferred — applied to an internal, distributed restoring force instead of a single external force on a point mass. In ordinary mechanics you often deal with a constant force doing work $W = F d$ ; here the force is not constant, it changes continuously as the wire stretches, which is exactly why you need the "area under the graph" method (or, equivalently, integration) rather than a single multiplication. The elastic potential energy stored is simply this variable-force work done, relabelled to emphasise that the wire itself is storing it, ready to give it back.