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Most elasticity problems stop at the extension of a wire. The next question is how much energy the wire now holds, and that is where students reach for Fl and get double the right answer. This post shows where the half comes from, what happens to the other half, and how to use the energy formula in three worked problems.
Is elastic energy still in NCERT and the exams?
Yes. The 2026-27 reprint of the NCERT Class 11 chapter Mechanical Properties of Solids keeps Section 8.5.5, "Elastic Potential Energy in a Stretched Wire", and ends it with Eq. 8.14, u = ½σε, the energy per unit volume.
The Syllabus for JEE (Main) 2026 lists "elastic behaviour, stress-strain relationship, Hooke's Law, Young's modulus" in Unit 7 and "the potential energy of a spring" in Unit 4. It does not name wire energy separately, and the NEET (UG) 2026 syllabus has the same elasticity wording. The JEE (Advanced) 2026 brochure lists "Hooke's law, Young's modulus". The energy result follows from these, so it is fair game. Check your own year's documents.
Why is the stored energy ½Fl and not Fl?
NCERT's derivation goes like this. While the wire is stretched by an amount x, the tension is F = YAx/L, where Y is Young's modulus, A the cross-section and L the original length. Stretching it a further dx takes work F dx. The force grows from zero, so you add up a growing force:
W = ∫₀ˡ (YA/L) x dx = ½ (YA/L) l²
Since the final force is F = YAl/L, this is W = ½Fl. Divide by the volume AL and use stress σ = F/A and strain ε = l/L:
u = ½σε
You can rewrite U in three ways. Pick the one that matches the data:
U = ½Fl, if you know the load and the extension.
U = F²L/(2AY), if you know the load but not the extension.
U = YAl²/(2L), if you know the extension but not the load.
Worked example 1: a copper wire under a load
A copper wire 2.5 m long and 1.2 mm in diameter carries an 8.0 kg load. Use NCERT's Table 8.1 value, Y = 1.1 × 10¹¹ Pa, and g = 9.8 m/s².
Force: F = 8.0 × 9.8 = 78.4 N.
Area: r = 0.60 mm = 6.0 × 10⁻⁴ m, so A = π(6.0 × 10⁻⁴)² = 1.13 × 10⁻⁶ m².
Stress: σ = 78.4 / 1.13 × 10⁻⁶ = 6.93 × 10⁷ Pa. Table 8.1 gives copper's yield strength as 200 × 10⁶ Pa, so Hooke's law holds.
Strain: ε = σ/Y = 6.30 × 10⁻⁴.
Extension: l = εL = 1.58 × 10⁻³ m.
Energy: U = ½Fl = 0.5 × 78.4 × 1.58 × 10⁻³ = 0.062 J.
Check with the density form: u = ½σε = 0.5 × 6.93 × 10⁷ × 6.30 × 10⁻⁴ = 2.18 × 10⁴ J/m³. The volume is AL = 2.83 × 10⁻⁶ m³, and u × volume = 0.062 J. The two routes agree.
Worked example 2: where does the other half go?
In example 1, the load falls 1.58 mm as the wire stretches, so gravity does mgl = 78.4 × 1.58 × 10⁻³ = 0.124 J of work. The wire stores only 0.062 J.
If you lower the load by hand, your hand pushes up on it the whole way and takes the other 0.062 J.
If you let go of the load suddenly at the unstretched length, nothing takes that energy at first. At the lowest point the load is momentarily at rest, so gravity's work equals the stored energy: mgx = YAx²/(2L). That gives x = 2mgL/(YA) = 2l. The load then oscillates about l, and in a real wire the extra energy ends up as heat.
So "released suddenly" doubles the maximum extension and the maximum stress. That is why a load that a wire can hold gently may break it when dropped on.
Worked example 3: which material stores more at the same stress?
From u = ½σε and ε = σ/Y, u = σ²/(2Y). At a fixed stress of 5.0 × 10⁷ Pa, using NCERT's Table 8.1 values:
Steel, Y = 2.0 × 10¹¹ Pa: u = 6.3 × 10³ J/m³.
Copper, Y = 1.1 × 10¹¹ Pa: u = 1.1 × 10⁴ J/m³.
Aluminium, Y = 0.70 × 10¹¹ Pa: u = 1.8 × 10⁴ J/m³.
The stiffer material stores less at the same stress, because it stretches less. At the same strain the order reverses, since u = ½Yε². Always ask which quantity is held fixed.
Common mistakes and how to avoid them
Writing U = Fl. That is gravity's work, not the stored energy.
Diameter used as radius. This makes A four times too big and U four times too small in the F²L/(2AY) form.
Comparing materials without fixing stress or strain. The answer flips depending on which is equal.
Forgetting the volume. ½σε is per cubic metre. Multiply by AL for the total.
For stress, strain and the practical's error formula, read our post on elasticity and Young's modulus problems. The same half appears in a spring, covered in our guide to spring-block energy problems. If a derivation step does not click, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Derive U = ½Fl from the integral once without notes.
Redo example 1 for a steel wire of the same size and check that U is smaller.
Solve two "released suddenly" problems and find the maximum extension.
Do one comparison at equal stress and one at equal strain.
Today, find the energy stored when a steel wire 1.0 mm in diameter and 2.0 m long stretches by 1.0 mm, using U = YAl²/(2L), and check you get about 0.039 J.
