Elastic Energy of a Spring

Insert spring constant and displacement into E = ½·D·s². The displacement must be in metres: a spring with D = 400 N/m stretched by 10 cm = 0.1 m stores E = ½ × 400 × 0.01 = 2 J. The most common mistake is the displacement in centimetres, which because of the square distorts the result by a factor of 10,000. Second most common: forgetting the factor ½. Whether the spring is stretched or compressed does not matter; s² makes the sign irrelevant. The formula holds as long as the spring follows Hooke law, i.e. in the elastic range.

From the linear force profile during stretching. The spring force is not constant; by Hooke it grows from zero at the start to D·s at the end. The stored work is therefore not final force times distance but the integral of force over distance, geometrically the area under the F-s line. This area is a triangle with base s and height D·s, so W = ½·(D·s)·s = ½·D·s². You can also argue with the average force: it equals (0 + D·s)/2 = ½·D·s, and average force times distance yields the same result. Taking the final force overestimates the energy by exactly a factor of 2.

They describe the same spring from two angles: F = D·s is the force at a given displacement, E = ½·D·s² the total work invested up to that point. Mathematically the energy is the integral of the force over distance; conversely the force is the derivative of the energy with respect to displacement. Both formulas are therefore valid in exactly the same range: as long as the characteristic curve is a straight line. Beyond the proportional limit the spring deforms plastically, part of the work goes into permanent deformation and heat, and both formulas fail together. In practice: from a measured F-s curve you read D as the slope and obtain the energy as the area.

Via energy conservation: as the spring relaxes, the elastic energy converts into kinetic energy, ½·D·s² = ½·m·v², solved as v = s·√(D/m). Spring gun example: D = 500 N/m, cocked by s = 6 cm, ball m = 20 g. Stored is E = ½ × 500 × 0.0036 = 0.9 J, so v = √(2 × 0.9/0.02) = √90 ≈ 9.5 m/s. In reality the muzzle speed is slightly lower because friction and the co-accelerated spring mass absorb energy. The same scheme works vertically with potential energy (trampoline: ½Ds² = mgh) and is the standard pattern for exam problems with springs.

Because further stretching works against an ever larger force. Over the first centimetres the opposing force is small, at the end it is large; the second part of the path therefore costs more work than the first. Quantitatively this sits in the square: E ∝ s², so 2s means four times and 3s nine times the energy. The triangle under the force-displacement line shows it immediately: doubling base and height quadruples the area. Consequence for oscillations: a spring pendulum with twice the amplitude carries four times the total energy, and in archery the last centimetres of draw add the largest energy gain.