Elastic Potential Energy

Fix one end of a spring, note its natural length, then pull the other end. For a well-made steel coil the force needed grows in proportion to the deformation, and the spring pulls back with Fₛ = −kx. Three conventions hide in that line. x is the change from natural length, not the length itself: a spring 25.0 cm long at rest and pulled to 31.0 cm has x = +0.060 m. The minus sign keeps the spring's force pointing back toward natural length, so it opposes a stretch and opposes a compression. And k, in N m⁻¹, is the slope of the force–extension graph: if 12.0 N holds that stretch, k = 12.0 N ÷ 0.060 m = 200 N m⁻¹. That number belongs to the spring, not to the steel, because more coils or a wider coil make the same wire softer. Calling the spring ideal then adds three testable claims: the line stays straight over the deformations you use, the spring has no mass, and unloading retraces loading.

To hold the spring at stretch x you must apply +kx. Pull slowly, so no kinetic energy builds up, and all of your work goes into the spring. The force is not constant, so the work is an integral, W = ∫₀ˣ kx′ dx′ = ½kx², the area of the triangle under the line. Read it as average force times distance: your force rose steadily from 0 to kx, so its average was ½kx, and multiplying the final force by the whole stretch counts the triangle twice. Because the spring hands the work back as it returns, it defines a potential energy U(x) = ½kx² with U = 0 at natural length, and the spring's own force does work Wₛ = −ΔU, negative while you stretch it and positive while it relaxes. For the 200 N m⁻¹ spring at 0.060 m, U = ½ × 200 × 0.060² = 0.36 J, not 12.0 N × 0.060 m = 0.72 J. A 0.060 m compression stores the same 0.36 J, because x² cannot tell the two apart.

Substitute x = F/k and the same energy reads U = ½kx² = F²/(2k). Both are true, and choosing between them is the step that settles most combination problems. When two springs are forced to share an extension, compare ½kx²: the stiffer one stores more. When they carry the same force, compare F²/(2k): the softer one does. Take k = 500 N m⁻¹ and k = 1500 N m⁻¹. Stretched by the same 4.0 cm they store 0.40 J and 1.20 J; pulled by the same 30 N they store 0.90 J and 0.30 J. Neither spring is better at storing energy in general; the constraint picks the winner. The same comparison guides design. If a mechanism limits the force a spring may exert, a soft spring with a long travel banks more energy; if it limits the travel, a stiff spring does.

Two springs hooked to a bar that stays level share one extension x, and their forces add: F = k₁x + k₂x, so kₚ = k₁ + k₂. Sharing x, each stores energy in proportion to its own k, so the stiffer spring carries more of both the load and the energy. Hook the same springs end to end and each carries the full force F while the extensions add: x = F/k₁ + F/k₂, so 1/kₛ = 1/k₁ + 1/k₂, and the chain is softer than either spring alone. Sharing F, each stores energy in proportion to 1/k, so now the softer spring holds more. For 200 N m⁻¹ and 800 N m⁻¹ the two rules give 1000 N m⁻¹ and 160 N m⁻¹. The series rule also explains cutting. A uniform spring is its own n equal pieces in series, so each piece is n times as stiff as the whole: a 90 N m⁻¹ spring cut into thirds gives three 270 N m⁻¹ springs. Parallel and series name what is shared, stretch or force, not how the drawing looks.

Stretch a steel spring past its elastic limit and it takes a permanent set: it no longer returns to its old natural length, and part of your work went into deforming the metal for good. Well before that, many elastic objects are not straight lines. A rubber band softens and then stiffens as its long molecules straighten, and a shock cord stiffens sharply once its braided sleeve pulls tight. While unloading still retraces loading, the stored energy remains the area, U = ∫₀ˣ F(x′) dx′, and ½kx² with any single k only approximates it. Rubber breaks that condition too: it unloads along a lower curve than it loaded on, and the area between the two curves becomes thermal energy every cycle, so only the area under the unloading curve comes back. The massless assumption fails when a spring's mass is not small beside its load, because its moving coils carry kinetic energy the model has nowhere to put; for a uniform spring the usual correction adds a third of the spring's mass to the moving load.

Hang a mass m on a vertical spring and let s be the stretch below natural length. The spring stores ½ks², and measured along the same downward s, the gravitational potential energy changes by −mgs. The load rests where ks₀ = mg. Write s = s₀ + y and expand: ½ks² − mgs = ½ky² + (ks₀ − mg)y + constant, and the middle term vanishes. Measured from equilibrium, spring and gravity together act like one spring with energy ½ky² plus a constant. That ½ky² is not the spring's own energy, so adding a separate mgy term on top counts gravity twice. With k = 50.0 N m⁻¹ and m = 0.400 kg, s₀ = 0.400 × 9.81 ÷ 50.0 = 0.0785 m. Pull the load a further 5.0 cm down: the spring's own energy rises by 0.259 J while gravity's falls by 0.196 J, so the combined energy climbs by only ½ × 50.0 × 0.050² = 0.0625 J.