Elastic Potential Energy

Draw back a bow and hold it. Nothing is moving, yet you can feel that something is waiting there — a whole arrow's worth of speed, coiled up and ready. Let go and the string snaps forward, hurling the arrow across the field. The energy that flew off with the arrow did not appear from nowhere: it was stored in the stretched bow all along, quietly banked while you held the draw.

That banked energy is elastic potential energy — the energy stored in anything stretched or squashed that springs back: a spring, a rubber band, a trampoline mat, a bungee cord, the wound coil of a clockwork toy. Whenever you do work to deform an elastic object, that work is not lost — it is saved inside, ready to be handed straight back the instant the object is released. This page works out exactly how much is stored, and where the neat little factor of one-half comes from.

Work stored = the area under the force–extension graph

Stretching a spring takes work, and that work is banked as elastic potential energy. But here is the subtlety: the force you need is not constant. By Hooke's law, the force in a spring grows in proportion to how far it is stretched,

F = k\,x,

where x is the extension and k the spring constant. So when the spring is barely stretched the force is tiny, and it only reaches its full value F = kx at the very end of the stretch. You cannot just multiply "force × distance" with a single force — the force keeps changing all the way.

The way round this is to read the energy straight off the force–extension graph. For any force that moves through a distance, the work done — the energy transferred — is the area under the graph of force against distance. Because Hooke's law draws a straight line through the origin, that area is a triangle: base x along the bottom, height F = kx up the side.

The area of a triangle is \tfrac12 \times \text{base} \times \text{height}, so the stored energy is

E = \tfrac12 \times x \times F = \tfrac12\,F\,x.

And since F = kx, we can substitute to write it with only k and x:

E = \tfrac12\,F\,x = \tfrac12\,(kx)\,x = \tfrac12\,k\,x^{2}.

That factor of ½ is the whole story of this page. It is not a fudge — it is the triangle. The force builds up from zero, so on average the spring is only pulling with half its final force F as it stretches; the energy is that average force \tfrac12 F times the distance x.

The energy stored in a spring stretched (or compressed) by an extension x, while it obeys Hooke's law, is the triangular area under its force–extension graph:

E = \tfrac12\,F\,x = \tfrac12\,k\,x^{2}.

E — the elastic potential energy stored, in joules (J)
F — the force in the spring at that extension, in newtons (N), with F = kx
k — the spring constant (stiffness), in newtons per metre (N/m)
x — the extension, i.e. the extra length, in metres (m)

Watch the triangle grow

Below is the force–extension graph for a spring with k = 40\ \text{N/m}. Drag the extension slider and watch two things at once: the straight Hooke's-law line climbs to the current force F = kx, and the shaded triangle beneath it — whose area is the stored energy E = \tfrac12 k x^{2} — fills in. Notice how the energy readout leaps upward far faster than the extension does: that is the tell-tale sign of the x^{2} in the formula.

Why it grows with the square

The x^{2} is the single most important feature of this formula, and the thing exam questions love to test. Doubling the extension does not double the stored energy — it quadruples it:

E(2x) = \tfrac12\,k\,(2x)^{2} = \tfrac12\,k\,(4x^{2}) = 4\left(\tfrac12\,k\,x^{2}\right) = 4E(x).

Stretch it three times as far and it stores 3^{2} = 9 times the energy; ten times as far, a hundred times the energy. This is why drawing a bow back that little bit further sends the arrow so much faster, and why a longer trampoline stretch flings you so much higher. The energy lives in the square of the stretch.

You can see the same fact geometrically in the graph above. Double the base of the triangle and the line — being straight through the origin — reaches double the height too, so both the base and the height double, and the area (½ × base × height) goes up by 2 \times 2 = 4. The triangle grows in two directions at once.

Worked examples

Example 1 — energy from k and x. A spring of constant k = 200\ \text{N/m} is stretched by x = 0.1\ \text{m}. How much energy is stored?

E = \tfrac12\,k\,x^{2} = \tfrac12 \times 200 \times 0.1^{2} = \tfrac12 \times 200 \times 0.01 = 1\ \text{J}.

Example 2 — energy from the graph area. A force–extension graph is a straight line reaching a force of F = 12\ \text{N} at an extension of x = 0.5\ \text{m}. The stored energy is the triangle's area:

E = \tfrac12\,F\,x = \tfrac12 \times 12 \times 0.5 = 3\ \text{J}.

(Check with k: here k = F/x = 12/0.5 = 24\ \text{N/m}, so \tfrac12 k x^{2} = \tfrac12 \times 24 \times 0.25 = 3\ \text{J} — the two forms always agree.)

Example 3 — the effect of doubling. The spring in Example 1 is now stretched to x = 0.2\ \text{m} instead of 0.1\ \text{m}. Rather than recompute from scratch, use the square law: the extension doubled, so the energy is 2^{2} = 4 times as big:

E = 4 \times 1\ \text{J} = 4\ \text{J}.

Confirm directly: \tfrac12 \times 200 \times 0.2^{2} = \tfrac12 \times 200 \times 0.04 = 4\ \text{J}. ✓

Launching things: energy conservation

The whole point of storing energy in a spring is to get it back out again as motion or height. Release the spring and, if nothing is wasted, every joule of elastic PE is transferred — E = \tfrac12 k x^{2} in, the same number of joules out. Setting the stored energy equal to the energy it becomes is one of the most powerful tricks in mechanics, because it lets you skip all the messy detail of the motion.

To a speed. A spring-loaded toy of spring constant k = 800\ \text{N/m} is compressed by x = 0.05\ \text{m} and fires a 0.02\ \text{kg} pellet. Ignoring friction, all the elastic PE becomes kinetic energy:

\tfrac12\,k\,x^{2} = \tfrac12\,m\,v^{2}.

The stored energy is \tfrac12 \times 800 \times 0.05^{2} = 1\ \text{J}, so \tfrac12 \times 0.02 \times v^{2} = 1, giving v^{2} = 100 and

v = \sqrt{100} = 10\ \text{m/s}.

To a height. A jack-in-the-box spring stores E = 6\ \text{J} and launches a 0.3\ \text{kg} toy straight up. Taking g = 10\ \text{N/kg}, all the elastic PE turns into gravitational PE, mgh:

\tfrac12\,k\,x^{2} = mgh \;\Rightarrow\; 6 = 0.3 \times 10 \times h \;\Rightarrow\; h = \frac{6}{3} = 2\ \text{m}.

Notice we never needed to know the spring's k or x separately here — only that it banked 6 J. Conservation of energy handed us the height directly.

The four traps that catch out nearly everyone with elastic PE:

It's \tfrac12 k x^{2}, not k x^{2}. Forgetting the ½ is the single most common slip. The ½ is the triangle: the force climbs from zero to kx, so the energy is the area under a sloping line, half of the kx \times x rectangle you might expect. A constant force would give Fx; a spring's rising force gives \tfrac12 Fx.
The energy scales with the square of the extension. Double the stretch and the energy goes up four times, not two. Treble it and it goes up nine times. Never assume energy is proportional to extension — it is proportional to extension squared.
Use the extension, in metres — not the total length, not centimetres. x is the extra length the spring has gained, and it must be in metres. A spring stretched from 10 cm to 14 cm has x = 0.04\ \text{m}, not 0.14 m and not 4.
The formula only holds up to the elastic limit. E = \tfrac12 k x^{2} is the area under a straight line, and the line is only straight while Hooke's law holds. Stretch past the limit of proportionality and the graph bends — you must then find the area some other way (e.g. counting squares), and past the elastic limit the spring is permanently deformed and won't give the energy back at all.

A bungee jumper's cord looks alarmingly long and stretchy on purpose — and the reason is buried in that x^{2}. When you leap, your fall gives you a fixed amount of kinetic energy at the bottom, and the cord must soak up all of it as elastic PE, \tfrac12 k x^{2}, to bring you to a gentle stop. The force the cord yanks back on you with is F = kx — and it is that force, if it spikes too high, that would injure you.

Here is the clever part. To store a given energy, a cord that stretches a long way (big x) can be made floppy (small k), and a floppy cord that stretches far pulls back with a much gentler peak force than a stiff cord that barely gives. A long, soft stretch spreads the same energy over a long, soft deceleration. A stiff steel cable would store the same joules but halt you with a bone-shattering jerk. The same insight tunes a car's suspension, a pole-vaulter's flexing pole, and the springy sole of a running shoe: let it give a long way, and it takes the energy kindly.