Elastic potential energy stored in a spring — Physics

Before this

This leads to

Young's modulus
Energy in simple harmonic motion · coming soon
Capacitors — storing charge and energy · coming soon

Where did the work go?

The last lecture ended with a question. You do work to stretch a spring. Where does that energy go?

Watch what a spring does when you let go. Pull back the elastic of a catapult and release it: the stone flies off. Push down a spring-loaded toy and let go: it jumps into the air. The energy did not disappear while you held the spring still. It was waiting inside the stretched or squashed spring, ready to come back out. This stored energy is called elastic potential energy. Some books call it elastic strain energy.

So the answer to the question is: the work you did is stored in the spring. The harder question is how much. Here the formula you know for work seems to break.

A force that will not stay still

From the work lecture, the work done by a force is the force times the displacement: $W = F s$ . Take the spring from the last lecture, with $k = 50$ N/m. Stretch it by 12 cm. At that extension, Hooke's law gives a force of $F = k x = 50 \times 0.12 = 6$ N. So is the work $6 \times 0.12 = 0.72$ J?

No. $W = F s$ works for a force that stays the same all the way. The spring's force does not. At the start, with no extension, you need no force at all. After 1 cm, you need 0.5 N. After 6 cm, 3 N. Only at the very end do you need the full 6 N. Using 6 N for the whole stretch pretends you pulled hard the whole time. You did not. So 0.72 J is too much.

This is the hard part of the lecture, so spend time here. We need a way to add up the work when the force keeps changing. There are two ways to see it, and they give the same answer.

Way 1: the average force

The force grows steadily, in a straight line, from 0 to 6 N. When something grows steadily, its average is halfway between the start and the end. So the average force is $\frac{1}{2} \times 6 = 3$ N. The work is the average force times the extension: $3 \times 0.12 = 0.36$ J. That is exactly half of the wrong answer.

Way 2: the area under the graph

Draw force against extension. For a constant force, the graph is flat, and $F \times s$ is the area of a rectangle under it. For a changing force, cut the stretch into many thin steps. In each thin step, the force hardly changes, so the work in that step is almost a thin rectangle: force × small extension. Add up all the thin rectangles, and you get the area under the graph. So, for any force:

work done = area under the force-extension graph

For a spring that obeys Hooke's law, the graph is a straight line through the origin. The area under it is a triangle. See it below. Change the number of steps and watch the thin rectangles close in on the triangle.

Explore — adding up thin steps

The spring (k = 50 N/m) is stretched to 12 cm. The stretch is cut into equal steps. The dark rectangles use the force at the start of each step, so they add up to too little. The outlines use the force at the end of each step, so they add up to too much. Increase the number of steps.

Number of steps: 4

The work to stretch the spring (k = 50 N/m) to 12 cm, estimated with thin steps. "Too little" uses the force at the start of each step; "too much" uses the force at the end.

Steps	Too little / J	Too much / J	Average / J
1	0	0.72	0.36
4	0.27	0.45	0.36
8	0.315	0.405	0.36
32	≈ 0.349	≈ 0.371	0.36

Both estimates close in on 0.36 J, the area of the triangle under the line.

What should you have seen? With one step, the "too much" estimate is the wrong answer from before, 0.72 J. As the steps get thinner, the gap between the two estimates shrinks. With 32 steps, they are 0.349 J and 0.371 J. Both are closing in on the area of the triangle, 0.36 J, which agrees with Way 1. The thinner the steps, the closer you get. That is it. That is the whole idea: work done by a changing force is the area under its graph.

From the triangle to a formula

The triangle has a base of $x$ and a height of $F$ . Its area is half the base times the height:

E_{e} = \frac{1}{2} F x

For a spring that obeys Hooke's law, $F = k x$ . Put that in place of $F$ :

E_{e} = \frac{1}{2} (k x) x = \frac{1}{2} k x^{2}

Elastic potential energy

Up to the limit of proportionality, the energy stored in a stretched or compressed spring is

$E_{e} = \frac{1}{2} k x^{2} = \frac{1}{2} F x$

$E_{e}$ is in J, $k$ in N/m, $x$ (the extension or compression) in m, and $F$ is the force at that extension, in N.

The ½ is not a number to remember. It is the triangle. It is there because the force started at zero and grew steadily. Notice also that the formula is the same for compression. Squash the spring by $x$ , and it stores $\frac{1}{2} k x^{2}$ as well, because the force–compression graph is the same straight line.

Now look at the $x^{2}$ . It hides a surprise.

The spring with k = 50 N/m. The dark triangle (1) is the energy stored at 6 cm: ½ × 0.06 × 3 = 0.09 J. Stretch to 12 cm (point P), and the triangle under the line splits into four copies of triangle 1: 4 × 0.09 = 0.36 J.

Stop and think

A spring stretched by 6 cm stores 0.09 J. You stretch it further, to 12 cm. How much energy does it store now? How much work did the second 6 cm take?

Show the answer

Not twice as much: four times, 0.36 J. Doubling $x$ doubles both the base and the height of the triangle, so its area is multiplied by $2 \times 2 = 4$ . The figure shows the four copies. So the first 6 cm took 0.09 J, and the second 6 cm took $0.36 - 0.09 = 0.27$ J, three times as much. The second half is harder because you are pulling against a larger force the whole way.

This is why a spring that is stretched a little further stores much more energy. Triple the extension, and the energy is nine times as large.

Getting the energy back

A spring that stays within its elastic limit gives back the stored energy when it returns to its natural length. That is how a catapult or a spring launcher works. If no energy is wasted, the elastic potential energy becomes kinetic energy:

\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2}

This is conservation of energy, with the spring and the ball as the system. Try it below. Choose a compression and a mass, then launch.

Simulation — a spring launcher, in slow motion

The spring has k = 400 N/m. Set the compression and choose a ball. Press Play to release it, or Step to move one stage at a time. Watch the energy move from the spring into the ball. Then double the compression and launch again.

Ball:

Compression: 8 cm

A spring with k = 400 N/m is compressed and then releases a ball. All the stored energy becomes kinetic energy (no friction, and the spring's own mass ignored).

Compression / cm	Ball / g	Energy stored / J	Launch speed / (m/s)
4	50	0.32	≈ 3.58
8	50	1.28	≈ 7.16
8	100	1.28	≈ 5.06
10	20	2.00	≈ 14.14

Halfway through the release (compression 8 cm → 4 cm left), the spring still holds 0.32 J and the ball already has 0.96 J. Doubling the compression gives four times the energy but only twice the speed.

What should you have seen? At every moment, the two bars add up to the same total. The spring loses exactly what the ball gains. Notice how fast the ball gets its energy. When half the compression is still left, the spring holds only a quarter of the energy, so the ball already has three quarters. Then compare 4 cm and 8 cm with the same ball. Doubling the compression gives four times the energy, but only twice the speed, because the kinetic energy also has a square in it: $v$ is proportional to $x$ .

The simulation leaves things out on purpose. A real spring has its own mass, so some energy stays in the moving coils. Friction and sound take a little too. Real launchers are always slower than this ideal answer.

Past the limit: the energy that does not come back

The formula $\frac{1}{2} k x^{2}$ needs Hooke's law, so it holds only up to the limit of proportionality. Beyond that point the graph curves, and you cannot use the formula. But the idea behind it still works: the work done is the area under the graph, whatever its shape.

Return to the spring from the last lecture. It was loaded to 10 N, which stretched it to 26.4 cm, well past its elastic limit. Then the load was removed, and the spring came back along a straight line, ending 6.4 cm longer than before.

The spring loaded past its elastic limit to 10 N (solid curve), then unloaded (dashed line). The total area under the loading curve (A + B) is the work done stretching it, about 1.5 J. Only triangle B, ½ × 0.20 × 10 = 1.0 J, comes back on unloading. Area A, about 0.5 J, does not.

Count the energy. Stretching the spring took about 1.5 J, the whole area under the loading curve. On unloading, the spring pushed back only along the dashed line. The energy it gave back is the triangle under that line: its base is $26.4 - 6.4 = 20$ cm and its height is 10 N, so $\frac{1}{2} \times 0.20 \times 10 = 1.0$ J. About 0.5 J, area A, never came back. It was used to rearrange the metal permanently, and it ended up as thermal energy: the wire became slightly warmer. Energy was not destroyed. It was transferred to a store you cannot get back out as motion.

So "the stored energy can be fully recovered" is true only up to the elastic limit. Beyond it, part of the work is always dissipated.

Worked examples

Worked Example 1 — Energy in a compressed spring

A spring with $k = 400$ N/m is compressed by 8 cm. How much elastic potential energy does it store?

Convert first: $x = 8$ cm $= 0.08$ m. Then $E_{e} = \frac{1}{2} \times 400 \times 0.0 8^{2} = 200 \times 0.0064 = 1.28$ J.

Check with the other form. The force at 8 cm is $F = 400 \times 0.08 = 32$ N, so $\frac{1}{2} F x = \frac{1}{2} \times 32 \times 0.08 = 1.28$ J. ✓

Worked Example 2 — When you know the force, not k

A force of 20 N stretches a spring by 5 cm, within its limit of proportionality. How much energy is stored?

You could find $k = 20 \div 0.05 = 400$ N/m first, then use $\frac{1}{2} k x^{2}$ . But the triangle gives it in one step: $E_{e} = \frac{1}{2} F x = \frac{1}{2} \times 20 \times 0.05 = 0.5$ J.

Check: $\frac{1}{2} \times 400 \times 0.0 5^{2} = 200 \times 0.0025 = 0.5$ J. ✓ Writing $F x = 1$ J would be the rectangle, not the triangle. It is wrong by a factor of 2, because the force was not 20 N for the whole stretch.

Worked Example 3 — Launch speed

A toy launcher has a spring with $k = 250$ N/m, compressed by 6 cm. It launches a 0.05 kg ball along a smooth, level track. Find the ball's speed as it leaves the spring.

Energy stored: $E_{e} = \frac{1}{2} \times 250 \times 0.0 6^{2} = 125 \times 0.0036 = 0.45$ J.

All of it becomes kinetic energy: $\frac{1}{2} \times 0.05 \times v^{2} = 0.45$ . So $0.025 v^{2} = 0.45$ , $v^{2} = 18$ , and $v = 18 \approx 4.24$ m/s.

Check: $\frac{1}{2} \times 0.05 \times 4.2 4^{2} \approx 0.45$ J. ✓ The answer is an upper limit: a real launcher loses a little energy to friction.

Worked Example 4 — Designing a launcher

You want a launcher to give a 20 g ball a speed of 10 m/s, using a compression of 10 cm. What spring constant do you need?

Work backwards from the energy. The ball needs $E_{k} = \frac{1}{2} \times 0.020 \times 1 0^{2} = 1$ J. The spring must store 1 J: $\frac{1}{2} k \times 0.1 0^{2} = 1$ , so $0.005 k = 1$ and $k = 200$ N/m.

Check: $\frac{1}{2} \times 200 \times 0.01 = 1$ J ✓. If you allow for some energy lost to friction, choose a slightly stiffer spring.

Three classic mistakes

Trap 1 — Leaving x in centimetres

The error is squared. With $k = 400$ N/m and $x = 8$ written as 8 instead of 0.08, you get $\frac{1}{2} \times 400 \times 64 = 12800$ J instead of 1.28 J, ten thousand times too big. Convert to metres before you square.

Trap 2 — Using Fx without the ½

$F x$ is the work for a force that is $F$ all the way. The spring's force grows from zero, so the work is the triangle, $\frac{1}{2} F x$ . Leaving out the ½ doubles the answer.

Trap 3 — Using ½kx² beyond the limit of proportionality

The formula assumes a straight-line graph. Past the limit of proportionality, find the energy from the area under the real curve. Past the elastic limit, remember that not all of it comes back.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A spring with $k = 120$ N/m is stretched by 15 cm. How much energy does it store?

Hint 1
Change 15 cm into metres first.

Hint 2
$E_{e} = \frac{1}{2} \times 120 \times 0.1 5^{2}$ .

Full answer
$0.1 5^{2} = 0.0225$ , so $E_{e} = 60 \times 0.0225 = 1.35$ J. Check: the force at 15 cm is $120 \times 0.15 = 18$ N, and $\frac{1}{2} \times 18 \times 0.15 = 1.35$ J ✓.
A spring stores 0.2 J when stretched by 4 cm. How much does it store when stretched by 12 cm (still within its limit of proportionality)? Find $k$ too.

Hint 1
The extension is multiplied by 3. What happens to $x^{2}$ ?

Hint 2
For $k$ : $0.2 = \frac{1}{2} k \times 0.0 4^{2}$ .

Full answer
Three times the extension gives $3^{2} = 9$ times the energy: $9 \times 0.2 = 1.8$ J. For $k$ : $0.2 = \frac{1}{2} k \times 0.0016 = 0.0008 k$ , so $k = 250$ N/m. Check: $\frac{1}{2} \times 250 \times 0.1 2^{2} = 125 \times 0.0144 = 1.8$ J ✓.
The launcher from Worked Example 3 ( $k = 250$ N/m, compressed 6 cm) fires the 0.05 kg ball straight up. Ignoring air resistance, how high above its release point does the ball rise? Take $g = 9.8$ N/kg.

Hint 1
At the top, the ball has stopped. Which store has the energy now?

Hint 2
$0.45 = m g Δ h = 0.05 \times 9.8 \times Δ h$ .

Full answer
Elastic energy becomes kinetic, then gravitational potential energy: $0.45 = 0.05 \times 9.8 \times Δ h = 0.49Δ h$ , so $Δ h \approx 0.92$ m. Check: $0.05 \times 9.8 \times 0.918 \approx 0.45$ J ✓. We ignored the small rise of the ball while the spring was still pushing, and any energy lost to friction, so the real height is a little less.
From the last lecture: two springs, each with $k = 60$ N/m, carry a 6 N load, first in series and then in parallel. Find the energy stored in each arrangement. Which stores more, and why, when the stiffer arrangement has the larger $k$ ?

Hint 1
Series: $k = 30$ N/m, extension 0.20 m. Parallel: $k = 120$ N/m, extension 0.05 m.

Hint 2
Use $E_{e} = \frac{1}{2} F x$ with the same $F = 6$ N. What is different?

Full answer
Series: $\frac{1}{2} \times 6 \times 0.20 = 0.6$ J. Parallel: $\frac{1}{2} \times 6 \times 0.05 = 0.15$ J. The series pair stores four times as much. With the same force, the softer arrangement stretches further, so the triangle is wider. Combine $F = k x$ with $E_{e} = \frac{1}{2} F x$ to get $E_{e} = \frac{F ^{2}}{2 k}$ : for a fixed force, a smaller $k$ stores more. But for a fixed extension, $E_{e} = \frac{1}{2} k x^{2}$ , and a larger $k$ stores more. Which spring "stores more" depends on whether you fix the force or the stretch. Check: $36 \div 60 = 0.6$ and $36 \div 240 = 0.15$ ✓.

The bigger idea

The most useful thing in this lecture is not $\frac{1}{2} k x^{2}$ . It is the method behind it: when a force changes, the work it does is the area under its graph. That method works for any shape of graph, and it will come back many times. The next lecture uses it for whole materials, where the area under a stress–strain graph gives the energy stored in each cubic metre of a stretched wire.

The ½ also comes back. Whenever one quantity grows in proportion to another, starting from zero, the area under the graph is a triangle, and a ½ appears. Later you will meet the energy stored in a capacitor, where the charge grows in proportion to the potential difference. It is a ½ for exactly the same reason.

And the launcher showed energy moving from the spring to the ball. Attach the ball to the spring instead, and the energy moves back and forth: spring, ball, spring, ball. That is a vibration. Its study, simple harmonic motion, rests on the two ideas of these lectures: a force proportional to the extension, and the energy $\frac{1}{2} k x^{2}$ .

Key concepts in this lesson

The work done to stretch or compress a spring is stored in it as elastic potential energy.
For a changing force, the work done is the area under the force–extension graph.
Up to the limit of proportionality the area is a triangle, so $E_{e} = \frac{1}{2} F x = \frac{1}{2} k x^{2}$ , with $x$ in metres.
Because the energy depends on $x^{2}$ , doubling the extension stores four times the energy.
When a spring releases a mass with no losses, $\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2}$ , so the launch speed is proportional to the compression.
Beyond the elastic limit, only part of the work comes back on unloading, and the rest is dissipated as thermal energy.