C. Energy and Momentum · Work, Energy and Simple Harmonic Motion

Elastic Potential Energy

C3.1 C2.1 C2.3About 65 minutes

Learning goals

I can derive Ee = ½kx² from the area under a force–stretch graph.
I can calculate the elastic potential energy stored in a spring and the work needed to change its stretch.
I can relate the work done by a spring to the change in its elastic potential energy.
I can use elastic potential energy to find the speed of an object launched by a spring.

Words to know

elastic potential energy
Energy stored in a stretched or compressed elastic object; for an ideal spring Ee = ½kx².

displacement from equilibrium (x)
How far a spring is stretched (+) or compressed (−) from its natural length.

conservative force
A force whose work depends only on the start and end positions; the ideal spring force is one.

parabola
The U-shaped graph of a quadratic relation such as Ee = ½kx².

Stored in the stretch

Pull back a slingshot band, wind up a toy car, or press a pinball plunger. Work goes in, and nothing seems to happen until you let go: then the stored energy comes back out as motion. Energy stored by deforming an elastic object is called elastic potential energy. In this lesson you will find its formula from the area under a Hooke's-law graph.

Elastic potential energy from the area

In Work Done by a Varying Force: The Area Under the Graph you learned that work is the area under a force–displacement graph. To stretch an ideal spring slowly from 0 to x, you apply F = kx, which rises in a straight line from 0 to kx. The area is a triangle:

W = ½ (x)(kx) = ½kx²

Because the spring force is conservative (the work it does depends only on the start and end stretch), this work is stored and can be fully recovered. It is the elastic potential energy:

Ee = ½kx²

x is the stretch or compression from the natural length, in metres.
Ee is in joules and is never negative: x² is positive whether the spring is stretched or compressed.
Ee grows with the square of x: double the stretch, four times the energy.
Going from stretch x₁ to x₂: ΔEe = ½k(x₂² − x₁²). The work done by the spring is W_s = −ΔEe, just as W_g = −ΔEg for gravity.

The area under the F = kx line from 0 to x is a triangle with base x and height kx. Its area, ½kx², is the energy stored in the spring.

Worked example

Example 1: Energy in a compressed spring

A spring with k = 400 N/m is compressed 5.0 cm. How much elastic potential energy is stored?

Convert: x = 5.0 cm = 0.050 m.
Ee = ½kx² = ½(400)(0.050)² = ½(400)(0.0025) = 0.50 J.

0.50 J.

Worked example

Example 2: Stretching further

A spring with k = 300 N/m is already stretched 10 cm. How much work is needed to stretch it to 20 cm?

W = ΔEe = ½k(x₂² − x₁²) = ½(300)(0.20² − 0.10²).
= 150(0.040 − 0.010) = 150(0.030) = 4.5 J.
Compare: stretching from 0 to 10 cm takes only ½(300)(0.10)² = 1.5 J. The second 10 cm takes three times as much work.

4.5 J. It is not ½k(Δx)² = 1.5 J; you must subtract the two energies.

Elastic potential energy against displacement is a parabola. It is zero at the natural length and rises the same way for stretching and compressing; half the stretch stores only a quarter of the energy.

Worked example

Example 3: A toy dart launcher

A toy launcher has a spring with k = 500 N/m. It is compressed 4.0 cm and fires a 20 g foam dart horizontally. Assuming all the stored energy becomes the dart's kinetic energy, how fast does the dart leave?

Ee = ½(500)(0.040)² = 0.40 J.
Ee = Ek: 0.40 = ½(0.020)v².
v² = 40, so v = 6.3 m/s.

About 6.3 m/s. Real launchers are slower, because the spring's own mass and friction take some of the energy.

Worked example

Example 4: A trampoline

A 50 kg student lands on a trampoline and pushes its centre down 0.40 m. Treating the trampoline as an ideal spring with k = 5000 N/m, how much energy is stored at the lowest point? What is the upward force there?

Ee = ½(5000)(0.40)² = ½(5000)(0.16) = 400 J.
F = kx = (5000)(0.40) = 2000 N, about four times the student's weight (490 N).

400 J stored; 2.0 × 10³ N upward. That large upward force is what launches the student back up.

Common traps with Ee

Always measure x from the natural length, not from wherever the spring happens to start.
Square x after converting to metres: 5 cm → 0.05 m → 0.0025 m².
½k(x₂² − x₁²) is not the same as ½k(x₂ − x₁)².
Ee is a property of the spring, stored in its deformation, not of the object attached to it.

Just for fun

Why did the spring apply for a job at the bank?

It had excellent experience storing energy, and it always bounced back after a rough day.

Hands-on activity

Measure the energy in a rubber band shot

You need: a rubber band, a ruler, a luggage scale or spring scale, a small soft target (a paper ball), a tape measure, and an open space. Safety: aim only at the floor or a wall, never at people or pets, and wear glasses if you have them.

Hook the rubber band over the end of the ruler. Pull the free end back with the scale and record the force at stretches of 2, 4, 6 and 8 cm beyond its slack length.
Plot F against x and estimate the stored energy at 8 cm as the area under the graph (a triangle if the graph is straight).
Compare with ½kx², using k from the slope.
Launch the paper ball from the 8 cm stretch, held level 1.0 m above the floor, and measure how far it lands. From projectile motion (sibling unit Dynamics: Forces and Motion in a Plane), the launch speed is v = horizontal distance ÷ fall time, where fall time = √(2 × 1.0 ÷ 9.8) = 0.45 s.
Compare ½mv² of the ball (weigh it on a kitchen scale) with the stored energy.

What to look for: the ball's kinetic energy will be much less than the stored energy, often only a small fraction, because the band itself moves fast and carries energy, and some energy becomes thermal energy in the rubber.

Your turn

Find Ee for a spring with k = 600 N/m stretched 5.0 cm.

A spring with k = 400 N/m stores 2.0 J. How far is it stretched or compressed?

By what factor does Ee change if the stretch is tripled?

A spring with k = 100 N/m is stretched from 0.10 m to 0.30 m. How much work is done?

A spring is compressed 3 cm. Is its elastic potential energy positive or negative?

A spring with k = 50 N/m returns from a stretch of 0.20 m to its natural length. How much work does the spring do on the object attached to it?

A force–stretch graph rises in a straight line to 40 N at 0.25 m. Find the energy stored and k.

More practice: warm-up

Write the formula for elastic potential energy.

Find Ee for k = 200 N/m, x = 0.10 m.

What does the area under a force–stretch graph represent?

If x doubles, Ee is multiplied by:

Why can the energy stored in an ideal spring be fully recovered?

More practice: on your own

A trampoline (k = 5000 N/m) is pushed down 0.30 m. Find the energy stored.

A car spring with k = 3.0 × 10⁴ N/m is compressed 9.8 cm. How much energy does it store?

A spring (k = 300 N/m) compressed 5.0 cm launches a 50 g ball horizontally on a smooth surface. Find the ball's speed.

Which stores more energy: a 100 N/m spring stretched 0.20 m, or a 400 N/m spring stretched 0.10 m?

The same 20 N force stretches a 100 N/m spring and a 400 N/m spring. Which stores more energy?

Challenge

Show that the energy stored in a spring held by a force F is Ee = F²/(2k).

A bungee cord does not obey Hooke's law exactly. Measured forces at stretches of 0, 1, 2 and 3 m are 0, 400, 900 and 1500 N. Estimate the energy stored at a 3 m stretch.

A spring launcher fires a ball horizontally across a smooth table. Show that the launch speed is proportional to the compression x. If x doubles, what happens to the speed and to the kinetic energy?

Check your understanding

Quiz

The elastic potential energy of a spring with k = 200 N/m stretched 0.30 m is:

The formula Ee = ½kx² comes from:

If the compression of a spring is halved, the stored energy is:

A spring with k = 100 N/m is stretched from 0.10 m to 0.20 m. The work done is:

Which statement about elastic potential energy is correct?

A 2.0 J spring launcher fires a 0.10 kg ball on a frictionless surface. The ball's speed is:

For parents and teachers

Big idea (C3.1, C2.1): stretching or compressing a spring stores elastic potential energy, Ee = ½kx², which equals the area under the Hooke's-law graph. Because the spring force is conservative, this energy can be recovered as motion.

How to help: ask your teen why Ee has a ½ in it (the force grows from zero, so the average force is half the final force). Talk about everyday stored-energy devices: wind-up toys, pinball plungers, trampolines, mouse-trap cars.

Common difficulties:

Not converting cm to m before squaring.
Using ½k(x₂ − x₁)² instead of ½k(x₂² − x₁²).
Measuring x from the wrong starting point (it is always measured from the natural length).
Thinking a compressed spring has negative energy.