Maths › Further Mechanics 1 › Elastic potential energy
Elastic potential energy
A stretched string holds energy, and the amount is the area under the Hooke's law line. Adding that term to the energy equation extends the work-energy principle to every spring problem on the paper.
Builds on Hooke's law and elastic strings and Work, energy and power.
IN THIS TOPIC
- Derive and use the formula for the energy stored in a stretched string.
- Include elastic energy in a work-energy equation.
- Solve problems where a particle is projected by a spring or oscillates on a string.
WHAT YOU PROBABLY THINK
The elastic energy stored is the tension times the extension.
The area under the line
Since the tension grows from zero to λx/l as the string stretches, the work done in stretching it is the area under the straight Hooke's law graph: a triangle of base x and height λx/l.
The factor of a half is what the opening claim leaves out. Tension times extension would be the work done by a constant force of the final size, and the force was not constant: it started at nothing. The energy is also proportional to x², so doubling the stretch stores four times as much.
WORKED EXAMPLE
Energy in a stretched string
An elastic string of natural length 1.5 m and modulus 60 N is stretched to 2 m. Find the energy stored.
Extension = 0.5 m, and the tension at that extension is 20 N.
EPE = 60 × 0.25/(2 × 1.5) = 15/3 = 5 J.
Checking by area: ½ × 0.5 × 20 = 5 J, which is the same triangle read a different way.
Adding it to the energy equation
With elastic energy in hand, the work-energy principle covers everything on this paper: kinetic energy, gravitational potential energy, elastic energy, and work done against friction. Write down the total at one instant, the total at another, and set the difference equal to any energy lost to friction.
The commonest mistakes are geometric rather than physical: using the total length where the extension is wanted, forgetting that a string goes slack once its ends are closer than the natural length, and measuring heights from different levels in the two totals. Draw the two instants and label a single zero level for height.
YOUR TURN
A spring launcher
A particle of mass 0.5 kg is held against a spring of natural length 0.8 m and modulus 40 N, compressed by 0.2 m, on a smooth horizontal table. Find its speed when the spring reaches its natural length.
Show the working
Energy stored = 40 × 0.2²/(2 × 0.8) = 1.6/1.6 = 1 J.
The table is smooth and horizontal, so no energy is lost and no height changes.
All 1 J becomes kinetic: ½(0.5)v² = 1, so v² = 4 and v = 2 m/s.
Doubling the compression would store 4 J and double the speed, since the energy goes as the square of the compression and the speed as the square root of the energy.
THE EXAM BIT
- Include the factor of a half; tension times extension is twice the energy.
- Use the extension, not the length, and square it.
- Choose one zero level for height and use it in both energy totals.
- Check whether a string is slack at either instant; a slack string stores nothing.
CHECK YOURSELF
An elastic string of natural length 2 m and modulus 49 N is stretched by 0.5 m. Find the energy stored.
Show a hint
Square the extension and halve.
Show the answer
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The energy stored is λx²/2l, the area of the triangle under the Hooke's law line, so it grows with the square of the extension.
Add it to kinetic and gravitational potential energy, and set the total change equal to any work done against friction.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Derive and use the formula for the energy stored in a stretched string.
- Include elastic energy in a work-energy equation.
- Solve problems where a particle is projected by a spring or oscillates on a string.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.