Chapter 2 of 8

All 8 chapters
  1. 1Calculating energy stores, readunderstood
  2. 2Springs and elastic energy, readunderstood
  3. 3Work done, readunderstood
  4. 4Power, readunderstood
  5. 5Efficiency, readunderstood
  6. 6Specific heat capacity, readunderstood
  7. 7Energy resources, readunderstood
  8. 8Energy at home, readunderstood

S4.9.2 Chapter 2

Springs and elastic energy

Within its limit, a spring stretches 2 cm under a 1 N weight. How far will a 3 N weight stretch it?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

  • You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Use Hooke's law to calculate the extension of a spring
  • Calculate the energy stored in a stretched spring

A weight hung on a spring makes it longer. Keep your guess in mind as you read, and see how far a spring stretches and how much energy it keeps.

A spring has a natural length. That is its length with nothing pulling on it. Hang a load on it and it grows longer. The extension is how much longer: extension = length now − natural length.

See it three ways

As a picture

A spring hung from a bar with a load of 4 N on its end. A line beside it is labelled natural length 10 cm. A second line below that is labelled extension 4 cm. A longer line to the right is labelled length 14 cm.4 Nnatural length10 cmextension4 cmlength14 cm

In words

A spring of natural length 10 cm is 14 cm long with a 4 N load. Its extension is 14 − 10 = 4 cm. Twice the load gives twice the extension.

In symbols

e = length now − natural length. F = k × e.

Robert Hooke found that the extension of a spring is proportional to the force on it. Double the force and the extension doubles. This is Hooke's law.

force = spring constant × extension. In symbols, F = k × e. The force F is in N. The extension e is in m. The spring constant k is in N/m. A stiff spring has a large k.

A table of spring S with the columns Load, Length and Extension. A load of 0 N gives a length of 10 cm and an extension of 0 cm. 2 N gives 12 cm and 2 cm. 4 N gives 14 cm and 4 cm. 6 N gives 16 cm and 6 cm. 8 N gives 18 cm and 8 cm.LoadLengthExtension0 N10 cm0 cm2 N12 cm2 cm4 N14 cm4 cm6 N16 cm6 cm8 N18 cm8 cm
Spring S has a natural length of 10 cm. Each 2 N of load adds 2 cm.

Mind the units. The sum needs the extension in metres. A stretch of 4 cm is 4 ÷ 100 = 0.04 m. Forgetting to change cm to m is the commonest slip in this chapter.

Worked example

Spring S has a natural length of 10 cm. A load of 4 N makes it 14 cm long. Find its spring constant k in N/m.

  1. The extension is 14 − 10 = 4 cm = 0.04 m.
  2. k = F ÷ e = 4 ÷ 0.04 = 100 N/m.

Answer: 100 N/m

Worked example

A spring has k = 250 N/m. A load of 15 N hangs on it. The spring stays within its limit. What is its extension in cm?

  1. F = k × e, so e = F ÷ k.
  2. e = 15 ÷ 250 = 0.06 m.
  3. e = 0.06 × 100 = 6 cm.

Answer: 6 cm

Hooke's law has a limit. If you pull a spring too far, the extension grows faster than the force. The line on a graph of force against extension stops being straight. The point where it bends is the limit of proportionality.

A graph of force in N against extension in cm for spring S. The axes are labelled extension (cm), from 0 to 12 in steps of 2, and force (N), from 0 to 12 in steps of 2. A straight line labelled spring S rises from 0 to a dot labelled limit at 8 cm and 8 N. After the dot the line bends and rises more slowly.limitspring S24681012246810120extension (cm)force (N)
The line is straight up to the limit, then it bends. Past the limit, F = k × e does not work.

Past the limit of proportionality a spring may also stay stretched when you let go. It does not return to its natural length. So keep loads small when you test a spring. The point past which a spring stays stretched is the elastic limit. It is at or a little past the limit of proportionality.

Stretching a spring puts energy into its elastic store. The force is 0 at the start and F at the end. The energy stored is the area under the line on the graph. That area is a triangle: ½ × base × height = ½ × e × F.

A graph of force in N against extension in cm for spring S, a straight line labelled spring S from 0 to 8 cm and 8 N. Dashed lines from a dot at 6 cm and 6 N run to both axes, and the triangle between the line, the 6 cm guide and the extension axis is hatched to show the energy stored. The axes are labelled extension (cm), from 0 to 8 in steps of 2, and force (N), from 0 to 8 in steps of 2.spring S246824680extension (cm)force (N)
At 6 cm the force is 6 N. The hatched triangle under the line from 0 to 6 cm has an area equal to the stored energy.

Put F = k × e into ½ × e × F. The energy stored is ½ × k × e². This is the elastic potential energy, written Ee. It works only up to the limit of proportionality, where the line is straight.

Note the square. The extension is squared. Double the extension and the energy is four times as big.

Worked example

Spring S has k = 100 N/m. It is stretched by 6 cm, which is within its limit. How much energy is stored in it?

  1. Ee = ½ × k × e².
  2. e = 6 cm = 0.06 m.
  3. e² = 0.06 × 0.06 = 0.0036.
  4. Ee = ½ × 100 × 0.0036 = 0.18 J.

Answer: 0.18 J

A wrong idea many people have

Stretch a spring twice as far and it stores twice the energy.

Why it feels right: Twice the stretch needs twice the force, so it seems the energy should double as well.

The test that breaks it: Take k = 100 N/m. A stretch of 2 cm stores ½ × 100 × 0.02 × 0.02 = 0.02 J. A stretch of 4 cm stores ½ × 100 × 0.04 × 0.04 = 0.08 J. That is four times as much.

What is true: Both the force and the distance double, so the energy goes up four times. The extension is squared in ½ × k × e².

A wrong idea many people have

F = k × e holds for any stretch of a spring, however far.

Why it feels right: A formula in a book looks like a law of nature that always holds.

The test that breaks it: Look at the graph. The line is straight only up to the limit of proportionality. After that it bends, so the force is no longer in proportion to the extension.

What is true: F = k × e and Ee = ½ × k × e² work only up to the limit of proportionality. Beyond it a spring may be damaged.

Try it

Test a spring with water

You need: a spring from an old pen, string, a small plastic cup, water, a teaspoon, a ruler with mm marks, a pencil laid across two cups, to hang it from.

Be safe: Never let go of a stretched spring: it can snap back into your face. Do not stretch the spring far. Keep your face away from it.

  1. Hang the spring from the pencil. Tie the cup to its lower end with string.
  2. With the empty cup hanging, measure the length of the spring in mm. Call it the starting length. The cup already stretches the spring a little, so the extensions you find are for the water alone.
  3. Add 10 level teaspoons of water, about 50 mL. This is a load of about 0.5 N. Measure the length again.
  4. Add 50 mL twice more, up to 150 mL, and measure each time.
  5. For each load, work out the extension: length now minus the starting length.

What you will see: The extension doubles when the water doubles, and triples when the water triples. The stretch is proportional to the load. Your numbers will differ with your spring, but the pattern is the same. With much more water the spring may stretch faster than the load grows. That is the limit of proportionality.

If you cannot try it, this is what happens.
A table of example results with the columns Water, Length and Extension. 0 mL gives 25 mm and 0 mm. 50 mL gives 30 mm and 5 mm. 100 mL gives 35 mm and 10 mm. 150 mL gives 40 mm and 15 mm.WaterLengthExtension0 mL25 mm0 mm50 mL30 mm5 mm100 mL35 mm10 mm150 mL40 mm15 mm

Key words

natural length
the length of a spring with nothing pulling on it.
extension
how much longer a spring is than its natural length. Change it to m before using F = k × e.
Hooke's law
the extension of a spring is proportional to the force, F = k × e, up to the limit of proportionality.
spring constant
the force for each metre of extension, k in N/m. A stiff spring has a large k.
limit of proportionality
the point where the force extension line stops being straight.
elastic potential energy
the energy in a stretched spring, Ee = ½ × k × e², up to the limit of proportionality.

Back to your guess

The answer is D: 6 cm

Within the limit of the spring, the stretch is proportional to the force. Three times the force gives three times the stretch: 3 × 2 = 6 cm.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A spring has a natural length of 12 cm. A load makes it 15 cm long. What is the extension?

    A spring hung from a bar with a load of 3 N on its end. A line beside it is labelled natural length 12 cm. A longer line to the right is labelled length 15 cm.3 Nnatural length12 cmlength15 cm
    Tap your answer.

    1. Not this one. That is the length of the spring now. The extension is only the part longer than the natural length. Try another.

    2. Not this one. You added the two lengths. The extension is the difference between them, so take away. Try another.

    3. Yes, that is right. extension = length now − natural length = 15 − 12 = 3 cm.

    4. Not this one. That is the natural length. The extension is how much longer the spring is than that. Try another.

    See the answer

    3 cm

    extension = length now − natural length = 15 − 12 = 3 cm.

    • If you chose A: That is the length of the spring now. The extension is only the part longer than the natural length.
    • If you chose B: You added the two lengths. The extension is the difference between them, so take away.
    • If you chose D: That is the natural length. The extension is how much longer the spring is than that.

  2. A spring obeys Hooke's law. A force of 2 N stretches it by 3 cm. What extension would a force of 6 N give?

    Tap your answer.

    1. Yes, that is right. The force is 6 ÷ 2 = 3 times as big, so the extension is 3 times as big: 3 × 3 = 9 cm.

    2. Not this one. You added the extra force to the stretch. The stretch is multiplied: 6 N is three times 2 N. Try another.

    3. Not this one. You multiplied 2 by 6. The force went up three times, so multiply the 3 cm by 3. Try another.

    4. Not this one. You multiplied 3 cm by 6. The force is three times as big, not six times. Try another.

    See the answer

    9 cm

    The force is 6 ÷ 2 = 3 times as big, so the extension is 3 times as big: 3 × 3 = 9 cm.

    • If you chose B: You added the extra force to the stretch. The stretch is multiplied: 6 N is three times 2 N.
    • If you chose C: You multiplied 2 by 6. The force went up three times, so multiply the 3 cm by 3.
    • If you chose D: You multiplied 3 cm by 6. The force is three times as big, not six times.

  3. A force of 10 N stretches a spring by 5 cm. What is its spring constant?

    Tap your answer.

    1. Not this one. You multiplied 10 by 5. The spring constant is the force divided by the extension. Try another.

    2. Yes, that is right. e = 5 cm = 0.05 m. k = F ÷ e = 10 ÷ 0.05 = 200 N/m.

    3. Not this one. You divided 10 by 5 but kept the 5 in cm. Change 5 cm to 0.05 m first. Go to M2.6.1. Try another.

    4. Not this one. You divided the extension by the force, which is upside down. Use force ÷ extension. Try another.

    See the answer

    200 N/m

    e = 5 cm = 0.05 m. k = F ÷ e = 10 ÷ 0.05 = 200 N/m.

    • If you chose A: You multiplied 10 by 5. The spring constant is the force divided by the extension.
    • If you chose C: You divided 10 by 5 but kept the 5 in cm. Change 5 cm to 0.05 m first. Go to M2.6.1.
    • If you chose D: You divided the extension by the force, which is upside down. Use force ÷ extension.

  4. A spring with k = 300 N/m is stretched by 0.1 m, within its limit. How much energy is stored in it?

    Tap your answer.

    1. Not this one. You left out the half. Ee = ½ × k × e². Try another.

    2. Not this one. You did not square the extension. Ee = ½ × k × e × e. Try another.

    3. Not this one. You multiplied k by e. That is the force in newtons, not the energy. Try another.

    4. Yes, that is right. Ee = ½ × k × e² = ½ × 300 × 0.1 × 0.1 = ½ × 300 × 0.01 = 1.5 J.

    See the answer

    1.5 J

    Ee = ½ × k × e² = ½ × 300 × 0.1 × 0.1 = ½ × 300 × 0.01 = 1.5 J.

    • If you chose A: You left out the half. Ee = ½ × k × e².
    • If you chose B: You did not square the extension. Ee = ½ × k × e × e.
    • If you chose C: You multiplied k by e. That is the force in newtons, not the energy.

  5. The graph shows a spring. What does the limit of proportionality mean?

    A graph of force in N against extension in cm for spring T. The axes are labelled extension (cm), from 0 to 10 in steps of 2, and force (N), from 0 to 20 in steps of 4. A straight line labelled spring T rises to a dot labelled limit and then bends.limitspring T246810481216200extension (cm)force (N)
    Tap your answer.

    1. Yes, that is right. Up to the limit, force and extension are proportional. After it the extension grows faster than the force, and the line bends.

    2. Not this one. The spring does not need to break there. It is only where the line stops being straight. Try another.

    3. Not this one. The natural length is the length with no load. It is the point where the extension is 0. Try another.

    4. Not this one. A spring can carry more force after this point. It just no longer follows the straight line. Try another.

    See the answer

    The point after which the line is no longer straight, so F = k × e stops working.

    Up to the limit, force and extension are proportional. After it the extension grows faster than the force, and the line bends.

    • If you chose B: The spring does not need to break there. It is only where the line stops being straight.
    • If you chose C: The natural length is the length with no load. It is the point where the extension is 0.
    • If you chose D: A spring can carry more force after this point. It just no longer follows the straight line.

  6. The graph shows two springs. Which is stiffer?

    A graph of force in N against extension in cm. The axes are labelled extension (cm), from 0 to 6 in steps of 1, and force (N), from 0 to 24 in steps of 4. Two straight lines start at 0. The steeper line is labelled spring A and the other is labelled spring B.spring Aspring B12345648121620240extension (cm)force (N)
    Tap your answer.

    1. Not this one. The flatter line stretches further for the same force. That is the softer spring. Try another.

    2. Not this one. Stretching further for the same force means a softer spring, not a stiffer one. Try another.

    3. Yes, that is right. The slope of the line is the force for each cm of extension. So a steeper line is a larger spring constant. That means a stiffer spring.

    4. Not this one. Both obey Hooke's law, but the slopes differ. The slope is the spring constant. Try another.

    See the answer

    Spring A, because it needs more force for the same extension.

    The slope of the line is the force for each cm of extension. So a steeper line is a larger spring constant. That means a stiffer spring.

    • If you chose A: The flatter line stretches further for the same force. That is the softer spring.
    • If you chose B: Stretching further for the same force means a softer spring, not a stiffer one.
    • If you chose D: Both obey Hooke's law, but the slopes differ. The slope is the spring constant.

  7. From an earlier chapter

    A ball is dropped from 2.5 m. Ignore air resistance. Use g = 9.8 N/kg. What is its speed just before it lands?

    Tap your answer.

    1. Yes, that is right. v² = 2 × g × h = 2 × 9.8 × 2.5 = 49, and v = √49 = 7 m/s.

    2. Not this one. You stopped at v² = 2 × g × h = 49. The speed is the square root of that. Go to S4.9.1. Try another.

    3. Not this one. You multiplied g by h and stopped. Use v² = 2 × g × h, then take the square root. Go to S4.9.1. Try another.

    See the answer

    7 m/s

    v² = 2 × g × h = 2 × 9.8 × 2.5 = 49, and v = √49 = 7 m/s.

    • If you chose B: You stopped at v² = 2 × g × h = 49. The speed is the square root of that. Go to S4.9.1.
    • If you chose C: You multiplied g by h and stopped. Use v² = 2 × g × h, then take the square root. Go to S4.9.1.

  8. A force of 12 N stretches a spring by 3 cm. The spring stays within its limit. What is its spring constant in N/m?

    See the answer

    400 N/m

    e = 3 cm = 0.03 m. k = F ÷ e = 12 ÷ 0.03 = 400 N/m.

  9. A spring has k = 500 N/m and is stretched by 4 cm, within its limit. How much energy is stored in it, in joules?

    See the answer

    0.4 J

    e = 4 cm = 0.04 m. Ee = ½ × 500 × 0.04 × 0.04 = ½ × 500 × 0.0016 = 0.4 J.

  10. A spring has k = 400 N/m. A load of 10 N hangs on it, within its limit. What is its extension in cm?

    See the answer

    2.5 cm

    e = F ÷ k = 10 ÷ 400 = 0.025 m = 2.5 cm.

Keep it

Copy this into your exercise book.

Extension e = length now − natural length. F = k × e, with e in m, up to the limit. Ee = ½ × k × e². Double e, four times the energy.

Draw: A spring with its natural length and extension marked, and a force extension line that bends after the limit.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (758 KB). It opens on any phone.