AQA · GCSE · Combined Science

    Course version: 8464 · Version from Sept 2018

    Changes in energy: The amount of elastic potential energy stored in a stretched spring can be calculated using the equation:

    When a spring is stretched, work is done against the spring's stiffness and energy is transferred to the elastic potential store of the spring.

    • 2 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    When a spring is stretched, work is done against the spring's stiffness and energy is transferred to the elastic potential store of the spring. The amount stored depends on how stiff the spring is and how far it has been stretched from its natural length. The equation that links these quantities is Ee = 0.5 × k × e², where Ee is elastic potential energy in joules (J), k is the spring constant in newtons per metre (N/m) and e is the extension in metres (m). Extension means the increase in length beyond the unstretched length, not the total length.

    For example: A spring with k = 200 N/m stretched by 0.10 m stores 0.5 × 200 × (0.10)² = 1.0 J. The equation applies provided the spring has not been stretched beyond its limit of proportionality, so the force–extension graph remains a straight line through the origin.

    What earns the marks

    • State the equation as Ee = 0.5 × k × e² and identify each symbol with its unit: Ee in joules (J), k in newtons per metre (N/m), e in metres (m).
    • Explain that extension e is the change in length from the spring's natural length, so e = stretched length − original length, not the total stretched length.
    • Substitute values correctly, including squaring the extension before multiplying by 0.5 and by the spring constant.
    • Recognise that the equation applies only while the spring obeys Hooke's law, that is up to the limit of proportionality.
    • Interpret the factor 0.5 as arising because the force increases from zero to k × e as the spring stretches, so the average force is half the final force.

    Common mistakes and how to put them right

    • Using the total stretched length instead of the extension.

      Put it right: Subtract the original length first so e is the increase in length.

    • Forgetting to square the extension or squaring the whole expression.

      Put it right: Square only e, then multiply by 0.5 and k.

    • Mixing units, such as using centimetres for extension or grams for mass.

      Put it right: Convert all lengths to metres and use k in N/m before calculating.

    Exam tips

    • Write the equation, then substitute values with units before evaluating, so method marks are visible even if the final arithmetic slips.
    • Check whether the question gives total length or extension; if it gives total length, calculate the extension first and show that step.
    • Give the final answer in joules with an appropriate number of significant figures, and include the unit.

    Try the lesson

    The start of the Changes in energy: The amount of elastic potential energy stored in a stretched spring can be calculated using the equation: lesson, made from checked teaching notes for this specification point.

    Have a go first

    A spring has a spring constant of 200 N/m and is stretched by 0.3 m. Using Ee = ½ke², what is the elastic potential energy stored in the spring?

    Listen

    When a spring is stretched, work is done against the spring's stiffness and energy is transferred to the elastic potential store of the spring. The amount stored depends on how stiff the spring is and how far it has been stretched from its natural length. The equation that links these quantities is Ee = 0.5 × k × e², where Ee is elastic potential energy in joules (J), k is the spring constant in newtons per metre (N/m) and e is the extension in metres (m). Extension means the increase in length beyond the unstretched length, not the total length. For example, a spring with k = 200 N/m stretched by 0.10 m stores 0.5 × 200 × (0.10)² = 1.0 J. The equation applies provided the spring has not been stretched beyond its limit of proportionality, so the force–extension graph remains a straight line through the origin. A strong answer earns marks for these points. State the equation as Ee = 0.5 × k × e² and identify each symbol with its unit: Ee in joules (J), k in newtons per metre (N/m), e in metres (m). Explain that extension e is the change in length from the spring's natural length, so e = stretched length − original length, not the total stretched length. Substitute values correctly, including squaring the extension before multiplying by 0.5 and by the spring constant. Recognise that the equation applies only while the spring obeys Hooke's law, that is up to the limit of proportionality. A common slip: Using the total stretched length instead of the extension. Instead, subtract the original length first so e is the increase in length. Exam tip: Write the equation, then substitute values with units before evaluating, so method marks are visible even if the final arithmetic slips.

    Key words

    Say what each word means before you turn the card.

    Elastic potential energy store

    Try answering aloud or in your head before turning the card.

    Extension

    Try answering aloud or in your head before turning the card.

    See every definition
    Elastic potential energy store
    The store that energy is transferred to when work is done against a spring's stiffness as the spring is stretched.
    Extension
    The increase in length beyond the unstretched length, not the total length.

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