Elastic Strain Energy and Force-Extension Graphs · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Banking energy in a stretch

Science · Matter & Materials · ages 17-18
Name ______________________   Date ____________
  1. A sample's force-extension curves enclose a loop during a repeatable closed loading-unloading cycle. What does the loop area represent?

    • Work put in during the loading stage
    • Work dissipated over the closed cycle
    • Work returned during the unloading stage
  2. An ideal Hooke's law spring has a straight force-extension graph from the origin to 10 N at an extension of 0.2 m. How much energy is stored, in joules?

    Answer: ______________

  3. During slow loading, area under a graph of applied stretching force against extension equals work done on the sample.

    Circle one:   True   False

  4. An ideal Hooke's law spring at final force F and extension x transfers all its energy to mass m, initially at rest. There is no height change or energy loss. Which equation determines speed v?

    • half F x = half m v²
    • half F x = half m v³
    • half F x = half m v
  5. In a repeatable closed loading-unloading cycle, 6 J of work is put into a sample and 4 J is returned. How much energy is dissipated, in joules?

    Answer: ______________

  6. A metal wire is modelled as ideal elastic with negligible energy dissipation. It is loaded and fully unloaded within its elastic range. What happens?

    • It regains its length but returns less input work.
    • It keeps a permanent extension while returning all input work.
    • It regains its original length and returns the input work.
  7. An ideal spring has a straight force-extension line from the origin to final force F and extension x. Lee calculates stored energy as F x. What is his error?

    • He calculated the slope instead of the area under the line.
    • He used the original length instead of the spring's extension.
    • He used the surrounding rectangle's area instead of the triangle's area.
  8. A spring transfers all 1 J of its stored energy to a 0.5 kg ball initially at rest. There is no height change or energy loss. What is the ball's final speed, in m/s?

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Banking energy in a stretch W1-mt_4WnFUmR13O-s1

  1. Work dissipated over the closed cycle · The loop area is loading work minus returned work: energy dissipated during the cycle.
  2. 1 · The triangle area is half × 10 × 0.2 = 1 J.
  3. True · With applied force vertically and extension horizontally, loading area gives input work. It need not all remain recoverable.
  4. half F x = half m v² · The spring stores half F x. Under the stated conditions, this becomes kinetic energy half m v².
  5. 2 · Dissipated energy = input work minus returned work = 6 - 4 = 2 J.
  6. It regains its original length and returns the input work. · Elastic recovery restores length. The separate negligible-dissipation assumption means the input work is returned.
  7. He used the surrounding rectangle's area instead of the triangle's area. · F x gives the rectangle area. The required triangle has half that area, so Lee's result is twice the stored energy.
  8. 2 · v² = 2E/m = 2 × 1 / 0.5 = 4. The positive square root gives speed 2 m/s.
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