Leaving Certificate Physics

5.3 Elastic potential energy — Practise

Work independently. The questions and answers are shown together at the end.

Write x2x^2 out in full before you substitute. The extension is squared, so a small arithmetic slip changes the answer a great deal.

  1. A spring of spring constant 200 N/m is stretched by 0.10 m. Calculate the energy stored in it.
  2. A spring is stretched by 0.20 m and stores 4.0 J. Calculate the energy stored when it is stretched by 0.40 m.
Answers and concise solutions

1. Energy stored

Answer: E=12kx2=0.5×200×0.102=1.0 JE = \tfrac{1}{2}kx^2 = 0.5 \times 200 \times 0.10^2 = 1.0\,\text{J}.

2. Energy stored at double the extension

Answer: 16 J16\,\text{J} — four times as much, not twice. The extension is squared in 12kx2\tfrac{1}{2}kx^2, so doubling xx multiplies the energy by 22=42^2 = 4.

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