Leaving Certificate Physics

5.1 Hooke's law — Practise

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

Give every answer with its unit, and to the number of significant figures the data justify. Before you use F=kxF = kx, check that the spring is within its limit of proportionality.

  1. A spring has a spring constant of 40 N/m. Calculate the force needed to stretch it by 0.15 m.
  2. A force of 12 N stretches a spring by 0.04 m. Calculate its spring constant.
  3. A spring of spring constant 250 N/m is pulled with a force of 8.0 N. Calculate its extension.
  4. State what is meant by the limit of proportionality of a spring, and describe what is observed beyond it.
Answers and concise solutions

1. Force needed

Answer: F=kx=40×0.15=6 NF = kx = 40 \times 0.15 = 6\,\text{N}.

2. Spring constant

Answer: k=Fx=120.04=300 N/mk = \dfrac{F}{x} = \dfrac{12}{0.04} = 300\,\text{N/m}.

3. Extension

Answer: x=Fk=8.0250=0.032 mx = \dfrac{F}{k} = \dfrac{8.0}{250} = 0.032\,\text{m}, which is 32 mm.

4. Limit of proportionality

  • the point beyond which the force-extension graph is no longer a straight line through the origin
  • below it, force is directly proportional to extension, so doubling the force doubles the extension
  • beyond it the graph curves, and each extra newton gives a larger extension than the one before

Answer: The limit of proportionality is the point beyond which the graph of force against extension is no longer a straight line through the origin. Below it, force is directly proportional to extension, so doubling the force doubles the extension. Beyond it the graph curves, and each extra newton produces a larger extension than the one before.

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