Exam Style Question - 1

QuestionA spring obeys Hooke's law. A force of 12 N stretches it by 4.0 cm.(a) Calculate the stiffness k of the spring. (2 marks)
(b) Calculate the elastic strain energy stored when the extension is 4.0 cm. (2 marks)
(c) The spring is stretched further, to an extension of 6.0 cm, still within its limit of proportionality. Calculate the new force and the total energy stored. (3 marks)
(d) State how the energy stored could be found if the spring had been stretched beyond its limit of proportionality. (1 mark)Worked answer(a) Convert the extension to metres: 4.0 cm = 0.040 mk = ΔF ÷ Δx = 12 ÷ 0.040 = 300 N m^-1(b) Energy stored:ΔE_el = ½FΔx = ½ × 12 × 0.040 = 0.24 J(c) New force and energy:F = kΔx = 300 × 0.060 = 18 N
ΔE_el = ½ × 18 × 0.060 = 0.54 JThe extension increased by a factor of 1.5, but the energy increased by a factor of 1.5² = 2.25, because the energy stored depends on the square of the extension.(d) Plot the force-extension graph and find the area under the curve, for example by counting squares.Common mistakes: forgetting the ½ (FΔx gives double the correct energy) and leaving the extension in centimetres, which gives an energy 100 times too large.