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Springs and Newton’s second law Springs and equilibrium Spring force and elastic energy Springs and graph interpretation Work transfers energy into a spring Work, energy and power3.3 Materials3.4 Dynamics3.2.1 Equilibrium3.2.3 Springs3.4.1 Elastic potential energy Force–extension graphs
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Spring force and elastic energy

OCR A AS Physics · Two ideas, one explanation

The force needed to deform a spring increases during loading. The work done accumulates as elastic potential energy when deformation is elastic.

E = ½Fx = ½kx²

Follow the reasoning

  1. Measure deformation from the natural length.
  2. Identify the spring constant or the force at that deformation.
  3. Use the triangular area under the force–extension graph.
  4. Track where the energy goes when the spring is released.
Specification outcomes (3)
  • 3.4.1(b) State and apply Hooke’s law.
  • 3.4.1(c) Define the force constant k of a spring or wire using F = kx.
  • 3.4.2(b) Calculate elastic potential energy using E = ½Fx or E = ½kx².
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H156 v2.0 · May 2026

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PhysicsUK QWC6 marks

A technician compares two spring bumpers for a horizontal trolley. Spring A has a straight force-extension loading graph through the origin. Spring B has a curved loading graph but returns fully to its original length when unloaded. Both are used within a range where no permanent deformation occurs. Assume there are no internal energy losses within either spring on loading or unloading. A pupil makes two claims: “Elastic always means force is proportional to extension, so 0.5kx² gives the stored energy for both springs” and “The trolley accelerates at a constant rate until it leaves either spring.” Evaluate both claims. Explain how the force-extension graph gives the work done loading each spring, how this relates to energy transferred on release, and the effect of a constant resistive force on the trolley's motion. Assume the spring masses are negligible. [6]

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