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How do work, energy and power relate, and how do elastic strings and springs store energy?

Work done by a force, kinetic and gravitational potential energy, the work-energy principle, power as the rate of working, Hooke's law for elastic strings and springs, and elastic potential energy.

A focused answer to the OCR A-Level Further Mathematics A Mechanics option content on work, energy and power, covering the work done by a force, kinetic and gravitational potential energy, the work-energy principle and conservation of energy, power as force times velocity, Hooke's law for elastic strings and springs, and elastic potential energy.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. Work and energy
  3. The work-energy principle and conservation of energy
  4. Power
  5. Hooke's law and elastic potential energy
  6. Try this

What this dot point is asking

OCR's Mechanics option wants you to calculate the work done by a force, kinetic and gravitational potential energy, apply the work-energy principle and conservation of energy, treat power as the rate of doing work (and as force times velocity), use Hooke's law for the tension in an elastic string or spring, and calculate the elastic potential energy stored.

Work and energy

Work transfers energy. A force doing work on a body changes its energy, and the basic measure is force times the distance moved in the direction of the force.

The work-energy principle and conservation of energy

The work-energy principle links the net work done to the change in kinetic energy, and is often the quickest route to a final speed. When only conservative forces (such as gravity and elastic tension) act, the total mechanical energy is conserved.

Power

Power is how fast work is done. For a force driving a body at speed vv, the instantaneous power is the product of the force and the speed, which is the key relationship for vehicle problems.

Hooke's law and elastic potential energy

An elastic string or spring resists extension with a tension proportional to the extension. Stretching it stores elastic potential energy, found by integrating the tension over the extension.

Work, energy and power provide the energy methods used across the mechanics option, complementing the momentum methods and the analysis of circular motion.

Try this

Q1. Find the kinetic energy of a 44 kg mass moving at 33 m s−1^{-1}. [2 marks]

  • Cue. 12mv2=12(4)(32)=18\tfrac{1}{2}mv^2 = \tfrac{1}{2}(4)(3^2) = 18 J.

Q2. An engine works at 1212 kW. Find the driving force at a speed of 2020 m s−1^{-1}. [2 marks]

  • Cue. F=Pv=1200020=600F = \dfrac{P}{v} = \dfrac{12000}{20} = 600 N.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20196 marksA block of mass 22 kg is pulled 55 m along a rough horizontal floor by a horizontal force of 2020 N against a constant friction force of 88 N. Use the work-energy principle to find the speed gained, starting from rest. (Take the floor as horizontal so gravity does no work.)
Show worked answer →

Work done by the applied force (M1): 20×5=10020 \times 5 = 100 J. Work done against friction (M1): 8×5=408 \times 5 = 40 J.

Net work done on the block (A1): 100−40=60100 - 40 = 60 J.

By the work-energy principle, net work equals the change in kinetic energy (M1): 60=12mv2−0=12(2)v2=v260 = \tfrac{1}{2}mv^2 - 0 = \tfrac{1}{2}(2)v^2 = v^2 (A1).

Solve (A1): v2=60v^2 = 60, so v=60≈7.75v = \sqrt{60} \approx 7.75 m s−1^{-1}.

Markers reward the two work calculations, the net work, equating to the kinetic energy change, and the final speed.

OCR 20226 marksAn elastic string of natural length 0.80.8 m and modulus of elasticity 4040 N is stretched to a length of 1.01.0 m. Find the tension and the elastic potential energy stored.
Show worked answer →

Extension (M1): x=1.0−0.8=0.2x = 1.0 - 0.8 = 0.2 m.

Tension by Hooke's law T=λxlT = \dfrac{\lambda x}{l} (M1): T=40×0.20.8=80.8=10T = \dfrac{40 \times 0.2}{0.8} = \dfrac{8}{0.8} = 10 N (A1).

Elastic potential energy EPE=λx22l\mathrm{EPE} = \dfrac{\lambda x^2}{2l} (M1): =40×0.222×0.8=40×0.041.6=1.61.6=1= \dfrac{40 \times 0.2^2}{2 \times 0.8} = \dfrac{40 \times 0.04}{1.6} = \dfrac{1.6}{1.6} = 1 J (A1, A1).

Markers reward the extension, Hooke's law for the tension, the EPE formula, and the value 11 J.

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