Further Mathematics · OCR H245 · Option M · Topic 6.02

Work, energy and power

Work, energy, Hooke's law, conservation of energy, power.

  • 5 sub-topics
  • 2 in Year 12, 2 in Year 13
  • 85 practice questions
  • 5 animated lessons
Practice questions (85) Lessons Worksheets coming soon All OCR Further Mathematics topics

What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 6.02a-cWorkY12
  2. 6.02d-fEnergyY12
  3. 6.02g-hHooke's lawY13
  4. 6.02i-jConservation of energyY13
  5. 6.02k-mPowerY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 85 questions in this topic, free, no account needed.

Work, energy and power

Easy

Power · Year 12 · 1 mark

  1. A car's engine works at 45 kW while the car moves at 25 m s⁻¹. What is the driving force?
    (1)

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  1. P=FvP = Fv, with the power in watts.CC: 1800 N1800\ \mathrm{N}

Total1

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Elastic strings and springs and elastic energy

Not needed for Year 12Medium

Hooke's law · Year 13 · 5 marks

A light elastic string has natural length 0.5 m and modulus of elasticity 49 N. One end is fixed at a point OO and a particle of mass 2 kg hangs in equilibrium from the other end.
  1. (a)
    Find the extension of the string.
    (3)
  2. (b)
    Find the elastic potential energy stored in the string.
    (2)

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Part (a)3 marks

  1. Hooke's law.T=λxlT = \frac{\lambda x}{l}
  2. Resolve vertically: T=mgT = mg.49x0.5=2×9.8\frac{49x}{0.5} = 2 \times 9.8
  3. x=0.200 mx = 0.200\text{ m}

Part (b)2 marks

  1. λx22l\frac{\lambda x^2}{2l}
  2. =1.96 J= 1.96\text{ J}

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Work, energy and power

Hard

Work, energy and power · Year 12 · 8 marks

A car of mass 1000 kg moves up a straight road inclined at an angle α\alpha to the horizontal, where sin⁡α=120\sin\alpha = \frac{1}{20}. The resistance to motion is constant, of magnitude 750 N, and the engine works at a constant rate of 24 kW.
  1. (a)
    Find the maximum speed of the car up the hill.
    (4)
  2. (b)
    Find the acceleration of the car when its speed is 12 m s⁻¹.
    (3)
  3. (c)
    The car reaches the top of the hill and moves onto level road, with the same resistance and power. Find its new maximum speed.
    (1)
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Part (a)4 marks

  1. At maximum speed the driving force balances resistance and the weight component.24000v\frac{24000}{v}=750{}= 750+1000×9.8×120{}+ 1000 \times 9.8 \times \frac{1}{20}
  2. v=19.4 m s−1v = 19.4\text{ m s}^{-1}

Part (b)3 marks

  1. F=maF = ma along the slope.2400012−750−490\frac{24000}{12} - 750 - 490=1000a{}= 1000a
  2. a=0.760 m s−2a = 0.760\text{ m s}^{-2}

Part (c)1 mark

  1. 24000750\frac{24000}{750}=32.0 m s−1{}= 32.0\text{ m s}^{-1}

Total8

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