This question is about energy transfers. A funicular railway carriage with a mass of 1200 kg1200\text{ kg}1200 kg descends a mountain track with its cable drive disengaged.
The vertical height of the track is 45 m45\text{ m}45 m. The gravitational field strength is 9.8 N/kg9.8\text{ N/kg}9.8 N/kg. Calculate the gravitational potential energy of the carriage at the top of the track. Use the equation:
potential energy=mass×height×gravitational field strength \text{potential energy} = \text{mass} \times \text{height} \times \text{gravitational field strength} potential energy=mass×height×gravitational field strengthThe speed of the carriage at the bottom of the track is 24 m/s24\text{ m/s}24 m/s. Calculate the kinetic energy of the carriage at the bottom of the track. Use the equation:
kinetic energy=12×mass×(speed)2 \text{kinetic energy} = \frac{1}{2} \times \text{mass} \times (\text{speed})^2 kinetic energy=21×mass×(speed)2The kinetic energy at the bottom of the track is less than the potential energy at the top of the track. Explain why. Write about energy stores.
The test is repeated with a different funicular carriage. The potential energy of this carriage at the top of the track is 640 000 J640\,000\text{ J}640000 J. The kinetic energy of this carriage at the bottom of the track is 416 000 J416\,000\text{ J}416000 J. Calculate the efficiency of the transfer of energy from the potential store to the kinetic store. Use the equation:
efficiency=useful output energy transferinput energy transfer \text{efficiency} = \frac{\text{useful output energy transfer}}{\text{input energy transfer}} efficiency=input energy transferuseful output energy transfer37 exam-style questions on OCR GCSE Physics Power and efficiency. Each one has a worked solution and a mark scheme showing where the marks go.