Cambridge International AS & A Level · 5 · structured
Account for energy in a powered lift
Account for energy in a powered lift · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 11
- Topics
- 5
- Answer
- Complete
A motor lifts a20.0 kg load from rest through3.00m and brings it to speed 2.00m s⁻¹. During this interval it consumes900 J electrically. Take g=9.81m s⁻². Define useful output here as the load's gained gravitational plus kinetic energy.
- (a)
Calculate the increases in gravitational potential and kinetic energy.
4 marks - (b)
Write a total-energy balance and find energy transferred to unwanted stores or surroundings. Explain why this is not energy destroyed.
3 marks - (c)
Define efficiency, calculate it as a percentage, and find the input needed for a second lift requiring1200 J useful output at the same efficiency.
4 marks
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(a)
Calculate the increases in gravitational potential and kinetic energy.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
Use vertical height:mgΔh=20×9.81×3=588.6 J.
- 2
InitialKE=0; finalKE=½×20×2²=40 J.
(b)
Write a total-energy balance and find energy transferred to unwanted stores or surroundings. Explain why this is not energy destroyed.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
Choose system including load, motor and relevant surroundings.
- 2
Subtract useful gains from input:900−628.6=271.4 J.
- 3
Identify additional stores/transfers instead of calling energy lost from existence.
(c)
Define efficiency, calculate it as a percentage, and find the input needed for a second lift requiring1200 J useful output at the same efficiency.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
State ratio of like energy quantities.
- 2
Use 628.6 J useful output, not only gravitational energy.
- 3
Compute 69.8%.
- 4
Rearrange E_in=E_useful/η using unroundedη for1718.1 J.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- (a) Use vertical height:mgΔh=20×9.81×3=588.6 J.
- (a) InitialKE=0; finalKE=½×20×2²=40 J.
- (b) Choose system including load, motor and relevant surroundings.
- (b) Subtract useful gains from input:900−628.6=271.4 J.
- (b) Identify additional stores/transfers instead of calling energy lost from existence.
- (c) State ratio of like energy quantities.
- (c) Use 628.6 J useful output, not only gravitational energy.
- (c) Compute 69.8%.
- (c) Rearrange E_in=E_useful/η using unroundedη for1718.1 J.
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