An electric motor is required to haul a cage of mass 400 kg up a mine shaft through a vertical height of 1200 m in 2 minutes. What will be the electrical power required if the overall efficiency is 80%? (g = 10ms-2)
Correct answer: D. 50 kW
- A. 3.2 kW
- B. 32 kW
- C. 5 kW
- D. 50 kW
Explanation
Energy of motor = Gain in gravitational P.E80%W = mgh0.8(Pt) = mgh0.8P(120) = (400)(10)(1200)P = (400)(10)(1200) / (0.8)(120)=40000 / 0.8 = 50000 J = 50 kJ
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Work transfers energy when a force causes displacement, while kinetic energy, gravitational potential energy and power describe motion, position and the rate of energy transfer. The work-energy theorem links net work with change in kinetic energy, and efficiency accounts for energy losses rather than treating them as energy destruction.
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