Question

Suppose that the average U.S. household uses 13600 kWh (kilowatt-hours) of energy in a year. If the average rate of energy consumed by the house was instead diverted to lift a 2350 kg car 11.3 m into the air, how long would it take?

Using the same rate of energy consumption, how long would it take to lift a loaded 747, with a mass of 4.10 × 105 kg, to a cruising altitude of 9.67 km?

Answer 1

The average U.S. household uses 13600 kWh = 13600*10³*3600 W.sec = 4.9*10¹⁰ W.sec of energy in a year.

Power consumed in a year =   = 37808.6 watt

If the same rate of energy is used to lift a car of mass(say m) 2350 kg by 11.3 m(say d) then,

37808.6 = Work done/Time = (Force*Distance)/Time = (mg*d)/Time = (2350*9.8*11.3)/Time

=> Time = (2350*9.8*11.3)/37808.6 = 6.88 ~ 7 seconds

It would take almost 7 seconds to lift the car with the same amount of consumed rate of energy.

Using the same method, the time needed to lift loaded 747 of mass 4.1*10⁵ kg to altitude 9.67 km

Time = (4.1*10⁵*9.8*9.67*1000)/37808.6 = 1027651.4 seconds = 285.5 ~ 286 hours

It would take almost 286 hours to lift loaded 747 with the same amount of consumed rate of energy.