Question

Suppose that the average U.S. household uses 12100 kWh (kilowatt‑hours) of energy in a year. If the average rate of energy consumed by the house was instead diverted to lift a 1470 kg car 11.8 m into the air, how long would it take?

car:

Using the same rate of energy consumption, how long would it take to lift a loaded Boeing 747 airplane, with a mass of 4.15×105 kg , to a cruising altitude of 9.17 km ?

airplane:

Answer 1

Part A.

Power = Energy consumption/time

Given that Average energy consumption in US Household = 12100 kWh = 12100*10^3 W-hr

time = 1 year = 365.25 day = 365.25*24 hr

So,

Power = 12100*10^3 W-hr/(365.25*24 hr)

Power = 1380.33 W

Now Using this above power to lift a car to given height, required time will be:

time = Work-done in lifting the car/Power Applied

time = m*g*h/P

time = 1470*9.81*11.8/1380.33

time = 123.3 sec = time required to lift the car

Part B.

Now when same amount of power is used to lift a loaded Boeing 747 airplane, then time required will be:

time = m*g*h/P

time = 4.15*10^5*9.81*9.17*10^3/1380.33 = 27046029 sec

time = 2.70*10^7 sec

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