Question

Suppose that the average U.S. household uses 15600 kWh (kilowatt-hours) of energy in a year. If the average rate of energy consumed by the house was instead diverted to lift a 2130 kg car 13.3 m into the air, how long would it take? (in seconds)

Using the same rate of energy consumption, how long would it take to lift a loaded 747, with a mass of 3.90 × 105 kg, to a cruising altitude of 9.42 km? (in seconds)

Answer 1

Part A.

We know that

Power = Energy/time

Now average rate of energy consumed is

Energy used = 15600 kWh = 15600*10^3 W-hr

t = 1 year = 365 day = 365*24 hr

P = 15600*10^3 W-hr/(365*24 hr)

P = 1780.8 W

Now if this power is used to left a car then time taken for lift will be

Power = Energy/time

time = E/P = m*g*h/P

time = 2130*9.81*13.3/1780.8 = 156.1 sec

Part B.

time required to left loaded 747 will be

time = m*g*h/P

time = 3.90*10^5*9.81*9420/1780.8

time = 20238082.9 sec

(20238082.9 sec = 5621.7 hr)

Please Upvote.