Question

Suppose that the average U.S. household uses 13100 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 12.3 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 9.92 km?

Answer 1

Answer 2

SOLUTION :

Average power used in houses, P avg  

= Electric energy used in a year / No. of hours in a year

= 13100 kWh / (365*24) h 

= 1.4954 kW 

= 1495.4 W

a. 

Same power is used in lifting the car 

So,

Work done in lifting the car, W = Average power used for t secs.

=> m g h = 1495.4 * t

=> 1470 * 9.8 * 12.3 = 1495.4 * t  

=>  177193.8 = 1495.4 t

=> t = 177193.8 / 1495.4

=> t = 118.49 sec. (ANSWER)

b.

Work done in lifting the loaded 747, W = Average power used for t secs.

=> m g h = 1495.4 * t

=> 4.10* 10^5  * 9.8 * 9.92*10^3 = 1495.4 * t  

=>  3.985856 * 10^10  = 1495.4 t

=> t = 3.985856 * 10^6 / 1495.4

=> t = 2.6654 * 10^7  sec. (ANSWER)

=> t = 2.6654 * 10^7 / (60*60*24) days

=> t = 308.5 days (ANSWER)