Question

An electric motor is required to haul a cage of mass 400kg up a mine shaft through the vertical height of 1200m in 2 minutes. What will be the electrical power required if the overall efficiency is 80%

Answer 1

Answer:

5 kilo watts

Explanation:

i have attached a solution

since the the efficiency is not 100 percent you need more power .

Answer 2

Answer:

Work done = P.E = mgh = 4000 x 10 x 1200. ( take g= 10mls)

Now W= P.E = 48x 10^ 6J

As, Motor is 80% efficient so

W=P.E= 48x 10^ 6/80

W= P.E= 6 x 10^ 5 J

Power = W/t

Power = 600 x 10^3/ 120

Power= 5 x 10^ 3

Power= 5 KW