How much current does the elevator consume?

[AlexPilk]

Homework Statement

Energy conversion efficiency of the elevator is 80%, it is working from 220 volts. How much current does the electro motor consume while it is constantly lifting the 500 kg cabin at 2 m/s.

The attempt at a solution

I=P/U, so I figured I need to know the power. P=F*v. I know the velocity (2 m/s) so I need the force.
F=m(g+a). Now this looks simple but for some reason I'm stuck. The acceleration = 0, so F must = mg?
If so - F=500*9.81=4905. (I suppose kilograms and m/s shouldn't be converted to anything).
P=9810W. If the efficiency is 80% - the whole power consumed is 9810/0.8=12262.5W.
And then I=12262.5/220=55.7 amps.
I'm not sure if it's too much or too little for an elevator, so I'm not really sure if I solved it correctly.
What do you think? Thanks!

 

[SteamKing]

Homework Statement

Energy conversion efficiency of the elevator is 80%, it is working from 220 volts. How much current does the electro motor consume while it is constantly lifting the 500 kg cabin at 2 m/s.

The attempt at a solution

I=P/U, so I figured I need to know the power. P=F*v. I know the velocity (2 m/s) so I need the force.
F=m(g+a). Now this looks simple but for some reason I'm stuck. The acceleration = 0, so F must = mg?
If so - F=500*9.81=4905. (I suppose kilograms and m/s shouldn't be converted to anything).
P=9810W. If the efficiency is 80% - the whole power consumed is 9810/0.8=12262.5W.
And then I=12262.5/220=55.7 amps.
I'm not sure if it's too much or too little for an elevator, so I'm not really sure if I solved it correctly.
What do you think? Thanks!

You're somewhat careless with units.

W = mg has units of kg-m/s², which are otherwise known as Newtons. Energy has units of Newton-meters, and a power of 1 N-m/s is known as a watt. You should learn the definitions of these derived units in SI.