Answer to Question #19188 in Electric Circuits for Kiyanna
A 2344 kg elevator can safely carry a maxi- mum load of 719 kg.
The acceleration of gravity is 9.81 m/s2 .
What is the power provided by the electric motor powering the elevator when the ele- vator ascends with a full load at a speed of 2.7 m/s?
Answer in units of kW
The force of gravity that acts on the elevator witha full load is: F=Ma=(m1+m2)a=(2344+719)*9.81=30048N The power provided by electric motor must be: P=F*v=30048*2.7=81130Wt