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
1
Expert's answer
2012-11-29T08:01:16-0500
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
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Comments
Leave a comment