Engineering Mechanics - KOP: Work and Energy - Discussion

Discussion :: KOP: Work and Energy - General Questions (Q.No.1)

1. 

The elevator E and its freight have a total mass of 400 kg. Hoisting is provided by the motor M and the 60-kg block C. If the motor has an efficiency of e = 0.6, determine the power that must be supplied to the motor when the elevator is hoisted upward at a constant speed of vE = m/s.

[A]. P = 22.2 kW
[B]. P = 13.34 kW
[C]. P = 26.2 kW
[D]. P = 30.1 kW

Answer: Option A

Explanation:

No answer description available for this question.

Gokul Krsishnan said: (Nov 7, 2015)  
Please tell me the method used for solving this question.

Chirag Sethi said: (Nov 24, 2015)  
Mc = 60.
g = 9.8.
Me = 400.
E = 0.6.

Ve = 4 must be to get an answer.

F = (ME-MC)g.
Power = FV/E.
So Power = 22.2 KW.

Binth Ismail said: (Oct 27, 2016)  
@Chirag.

How do you get Ve = 4?

Lakshmipathi said: (Aug 16, 2017)  
How you got Ve = 4?

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