How Is Elevator Motor Efficiency Calculated?

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Homework Help Overview

The discussion revolves around calculating the efficiency of an electric motor used to pull an elevator at a constant velocity. The subject area includes concepts of work, power, and efficiency in the context of mechanical systems.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of efficiency using the formula "work out/work in." There are inquiries about how to determine the necessary values for work and power, as well as unit conversions between kilowatts and watts.

Discussion Status

The discussion is active, with participants offering guidance on finding force and power, and emphasizing the importance of unit consistency. There is no explicit consensus yet, but helpful suggestions have been made regarding the calculation process.

Contextual Notes

Participants are navigating unit conversions and the relationship between work and power in the context of the problem. There is an acknowledgment of the need for consistent units to proceed with calculations.

larry21
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An 8.5x10^2kg elevator is pulled up at a constant velocity of 1.00m/s by a 10.0kW electric motor. Calculate the efficiency of the motor.

Efficiency = work out/work in * 100

How do I calculate "work out/work in" using the information given?
 
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hi larry21! :smile:

(try using the X2 icon just above the Reply box :wink:)

find the force, then find the power, then divide by 10.0 …

what do you get? :smile:
 
thanks, that was a great help. However, do I need to convert the "kW" to "W"
 
oh yes … you'll need both to be in the same units :smile:
 

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