Calculate Efficiency of 10.0 kW Motor for 8.5 kg Elevator

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Homework Help Overview

The discussion revolves around calculating the efficiency of a 10.0 kW motor used to pull an 8.5 kg elevator at a constant velocity of 1.0 m/s. Participants explore the relationship between power, force, and work in the context of this problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various methods to calculate work and efficiency, questioning how to derive force from the given power and velocity. Some suggest using potential energy changes, while others express confusion about the relevance of the motor's power in their calculations.

Discussion Status

There is ongoing exploration of different approaches to the problem, with some participants providing clarifications and guidance on how to relate power, force, and efficiency. However, there is no explicit consensus on the correct method or final answer yet.

Contextual Notes

One participant noted a discrepancy in the mass of the elevator, initially stating 8.5 kg and later correcting it to 85 kg. This change may impact the calculations discussed.

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Homework Statement



An 8.5 kg elevator is pulled at a constant velocity of 1.0 m/s by a 10.0 kW motor. Calculate the efficiency of the motor.


Homework Equations



W = change in E = Fd

e = Work out/Work in * 100 = energy out/energy in *100

Epotential = mgh
Ekinetic = 1/2mv^2

P = W/t

The Attempt at a Solution



Ekinetic = 1/2mv^2 = 8.5 J (Energy in)

I don't know how to get the Work from the motor as I don't have the time?
 
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petuniac said:
I don't know how to get the Work from the motor as I don't have the time?

Power is also written as: Power = Force x Velocity. You have two known quantities viz. Power & velocity. Can you find the force?
 
if you want to approach this using time and simply potential energy change, you can set an initial condition and a final condition:

ex)
initial condition:
height=0m
t=0

final condition:
height=10m
t=10s

Then play with (delta)E in a much simpler way.
 
Sorry Reshma but I am still confused ... if I solve for F using the P = Fv equation what does this accomplish as I need to calculate the efficiency, and why would I use the 10 kW power given for the motor in that equation and the speed of the elevator?? Please help.
 
petuniac, you should not be using 10kW and the elevator's speed in the same equation.

It sounds like 10kW is your Power in, the power supplied to the motor.

efficiency is also Pout/Pin.

So Pout is the amount of power used while lifting the elevator. For this calculation you can in fact use P=Fv. Your unknown here is NOT F, but rather P. If you're stuck on figuring out F, remember gravity's effect on the elevator's mass.
 
Thank you Matheyrichs! You've made it much clearer to me!

I made an error in quoting the original question.. the elevator mass is 85 kg.

Ok, so for the solution I get...

Pout = Fv = mgh = (85*9.81*1) = 833.85 W

Pin = 10000 W

e = Pout/Pin = (833.85)/10000 *100 = 8.33%

The only issue is that the answer is 83.3% so I am off by an order of magnitude and can't seem to see why?
 
are there any people in the elevator?? :-p

I'm not sure. 833/10,000 is certainly only 8.33%. I also did the problem a different way, substituting the initial start and end and got the same answer.

Check the problem again and make sure V_elevator and P_motor are actually 1 m/s and 10kW as well.

83% sounds like a reasonable efficiency, 10kW is ~13horsepower, and 85kg is ~187lbs.

I'd say that the most unreasonable piece of that is for an elevator to only weigh 85kg.
 
Thanks again!
 

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