What Is the Correct Calculation for the Power Delivered by an Elevator Cable?

[shawonna23]

The loaded cab of an elevator has a mass of 3.0*103^ kg and moves 245 m up the shaft in 23 s at constant speed. At what average rate does the force from the cable do work on the cab? (in kW)

I know that Power= Work/Time or Change in Energy/Time

I did this but it was wrong:

(1/2)(3.0*10^3)(245/23)^2 divided by 23
=7400000kW

Can someone tell me what i did wrong. I think I calculated the work wrong, but I don't know what else too do.

 

[whozum]

You got your equations mixed up.

P = Work / Time, but you misinterpreted kinetic energy for the energy exerted on the elevator.

If it is moving at a constant speed, the force pushing upwards is the same as the force pushing downards, net force = 0. The force pushing downwards, gravity, is 9.8*3000 = 29400N, so this is the force pulling upwards. Multiply this by distance and divide by time to find power.

 

[whozum]

Both approaches are correct, I don't know why were getting different answers.

I got ~7000J and ~13000J

 

[xanthym]

The loaded cab of an elevator has a mass of 3.0*103^ kg and moves 245 m up the shaft in 23 s at constant speed. At what average rate does the force from the cable do work on the cab? (in kW)

I know that Power= Work/Time or Change in Energy/Time

I did this but it was wrong:

(1/2)(3.0*10^3)(245/23)^2 divided by 23
=7400000kW

Can someone tell me what i did wrong. I think I calculated the work wrong, but I don't know what else too do.

The problem requires determination of the power CURRENTLY being delivered by the cable. Since the cab's velocity is CURRENTLY constant, there is NO change in Kinetic Energy, and therefore Kinetic Energy cannot be used for this problem. Only Potential Energy is changing for this problem:
{Power Delivered by Cable} = {ΔP.E.}/{ΔT} = m*g*Δh/ΔT =
= (3.0e(+3) kg)*(9.81 m/sec^2)*(245 m)/(23 s) =
= (313.5 kW)

(Note: At one time, the cab's velocity changed from 0 to what it is now, and thus the cable delivered power to increase the cab's Kinetic Energy at that time. However, that occurred BEFORE this problem began and is no longer applicable to the power currently being delivered at constant velocity.)


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