How Fast Does the Elevator Cable Perform Work?

[needhelp1234]

The loaded cab of an elevator has a mass of 3.0 103 kg and moves 165 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?


I don't know where to begin. The answer is sin kW.

 

[LowlyPion]

Work is what?

Did I hear you say change in potential energy? Because if I did then I'd be thinking you were on the right track already.

 

[thomate1]

I can help you solve this problem by using the formula for work, W = Fd, where W is the work done, F is the force applied, and d is the distance moved. In this case, we are looking for the average rate at which the force from the cable does work on the cab, which can be calculated by dividing the work done by the time taken.

First, we need to calculate the work done by the force from the cable. Since the cab moves at a constant speed, we can assume that the net force on the cab is zero, and therefore the force from the cable is equal to the weight of the cab, which can be calculated using the formula F = mg, where m is the mass of the cab and g is the acceleration due to gravity (9.8 m/s^2).

F = (3.0 * 10^3 kg) * (9.8 m/s^2) = 2.94 * 10^4 N

Next, we need to calculate the distance moved by the cab, which is given as 165 m in the problem.

Now, we can calculate the work done by the force from the cable:

W = Fd = (2.94 * 10^4 N) * (165 m) = 4.85 * 10^6 J

Finally, we can calculate the average rate at which the force from the cable does work on the cab by dividing the work done by the time taken:

Average rate = W / t = (4.85 * 10^6 J) / (23 s) = 2.11 * 10^5 J/s or 211 kW

Therefore, the average rate at which the force from the cable does work on the cab is 211 kW.