How Much Power Does the Elevator Cable Exert in kW?

[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 even know where to begin. The answer is in kW

 

[Irid]

Employ a formula to calculate power:

$$P = Fv$$

 

[thomate1]

.

I would start by breaking down the problem into its components and using the appropriate equations and principles to find the solution. The first step would be to calculate the work done on the cab by the force from the cable, which can be found using the formula W = Fd, where W is the work done, F is the force, and d is the distance traveled. In this case, the distance traveled is 165 m.

Next, we need to find the force exerted by the cable on the cab. This can be done using Newton's second law, which states that force is equal to mass times acceleration (F=ma). In this case, the acceleration is zero since the cab is moving at a constant speed, so the force exerted by the cable would simply be equal to the weight of the cab, which can be found using the formula F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2).

Now, we can calculate the work done on the cab by the force from the cable by substituting the values we have found into the equation W = Fd. This gives us a work done of (3.0 x 10^3 kg)(9.8 m/s^2)(165 m) = 4.851 x 10^6 J.

Finally, to find the average rate at which the force from the cable does work on the cab, we can use the formula Power = Work/Time. In this case, the time is given as 23 s, so the power would be (4.851 x 10^6 J)/23 s = 210.5 kW.

Therefore, the average rate at which the force from the cable does work on the cab is approximately 210.5 kW.