How Is Elevator Motor Efficiency Calculated?

[larry21]

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?

 

[tiny-tim]

hi larry21!

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

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

what do you get? ​

 

[larry21]

thanks, that was a great help. However, do I need to convert the "kW" to "W"

 

[tiny-tim]

oh yes … you'll need both to be in the same units

 

[rwisz]

To calculate "work out" and "work in," we need to use the formula for work which is work = force * distance. In this case, the force is the weight of the elevator (8.5x10^2kg) and the distance is the height the elevator is being pulled (1.00m). Therefore, work out = (8.5x10^2kg * 9.8m/s^2) * 1.00m = 8,330 J.

Now, to calculate work in, we need to convert the power of the electric motor from kilowatts to watts (1 kW = 1000 W). So, the power of the motor is 10,000 W. Using the formula for power (power = work/time), we can rearrange it to find work, which gives us work in = power * time. In this case, the time is 1 second. Therefore, work in = 10,000 W * 1 s = 10,000 J.

Now, we can plug these values into the efficiency formula: Efficiency = (8,330 J/10,000 J) * 100 = 83.3%. This means that the electric motor is 83.3% efficient in converting the energy it receives into work to lift the elevator.