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Find the useful power output of an elevator motor that lifts a 2500-kg load a height of 35.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s.

A. (a) 35.0 kW, (b) $37.80
B. (a) 35.0 kW, (b) $0.126
C. (a) 43.5 kW, (b) $0.126
D. (a) 43.5 kW, (b) $37.80

User Drekka
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1 Answer

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Final answer:

To find the useful power output of an elevator motor that lifts a load and accelerates it, we need to calculate the work done and divide it by the time taken. The cost can be calculated by multiplying the power output with the electricity cost per kilowatt-hour.

Step-by-step explanation:

To find the useful power output of the elevator motor, we need to calculate the work done and divide it by the time taken. We can start by calculating the work done to lift the load using the formula:

Work = Force * Distance

The force can be calculated by multiplying the mass of the load by the acceleration due to gravity. The distance is given as 35.0 m. The work done to accelerate the load can be calculated using the formula:

Work = (1/2) * Mass * Velocity^2

By adding the two works together, we can find the total work done. Finally, dividing the work by the time taken gives us the useful power output of the elevator motor.

After finding the power output, we can calculate the cost by multiplying it with the electricity cost per kilowatt-hour.

User Mfcabrera
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