Final answer:
In Physics, to determine the useful power output of an elevator system, one must calculate the work done in lifting a load and accelerating the system, then divide by time to get power. Subsequently, to find the cost of electricity, the calculated energy consumption in kW·h is multiplied by the electricity rate.
Step-by-step explanation:
The question involves calculating the useful power output of an elevator motor system, which comes under the subject of Physics, specifically involving concepts such as work, power, and energy. To find the power output, one must consider the work done on two fronts: lifting the 2500-kg load to a height of 35.0 m against gravity, and accelerating the entire 10,000-kg system from rest to a speed of 4.00 m/s, all within a time frame of 12.0 seconds.
The cost of electricity used by the motor can be calculated by first determining the energy consumed in kilowatt-hours (kW·h) and then multiplying it by the cost of electricity, which is $0.0900 per kW·h. The provided information highlights the importance of considering the counterbalance system in an elevator to accurately calculate the power output and costs associated with operation.