(a) 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. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2500 kg is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost, if electricity is $0.0900 per kW ? h ?
Step-by-step solution Step 1 of 5 The equation to find the power is, Here is the power, is the work done and is the time period. Step 2 of 5 (a) In this case, the work done by the motor is equal to the sum of change in kinetic energy and the potential energy, Here is the height attained by the load. The initial kinetic energy of the load is zero. Step 3 of 5 Replace with in . Substitute for , for , for and for . Therefore, the power of the motor is .