Solution Found!
Solved: (a) Find the useful power output of an elevator
Chapter 7, Problem 39(choose chapter or problem)
Problem 39PE
(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 ?
Questions & Answers
QUESTION:
Problem 39PE
(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 ?
ANSWER:
Problem 39PE
Solution 39PE
Step 1 of 4:
In this we need to find power of output elevator motor
In the second part, we need to find the cost of electricity used
Data given
Mass to lift
Height to be lifted
Time taken
Final velocity
Total mass of system
Cost of electricity