### Solution Found!

# Solved: (a) Find the useful power output of an elevator

**Chapter 7, Problem 39**

(choose chapter or problem)

**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 ?

### 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