# Solving an Elevator Power Problem: Find the Average Power in kW

• BitterSuites
In summary, the elevator's acceleration can be calculated by dividing the final speed by the time, and the average speed can be found by adding the final and initial speeds and dividing by 2. The distance traveled can then be found by multiplying the average speed by the time. Using the mass and acceleration, the force exerted by the elevator can be calculated. To find the work, multiply the force by the distance. Finally, to find the power, divide the work by the time. In this case, the elevator's average power is 18.117 kW.
BitterSuites

## Homework Statement

A 523 kg elevator starts from rest. It moves upward for 3.79s with a constant acceleration until it reaches its cruising speed of 1.78 m/s.

The acceleration of gravity is 9.8 m/s^2.

Find the average power delivered by the elevator motor during the period of this acceleration. Answer in units of kW.

w = m(a+g)
p = w/t

## The Attempt at a Solution

m = 523
a = 1.78
t = 3.79

w=m(a+g)=523(1.78+9.8)=6056.34

p=w/t=6056.34/3.79=1597.98W

1597.98W becomes 1.59798kW

I was incredibly confident in this answer, but it is wrong. Where did I make my silly error?

BitterSuites said:

## Homework Equations

w = m(a+g)
m(a+g) is the force exerted, not the work. What's missing?

Hmm. I guess it is missing d, as in W=Fd. Am I capable of calculating d?

BitterSuites said:
Hmm. I guess it is missing d, as in W=Fd.
Right.
Am I capable of calculating d?
One easy way to find distance is to use average speed X time. What's the average speed?

BitterSuites said:

## The Attempt at a Solution

m = 523
a = 1.78
t = 3.79
Your value for the acceleration is incorrect. That's the final speed after 3.79 seconds. Use the change in speed and the time to calculate the acceleration.

Ok. So a = 1.78/3.79 = .469657
avg v = (V + Vo)/2 = (1.78 + 0)/2 = .89

So, d = avg v * t = .89 * 3.79

Am I at least on the right side of the highway?

You are back on track.

So F = m(a+g) = 523 (.469657 + 9.8) = 5371.03

d = .89 * 3.79 = 3.3731

W = Fd = 5371.03 * 3.3731 = 18117W becomes 18.117 kW

Where am I still going wrong? This was not correct.

You calculated the work (in Joules). Now find the power (in Watts).

## 1. What does "average power" mean in this context?

Average power refers to the average rate of energy consumption or production over a period of time. In the context of solving an elevator power problem, it refers to the average amount of power used by the elevator over a specific time period.

## 2. How is average power calculated for an elevator?

Average power can be calculated by dividing the total energy used by the elevator (in kilowatt-hours) by the time period over which it was used (in hours). This will give the average power in kilowatts (kW).

## 3. What factors affect the average power of an elevator?

The average power of an elevator can be affected by various factors such as the weight of the elevator, the number of passengers, the distance traveled, the speed of the elevator, and the efficiency of the elevator's motor and other components.

## 4. Why is it important to solve an elevator power problem?

Solving an elevator power problem is important because it can help identify any issues with the elevator's power consumption, which could lead to higher energy costs and potential mechanical problems. It can also help optimize the elevator's energy usage and improve its efficiency.

## 5. Can the average power of an elevator change over time?

Yes, the average power of an elevator can change over time. Factors such as changes in passenger traffic, maintenance and repairs, and upgrades to the elevator's components can all affect its average power usage.

• Introductory Physics Homework Help
Replies
31
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
930
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
475
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
2K