# Calculating average horsepower of an elevator help!

Hello. I hope someone here can give me some advice for solving this problem for an introductory mechanical engineering course. Below is the problem statement verbatim, and them my attempt at solving it. Advice and guidance is greatly appreciated!

## Homework Statement

Here is the problem statement:

"An elevator has a capacity of 2000 lbf including its own weight. It
can travel from the first floor to the fifth floor in 7 seconds. The
vertical distance between each floor is 15 ft. Estimate the average
power requirement in horsepower for this elevator. The elevator
starts with an initial velocity of 0, and reaches the fifth floor (just
before stopping) with a final velocity that is equal to the distance
traveled divided by the time required to get to the fifth floor."

## Homework Equations

Work = lbf X distance
Work = change in kinetic energy + change in potential energy
Force = lbm * 32.2
Units = English/American
1 Slug = lbm/32.2 ft/s
change in KE= 1/2mVf^2 - 1/2mVi^2
change in PE = force X change in height

## The Attempt at a Solution

Since the elevator is traveling from the 1st floor to the 5th floor, and each floor is 15 ft, I think the distance traveled is 60 feet.

D = 15ft

Since the problem is asking for average horsepower, I'm assuming that I should calculate WORK, which is the change in potential energy + change in kinetic energy, and divide by the time of 7 seconds.

Work = (deltaKE + deltaPE)/t
t = 7 sends

SO, here's what I have:

To calculate the change in kinetic energy, I did this:

First, find the mass in SLUGS since I'm using American units:

F=MA
2000lbf = lbm*32.2
lbm=2000lbf/32.2
lbm= ~62.112

1slug=1lbm/32.2

so:

62.112/32.2=1.929 slugs

Then i need to calculate the initial and final velocities:

Vi=0 ft/sec
Vf=60ft/7sec= ~8.571 ft/sec

now I can plug this info. into the KE equation:

[1/2m(Vf)^2]-[1/2m(Vi)^2] = deltaKE

=

1/2(1.929slugs)(8.571)^2 - 1/2(1.929slugs)(0)^2 = 70.854lbf*ft

NOW, since I'm looking for total work, i need to add that to deltaPE, which is:

2000lbf*60ft = 120,000 lbf*ft

so: Work = deltaKE + deltaPE = 70.854lbf*ft + 120,000 lbf*ft = 120,070.854 lbf*ft

NOW, to calculate POWER, I use:

Power = Work/Time

so

120,070.854 lbf*ft / 7 seconds = 17,152.979 ft*lbf

and to calculate Horsepower I use:

550lbf*ft = 1 HP

so

17,152.979lbf*ft / 550 lbf*ft = 31.187 HP

is that correct?!?!?!

Delphi51
Homework Helper
Welcome to PF!
I agree with your answer. I didn't bother with the kinetic energy, on the theory that it probably coasted the last few feet and converted its KE to PE. I just went with the weight x distance divided by time. My first time using feet and pounds in this way since 1967, I think!

Thank you so much! I was wondering why the KE was SO LOW, and that's exactly what made me think I was approaching this problem the wrong way. I just assumed I was supposed to include KE because the problem included initial and final velocities. Thank you again!

Delphi51
Homework Helper
Most welcome!