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Homework Help: Calculating average horsepower of an elevator help!

  1. Nov 14, 2011 #1
    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!

    1. The problem statement, all variables and given/known data

    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."

    2. Relevant 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

    3. 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:

    2000lbf = lbm*32.2
    lbm= ~62.112



    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


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

    and to calculate Horsepower I use:

    550lbf*ft = 1 HP


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

    is that correct?!?!?!
  2. jcsd
  3. Nov 14, 2011 #2


    User Avatar
    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!
  4. Nov 14, 2011 #3
    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!
  5. Nov 15, 2011 #4


    User Avatar
    Homework Helper

    Most welcome!
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