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Simple Integral Word Problem

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A 100 lb bag of sand is lifted for 2 seconds at the rate of 4 feet per second . find the work done in lifting the bag if the sand leaks out at the rate of 1.5 lbs per second.


    2. Relevant equations



    3. The attempt at a solution
    Here's my attempt at it:
    work = force x distance
    force = 100-1.5t
    distance = 4t
    work = 400t-6t2
    Now I integrate for t=2 and t=0
    ∫(400t-6t2)dt
    =200t2-2t3 for t=2 and t=0
    =800-16=784ft-lb

    Any mistakes?
     
  2. jcsd
  3. Feb 15, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    This is not correct. Work =force times distance is valid only when the force is constant during the displacement.
    If you integrate work with respect to time, you do not get work.

    Use that power = force times velocity, and integrate the power over time.

    ehild
     
  4. Feb 15, 2012 #3
    Ehild, thanks for pointing this out. I had a hunch that integrating work does not give you work, that should be common sense by now for me.

    Correction:
    power = force x velocity
    force = 100-1.5t
    velocity = 4
    power = 400-6t
    Now I integrate for t=2 and t=0
    work = ∫(400-6t)dt
    =400t-3t2 for t=2 and t=0
    =800-12=788ft-lb
     
  5. Feb 15, 2012 #4

    Dick

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    That looks much better.
     
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