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Calculating work done by gravity

  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known data
    A 10kg bucket of water is lifted vertically 3.0m at a constant speed. How much work did gravity do on the bucket during this process?

    2. Relevant equations
    Work=Force*Distance (what thought to use)
    Wext=Fextdcosθ (what the answer key says to use.

    3. The attempt at a solution
    My attempt got me as far as plugging in: Work=(9.8m/s2*10kg)*3.0m=(98N)*3.0kg=294J
    ***In the answer key they used 10m/s2 for gravity, but also had a negative answer of -300J. I assume that they used the negative as a direction, but is work a vector or scalar? Looking to understand why the way I did this does not yeild the same answer, and how the formula the keys says to use works?***
     
  2. jcsd
  3. Nov 5, 2016 #2

    mfb

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    Staff: Mentor

    The force is in the opposite direction of motion, therefore work done by gravity is negative. You should change the sign of either the acceleration or the 3 meter distance, because they point in opposite directions.
     
  4. Nov 5, 2016 #3
    Okay thanks I was thinking something along those lines, but why would the key ask to solve using that other equation? It seems more complicated than the way I did it. does my method only work in limited scenarios and the other has a broader range of uses?
     
  5. Nov 5, 2016 #4

    mfb

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    Staff: Mentor

    Well, the equation with the angle is more general, but as your problem is one-dimensional you don't need that.
     
  6. Nov 5, 2016 #5
    I see. In that case I will stick to my simple method :) Thanks!
     
  7. Nov 5, 2016 #6

    CWatters

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    In this case the angle is zero and Cos(0)=1 so both equations are the same.
     
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