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How much work is done by force

  1. Jun 4, 2015 #1
    1. The problem statement, all variables and given/known data


    On my exam, I got this problem wrong, which I believed was incredibly easy, and I don't know why it's wrong...

    "A constant force of 25 N is applied as shown to a block which undergoes a displacement of 7.5 m to the right along a frictionless surface while the force acts. What is the work done by the force?"
    Juhnd9L.jpg

    2. Relevant equations
    W = F * S

    3. The attempt at a solution
    There is a displacement to the right, while the force vector is to the left, therefore work is in the negative.

    displacement (s) = 7.5 m
    F = 25 N (cos 30 degrees)

    I chose cosine because we are only concerned for the x direction...which is adjacent to the angle of 30 degrees...

    therefore
    W = F * s = [25 N (cos 30)] * [-7.5 m]
    W = -162 J

    However the answe was -94 J...and this is because sine was used instead. Why? My professor didn't care to explain why...
     
  2. jcsd
  3. Jun 4, 2015 #2

    haruspex

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    I can only suggest you ask your professor what the answer would have been with an angle of 90 degrees.
     
  4. Jun 4, 2015 #3
    are you agreeing that I was correct? because if so that's a whole letter grade I get back to my test if I can stump my professor and prove that I was right...and I help out my classmates with the same answer as me.

    and that's true...because if it was 90 degrees no work would be done but since he used sine which equals 1 it's implying work was done...and a verticle force never causes a horizontal displacement.

    I thought you also never use sine for work. I never came across a problem where you had to besides this one.
     
  5. Jun 4, 2015 #4

    haruspex

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    Yes.
    It depends how you are defining the angle. What if F were dragging the block up against a vertical wall? Or if the given angle were measured against the vertical?
    If you are defining the angle to be that between the force vector and the displacement vector then I agree. ##|\vec x.\vec y| = |\vec x|.|\vec y| |\cos(\theta)|##
     
    Last edited: Jun 5, 2015
  6. Jun 4, 2015 #5
    I see from your examples, it would make sense to use sine then in those situations.

    thank you for the explanation. I suppose I will confront him about this, as nice as possible.
     
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