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A Physics problem

  1. Feb 9, 2010 #1
    Hello Peace be upon you

    http://img8.imageshack.us/img8/9198/workdone.jpg [Broken]

    The first part of this problem is OK but I'm asking about the second part
    We know that the work done is F *d*cosine the angle between the force direction and
    the displacement direction .So why the book calculated the work done in this way?

    I know that the work done in this case is the potential energy which is (m*g*d) and this does works!! But my problem is why the book calculated the work done in this way?

    Another way :
    Its given that the angle between force direction and the displacement direction is 60 then the result is the same.
    So Are all of these ways right?
    I hope you can understand my problem
    Thanks so much
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 9, 2010 #2


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    Yes, it's all the same, either it's work = force times displacement times cos 60, where the displacement, d, is the distance along the incline. Since d =15/cos 60, then work = (15mg/cos60)(cos60), which is just Work = 15mg. Thus, the work done against gravity can be calcualted from W=mg(vertical distance) = 15mg. I agree that the theta = 0 in the book solution is somewhat misleading. Note also that the work done by gravity is (- delta U_g).
  4. Feb 10, 2010 #3
    Yeah this was exactly what i'm asking about

    Thanks so much
    this really helped
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