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Stuck at this prob.on work done

  1. Sep 1, 2005 #1
    I got a question below:

    A force F = (2x i + 6y j) N acts on an object as it moves in the x direction from the origin to x = 5 m. Find the work W = Fdr done on the object by the force.

    I felt that the answer shld be 2x * 5 = 10 but it's too simple to be true is it?

    One more question:

    A mass (m = 3 * 10^-5 kg) drops vertically at constant speed under the influence of gravity and air resistance.
    (a) After the drop has fallen h metres, what is the work done by gravity?
    (b) what is the work done by the air resistance?

    The ans shld be mgh for part A and -mgh for part b right? cos both are in opposite direction?
  2. jcsd
  3. Sep 1, 2005 #2
    firstly, The question is telling you that the Force is a function of position NOT a constant force so yes it is not so simple

    secondly yes you appear to be correct, at terminal velocity air resistance equals weight and is in opposite direction, you might usefully consider where the energy is going and so get some insight into the process.
  4. Sep 1, 2005 #3
    pls give hints: thanx!

    hmm.. if thats the case, how do i proceed from here? I am really clueless as to what is this.. ...
  5. Sep 1, 2005 #4


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    The general form for work done is

    [tex]W = \int_{x_1}^{x_2} F \cdot dx[/tex]

    It is why you are able to only use the x-component of F to find work. The dot product of the y-component with the displacement is zero. So only the x-component of the force is doing any work along x.

    Since, as has been mentioned, F isn't a constant over the displacement, you have to "sum" up all the F's over each infinitesimal displacement. Luckly, calculus has taught us how to do that via an integration procedure. You should be able to do the rest.

  6. Sep 1, 2005 #5
    so after working out, i got... ...

    let me verify this: so i got to integrate 2x with respect to x and its upper limit is 5 and lower limit 0 is it?
  7. Sep 1, 2005 #6


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    As simple as that, yes.

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