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B Integration of a quantity when calculating Work

  1. May 25, 2017 #1
    While integrating work we write dw = INT. F.ds. But why can't we write dw = INT. dF.s ?
     
  2. jcsd
  3. May 25, 2017 #2

    jedishrfu

    Staff: Mentor

    The definition of work is ##work = force \times distance## and while the force can change over position and over time as we move along some path a distance s, we can generalize it as an integral over the path with ds being a very small distance to handle non-linear paths.

    However dF isn't a force its a change in force and if used breaks the definition of work which is ##work = force \times distance## not ##work = change in force \times distance ##.

    http://hyperphysics.phy-astr.gsu.edu/hbase/wint.html
     
  4. May 25, 2017 #3
    First what do you mean by dF multiplied by s....give the physical interpretation...
     
    Last edited by a moderator: May 25, 2017
  5. May 25, 2017 #4
    Isn't by definition ##\displaystyle w = \int_C f\cdot ds## ?
     
  6. May 25, 2017 #5
    Yes, it is so...
     
  7. May 25, 2017 #6
    Check integration by parts. You will see that f.ds is not the same as df.s but there is a relation between them. That is math. then you will have to see if that has physical sense.
     
  8. May 25, 2017 #7
  9. May 25, 2017 #8

    Mark44

    Staff: Mentor

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