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Work done by force (Vector notation)

  1. Apr 3, 2011 #1
    1. The problem statement, all variables and given/known data

    How much work is done by a force= (2x)i N + (3)j N with x in meters, that moves a particle from a position (initial) = (2m)i + (3m)j to a position (final) = (-4m)i + (-3m)j?

    Answer:
    - 6 J

    2. Relevant equations

    Work = integral of Force

    3. The attempt at a solution

    W = [x^2] i + [3x] j
    W = [4^2 - 2^2] i + [3(-3) - 3(3)] j
    W = 12 i -18 j
    W magnitude = 21.6J

    I realize I am doing this wrong becasue the Magnitude will always be positive and the answer (from back of book) is negative.

    I tried this in a similar way where I took the difference between the two position vector and then placed it in the integral
    rd = (-6) i + (-6) j
    Got W = 36 i + 18 j
    still wrong.

    Help, pretty please?
     
  2. jcsd
  3. Apr 3, 2011 #2
    To save typing all the i's and j's, I'll use the notation (2,3) = 2i + 3j. I'll also cut corners by leaving out the units.

    Assuming the movement is in a straight line, we can parameterize the particle's path from the initial point (2,3) to the final point (-4,-3) by an equation such as

    [tex]\mathbf{r}(t) = (x,y) = (2,3) - t(6,6), \enspace 0 \leq t \leq 1.[/tex]

    -(6,6) is the displacement vector from the initial point to the final point, that is, the difference between them.

    Then [itex]x=2-6t[/itex], and

    [tex]d\mathbf{r} = (dx,dy) = -(6,6) dt.[/tex]

    So

    [tex]\int_{\mathbf{r}(0)}^{\mathbf{r}(1)} \mathbf{F}(\mathbf{r}) \cdot d\mathbf{r}[/tex]

    [tex]=\int_{(2,3)}^{(-4,-3)}(2x,3) \cdot (dx,dy) = \int_{0}^{1}(42-72t) \; dt = -6[/tex]
     
  4. Apr 3, 2011 #3
    Thank you Rasalhague. But how did you get from (2x,3) to (42-72t)? I don't understand.
     
  5. Apr 3, 2011 #4
    Oopsh, sorry, I missed out a minus in front of the integral on the right. Are the following steps clear?

    [tex]x = 2 - 6t[/tex]

    [tex](2x,3) = (2(2-6t),3) = (4-12t,3)[/tex]

    [tex](4-12t,3) \cdot (-6,-6) = -6(4-12t) - 18 = -42 + 72t.[/itex]

    -(6,6) comes from differentiating [itex]\mathbf{r}[/itex].
     
  6. Apr 9, 2011 #5
    Hi Rasalhague. Thank you for helping me.

    My test is this Wednesday, but I have lots of other tests so sorry about the delay.

    A couple things:

    1) where did time come from? The question never mentioned it- only position.

    2) I understand this:
    r = (-6) i + (-6) j
    r = -(6,6)

    3) where does this come from: x=2-6t (I think this is an extension of being confused about the time)?

    Why is integral not (x^2)i + (3x) j ???

    Thank you again for your help and I am sorry if I am difficult.
     
  7. Apr 10, 2011 #6
    The reason time confuses me is that isn't POWER change in work over time? I thought the difference between Force and power is whether your taking the derivative of work with respect to time or with respect to position....?
     
  8. Apr 10, 2011 #7
    Okay, I figured out what I did wrong. I had this part right:

    W = [x^2] i + [3x] j
    W = [4^2 - 2^2] i + [3(-3) - 3(3)] j
    W = 12 i -18 j

    What I needed to do was subtract 18 from 12
    W = 12 - 18 = -6 (answer!)

    I wasn't doing this becasue I thought they were sepearte becasue of the separate components. Wow, I can't believe I was so close for so long!

    I still don't completely understand why you just add i and j together to find work instead of finding the magnitude... but it works for many similar book problems so I know this is the right way to do it.

    Thanks for the help. :)
     
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