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Homework Help: Simple Work Done by Force Problem

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    How much work is done by the force [tex]\vec F = (-4.0\hat\imath-6.0\hat\jmath)\;{\rm N}[/tex] on a particle that moves through displacement [tex]\Delta\vec{r}=(-3.0\hat\imath+2.0\hat\jmath)\;{\rm m}[/tex]?


    2. Relevant equations
    [tex]A_{x}B_{x} + A_{y}B_{y}[/tex]


    3. The attempt at a solution
    I decided to use the component form of the dot product method for this problem. (Or is the component form an alternative to the dot product?) Anyways, I set F as A and R as B, resulting in the following:

    [tex]A_{x}B_{x} + A_{y}B_{y}[/tex]
    [tex]F_{x}r_{x} + F_{y}r_{y}[/tex]

    [tex]\vec F = <-4, -6>[/tex]
    [tex]\vec r = <-3, 2>[/tex]

    [tex]-4(-3) + (-6)(2)[/tex]
    [tex]\vec C = <12,-12>[/tex]

    Then I took the square root of the sum of the sqaures:

    [tex]\sqrt{12^{2}+(-12)^{2}}[/tex]
    [tex]\sqrt{288} \approx 17[/tex]

    My answer of 17 was marked incorrect. Where did I go wrong? Any input is greatly appreciated.

    I just realized that I must have made some major mistake. I just kind of "created" vector C out of nowhere. Any direction on this problem would be great.

    EDIT: Nevermind, I figured out my error. Instead of creating vector C, I should have just stopped there, added 12 and -12, and got the correct answer of zero. Huzzah!
     
  2. jcsd
  3. May 4, 2010 #2

    ehild

    User Avatar
    Homework Helper

    The dot product is not a vector! It is scalar.

    ehild
     
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