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Work and vectors

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A force F = (2.7i + 3.2j) N is acting on a particle while it travels from the point (x, y) =
    (1, -1) m to the point (2, 1) m. Find the work done by the force F on the particle.

    2. Relevant equations
    W=FD and A=SqRt of Ax^2+Ay^2

    3. The attempt at a solution
    ok i used the right triangle to find the resultant vector of the force F by squaring both the 2.7 and 3.2 then adding them and take the square root i got 4.18 so that should be the F in W=FD and then i foudn teh reultant vector of the coordinates 1,-1 and 2,1 and got square root of 5. which should be the distance and then i multiplied 4.18 and radical 5 and got 9.34 which is wrong.. where did i go wrong? btw 9.10 is the answer
     
  2. jcsd
  3. Mar 20, 2009 #2

    alphysicist

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    Homework Helper

    Hi silentsaber,

    The problem is that the formula for work is not [itex]W=Fd[/itex] (product of magnitudes of force and displacement), it is

    [tex]W=\vec F\cdot\vec d[/tex]

    (dot product of force and displacement vectors). Are you familiar with the dot product? If so, then once you find the two vectors involved the problem is quite straightforward.
     
  4. Mar 20, 2009 #3
    You're working with vectors. So you need to find the dot product. W = F dot D.
     
  5. Mar 20, 2009 #4
    thankyou guys very much i totally forgot the dot product existed XD
     
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