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Work in 3D

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A 13 N force with a fixed orientation does work on a particle as the particle moves through displacement = (3i - 5j + 3k) m. What is the angle between the force and the displacement if the change in the particle's kinetic energy is (a) +25.6 J and (b) -25.6 J?


    2. Relevant equations

    Angle = √(x^2 + y^2 + z^2)
    Work = mad or F (dot) d

    3. The attempt at a solution

    I'm not sure how to approach this. I know that the dot product really just means to multiply the x parts by x parts, y parts by y parts, and z parts by z parts (or [itex]\hat{i}[/itex], [itex]\hat{j}[/itex], and [itex]\hat{k}[/itex] parts, if you prefer).

    If the particle starts at (0, 0, 0), then the force would have caused it to go to its current position (3, -5, 3). I'm still not sure that is what is being said though.
     
  2. jcsd
  3. Sep 30, 2011 #2

    SammyS

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    There is another way to express the dot product, [itex]\vec{F}\cdot\vec{d}\,.[/itex]

    It involves the magnitude of each of the vectors as well as the cosine of the angle between their directions.
     
  4. Sep 30, 2011 #3
    Yes, work is equal to Fx*dx+Fy*dy+Fz*dz where where Fx,Fy,Fz are the force components and dx,dy,dz are the displacement components...but there is another common way to write the the dot product of two vectors that should make quick work of this problem.
     
  5. Sep 30, 2011 #4
    Got it, thanks for your help.
     
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