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Need help with finding Work from given force and distance components

  1. Mar 5, 2010 #1
    1. The problem statement, all variables and given/known data
    A force ~F = Fx ˆı+Fy ˆ acts on a particle that
    undergoes a displacement of ~s = sx ˆı + sy ˆ
    where Fx = 8 N, Fy = −4 N, sx = 5 m, and sy = 1 m.
    Find the work done by the force on the
    particle.
    Answer in units of J.

    2. Relevant equations
    x^2 + y^2 = z^2
    Work = (Force)(Distance)(cos(theta))

    3. The attempt at a solution
    I found the magnitude of Force by squaring each component
    F = (8)^2 + (-4)^2 = 8.9442N
    I found the magnitude of Distance by squaring each component
    F = (5)^2 + (1)^2 = 5.099m
    I then created a triangle to find theta, using distance for my triangle sides.
    tan(theta) = 1/5 so theta = inverse tan(1/5). I then got a theta of 11.310.
    I plugged my force, distance and theta into the Work formula but I'm not getting the right answer.
    W = (8.9442N)(5.099)(cos(11.310))
    W = 44.721 Joules

    THIS IS NOT THE RIGHT ANSWER. WHAT AM I DOING WRONG. PLEASE HELP ME.

    1. The problem statement, all variables and given/known data
    Find the angle between ~F and ~s.
    Answer in units of ◦.


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 5, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi yang09! Welcome to PF! :smile:
    (have a theta: θ and try using the X2 and X2 tags just above the Reply box :wink:)
    Work also = Force "dot" Distance, which is the easiest way to do it if you have the coordinates.

    So what is (8,-4)·(5,1) ? :smile:
     
  4. Mar 6, 2010 #3
    Thanks for the tips but what do you mean by:
    What is (8,-4)·(5,1) ?
    Do you want me to multiply them together or what? Or are you asking me what they stand for because the 8 is the x-component of force, -4 is the y-component of force, 5 is the x-component of distance, and 1 is the y-component of distance.
    I'm pretty sure my magnitude is right, but am I doing something wrong with my θ?
     
  5. Mar 6, 2010 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi yang09! :smile:
    I meant the "dot" product of two vectors (also called the "scalar product" or the "inner product"), where you multiply the components and then add.

    Have you learnt about that?
    Yes, your θ is tan-1(1/5), but that is the angle between (5,1) and the x-axis: you need the angle between (5,1) and (8,-4).
     
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