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Work done by a force with a given equation

  1. Jan 2, 2013 #1
    1. The problem statement, all variables and given/known data

    A force acts on a particle and is given by the following expression:
    F(x,y)=2x^3 y^2 i+3xy^3 j
    What is the work done by this force in moving the particle from a position (x,y) = (0,0) to (4,2) along the path given by the curve y=√x ?



    2. Relevant equations
    I know the work is the integral of the Force


    3. The attempt at a solution

    Honestly, very confused. I don't know how to deal with the i and j values and I have no clue how to handle the work done along the path of a different curve. Any help is appreciated!
     
  2. jcsd
  3. Jan 2, 2013 #2

    haruspex

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    Yes, work is the integral of the force wrt distance, each being a vector. So the integrand and differential element look like F.dp, the dot being the scalar product of the two vectors. (I've used p for position vector to avoid confusion with the x scalar in the question.) So you need to express the vector distance element along the path in i and j coordinates. When the particle moves a distance dy in the y direction, how far does it go in the x direction?
     
  4. Jan 2, 2013 #3
    I kind of understand what you're saying, but not really. I don't even think I understand what the question is asking to be honest. I don't understand how there can be x and y but also i and j coordinate systems.
     
  5. Jan 2, 2013 #4

    haruspex

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    The i and j are unit vectors representing the x and y directions respectively. X and y themselves are magnitudes of position in those directions. I.e. the position vector of the particle at time t is x(t)i + y(t)j.
     
  6. Jan 2, 2013 #5
    I'm honestly trying to do it but since my math isn't that strong and I've never seen a problem like this I'm just having problems. Can you do the first step for me or walk me through a bit more step-by-step? I'm just having extreme difficulty
     
  7. Jan 2, 2013 #6

    haruspex

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    y2 = x; 2ydy = dx
    A small step in position = idx + jdy = 2iydy + jdy
    You know the vector form of F. Take the dot product of this with position differential above.
     
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