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Work as an integral problem

  1. Oct 19, 2006 #1
    Ok so my question is this,

    A force F=(4xi +3yj) acts on an object as the oject moves in the x direction from the origin to x = 5.00m Find the work
    w= (integral sign)F*dr done on the object by the force.

    I'm completely stuck on this one, I know that I should just integrate the dot product of the force vector and displacement vector but I completely forget how to integrate!!! Or am I wrong??
  2. jcsd
  3. Oct 19, 2006 #2
    When you are integrating over the displacement..in what direction is the displacement vector? what are the limits of integration? does that help?

    edit -- also think of an object with a force diagram on it... and what you work against to move it where you want it to go.
  4. Oct 19, 2006 #3
    Ok so I think the object only moves in the x direction, so the only component of the force acting on it is the 4xi component. and the interval of the integral is x=0 to x=5. This kind of helps a little bit, but I'm still confused as to the whole concept of integrals, I did them in calculus but now that its over I completely forget.
  5. Oct 19, 2006 #4
    I think the answer to the problem is that I have to find the area under the triangle formed by the graph of the Force vs Distance. which is 1/2(5*4x) or 1/2(5*20). But I would like to know the other way of doing it using integrals because I know eventually I will need to know them.
  6. Oct 19, 2006 #5


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    The work done by the force equals the change in potential U, where [tex]\vec{F}(x, y)=4x\vec{i} + 3y\vec{j}=\nabla U = \frac{\partial U}{\partial x}\vec{i}+\frac{\partial U}{\partial y}\vec{j}[/tex]. You can obtain the potential U(x, y) by integration now easily. Hence, the work will equal W = U(5, 0) - U(0, 0).
  7. Oct 19, 2006 #6
    I'm sorry I've never seen U before? Are you talking about energy? This is my first physics course and the only other time I've seen energy is in chemistry. How does work relate to U?? and is there a way to use integration in this problem without introducing energy?
  8. Oct 19, 2006 #7


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    Yes, U is potential energy. Since you mentioned integration, this is the only way that crossed my mind, and it is pretty simple. Btw, http://theoryx5.uwinnipeg.ca/mod_tech/node31.html.
  9. Oct 19, 2006 #8
    thanks for the link but i still don't follow what you did, is the symbol in front of U delta? and I still don't understand integration.. I think I'm on the wrong forum for this..
  10. Oct 19, 2006 #9
    your #4 message - correct!

    to integrate, you would integrate 4x, to get 1/2*4x^2 and evaluate from 0 to 5... 1/2*4(25)=50... just like you got.
  11. Oct 19, 2006 #10


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    ...which is correct, since [tex]U(x, y) = 2x^2 + \frac{3}{2}y^2[/tex], and so [tex]W = U(5, 0) - U(0, 0) = 2\cdot 5^2 = 50[/tex] [J].
  12. Oct 19, 2006 #11
    Thanks physics girl, and radou I understand where your coming from now, thanks.
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