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Potential energy of a force vector

  1. Mar 10, 2009 #1
    If F is a conservative force, written in terms of a vector:

    F = kx i + ky j + kz k where k is a constant,

    How do i find the potential energy?

    I know how to use partial derivatives, but im not sure how to integrate a vector. Please Help.
  2. jcsd
  3. Mar 10, 2009 #2
    I think you have somewhat the wrong perspective on this.

    By definition, the potential is a scalar function V(x,y,z) such that
    dV/dx = Fx
    dV/dy = Fy
    dV/dz = Fz
    where those are all partial derivatives, not ordinary derivatives.
    Finding V(x,y,z) is a matter of solving this system of partial differential equations, that is finding a function that has the correct partial derivatives.
  4. Mar 10, 2009 #3
    so i just integrate the components;

    Ux = k/2 x^2

    Uy = k/2 y^2

    Uz = k/2 z^2

    how do i put U as a scalar function????
  5. Mar 10, 2009 #4


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    Staff: Mentor

    1. Choose the point [itex]{\vec r}_0 = x_0 \hat i + y_0 \hat j + z_0 \hat j[/itex] at which you want the potential energy to be zero.

    2. Choose a path between [itex]{\vec r}_0[/itex] and [itex]\vec r = x \hat i + y \hat j + z \hat j[/itex] that makes it easy to evaluate the integral

    [tex]U = - \int^{\vec r}_{{\vec r}_0}{\vec F \cdot d \vec s}[/tex]

    where [itex]d \vec s = dx \hat i + dy \hat j + dz \hat k[/itex]. The force is conservative, so you should get the same result no matter which path you choose. There are no wrong paths, only easier and harder ones. :wink:

    Hint: Often it's best to break up the path into segments, each of which is individually easy to do, rather than use a single "straight-line" path that gives you a messier integral.

    3. Evaluate the product [itex]\vec F \cdot d \vec s[/itex] and perform the integral. If you chose a path in segments, you'll probably have to do this step separately for each segment. Then, of course, you find the sum for all the segments.
  6. Mar 10, 2009 #5
    Thanks a lot, but something that i dont understand is how to choose the segments of the path.... can i just evaluate the indefinite integral?????
  7. Mar 10, 2009 #6
    When you integrated the first part, you should have written
    U = (1/2)*k*x^2 + f(y,z)
    From the second,
    U = (1/2)*k*y^2+g(x,z)
    and from the third,
    U = (1/2)*k*z^2 + h(x,y)
    Then you ask yourself how to satisfy all of these and see what will do the job.
  8. Mar 10, 2009 #7
    yeah i forgot the constants......but i thought i wouldnt need them....because i dont have initial conditions to solve for them. thanks
  9. Mar 10, 2009 #8
    ohhhhhhhh i just need to derive these equations and substitute the constants and find U (x, y, z) Thank you so much guys!!!!!!!!!!!!
  10. Mar 10, 2009 #9
    U (x , y , z) = (1/2)*k*x^2 + (1/2)*k*y^2 + (1/2)*k*z^2 + C im i correct?????
  11. Mar 10, 2009 #10
    Well, you answer the question. Do the partial derivatives give the force components?

    As you observed, you don't have the conditions to evaluate C, so you can make it zero anywhere that is convenient (and meaningful) for the problem.
  12. Mar 10, 2009 #11
    Thank you very much.
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