Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A general way of determining the sign of work?

  1. Oct 16, 2009 #1
    Good evening,

    (NOTE: This post got a little long; I hope someone has enough patience to read it.)
    I self-teach physics to myself, but I don't have a complete book (I only have "Volume 2: Gravity, Waves and Thermodynamics" of "Physics" by Paul Tipler). Volume 1 discusses work and energy, but I don't have it.
    So, it becomes much more difficult to understand little details that would otherwise simply be read in a book.
    Another thing: my knowledge of calculus is also self-taught, so it is still basic.
    One of those details is the sign of the work, that is, what signs should be put in the definition of work, depending on the situation, so that the final result has the correct sign according to whether the work is resistent or motive. I understand work as being dW = Fds, where F is the component of force acting in the direction of displacement.
    Do physics books usually discuss the sign of work?
    For example, in the case of the force of gravity. Its absolute value is given by:
    [tex]F = \frac{GMm}{r^2}[/tex].
    But every site that discusses the work of the gravitational force gives this expression:
    [tex]W = \int^{r_2}_{r_1} -\frac{GMm}{r^2} dr = \frac{GMm}{r_2} - \frac{GMm}{r_1}[/tex]
    Where does the first minus sign come from? I can't accept that it is from the definition of gravitational force, because its definition in absolute value is the one I gave above, and it doesn't have a minus sign.
    What I can accept is that the minus sign arises when the body is moving away from the source of the gravitational field; since the force that acts in the moving body is opposite to the orientation of motion, then you have to put a minus sign in the force to indicate that the work will be resistent, negative. However, in the case of a body approaching the source of gravitational field, the gravitational force acts in the same orientation of motion; in this case, since the work is motive, positive, shouldn't it be dW = +Fds?
    After thinking about it for a long time, I came to the following conclusions:
    (I will represent work done by gravitational force along a radial displacement as dW = Fdr, where F is the absolute value of the gravitational force as I gave above, which, of course, always acts in the radial direction. Also, I will assume that POSITIVE displacement is to move away from the source of gravitational field, and this is an obvious choice, because, if r2 > r1, the displacement r2 - r1 will be positive):
    1. If the displacement is positive ([tex]r_2 - r_1 > 0[/tex]) and force acts in the orientation of motion, dW = +Fdr => W > 0
    2. If the displacement is positive ([tex]r_2 - r_1 > 0[/tex]) and force acts opposite to the orientation of motion, then dW = -Fdr => W < 0
    3. If the displacement is negative ([tex]r_2 - r_1 < 0[/tex]) and force acts in the same orientation of motion, then dW = -Fdr (because the displacement is negative, the force, having the same orientation, has to be negative too; no sign needs to be put in the displacement, because, since it is already negative and minus * minus = +, it will make the final result positive) => W > 0
    4. If the displacement is negative ([tex]r_2 - r_1 < 0[/tex]) and force acts opposite to the orientation of motion, dW = +Fdr (since force is opposite to displacement and displacement is negative, then force is positive) => W < 0
    Only situations 2 and 3 apply to the gravitational force, since it is always oriented towards the source of the gravitational field. And, in both cases, dW = -Fdr. So, that seems to explain very well where that minus sign comes from.
    Is this a correct way to determine the sign of work? Or is there other way?

    Thank you in advance.
     
    Last edited: Oct 16, 2009
  2. jcsd
  3. Oct 16, 2009 #2

    rcgldr

    User Avatar
    Homework Helper

    You explanation seems correct. The sign of work, velocity, ... depend on the frame of reference. In this case, the frame of reference is the center of the object producing gravity, and the distances r1 and r2 are relative to the center of that object.
     
  4. Oct 16, 2009 #3
    Thank you for the answer.
    But this doesn't seem to work very well with elastic force. Take, for example, an object that is connected to a spring. The other side of the spring is fixed to a wall.
    The frame of reference is the position where the spring is at rest, that is, its natural length (this is the point zero). The spring can be compressed or stretched out by the object; then, the spring exerts an elastic force in the object, and the value of this force is F = kx. Applying the same reasoning:
    (NOTE: I am considering the elastic reaction force that the spring applies on the body, and, thus, the work done on the body)
    1. When the spring is stretched out, it exerts in the object a force F that is in the opposite orientation of motion (-). The object goes from 0 to +x. So, dW = -Fdx => W < 0.
      [tex]W = \int^{x}_{0} -kx dx = \frac{-kx^2}{2}[/tex]
    2. When the spring is compressed (-), it exerts in the object a force F that is in the opposite orientation of motion (+). The object goes from 0 to -x. So, dW = +Fdx (since the displacement is already negative, no need for sign here) => W < 0.
      [tex]W = \int^{-x}_{0} kx dx = \frac{kx^2}{2}[/tex]
    As you can clearly see, in the first case, the work is negative, but, in the second case, the work is positive when it should be negative. It seemed that, since in the second case the displacement is negative, it should already make the result negative, as happened with the gravitational force.
    Any ideas?
    Thank you in advance.
    EDIT: error fixed in the elastic force work expression for case 1.
     
    Last edited: Oct 17, 2009
  5. Oct 16, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    No, the work done is positive in both of the cases you considered. You do positive work whether you compress or stretch a spring from its equilibrium position.

    The idea is simple: If the force and the displacement are in the same direction, the work done is positive.

    When you describe the gravitational force law in vector form, there is a minus sign. If the position vector of m with respect to M is r, then the force of gravity on m is in the -r direction. Thus the work done by gravity when m is moved away from M (thus in the +r direction) is negative, since the force and the displacement are in opposite directions.
     
  6. Oct 17, 2009 #5
    Actually, I was talking about the force the spring exerts on the object (I didn't make that very clear in the first post; I've edited it). In this case, the work done is negative, because the elastic force the spring applies in the body is opposite to the displacement of the body.

    Thank you for the responses.
    I thought the minus sign of F in the work expression depended on the orientation of force with respect to the displacement. Instead, the minus sign indicates F is opposite to the orientation of r, not the displacement of m provoked by F. I get it, now.
    In this case, I should always use F = -GMm/r² and F = -kx for calculating work (in the case of F = -kx, the minus sign indicates that F, the restoring force, is always opposite to the direction of x).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A general way of determining the sign of work?
  1. General Work Functions (Replies: 1)

Loading...