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The product rule applied to the equation for work

  1. Apr 19, 2009 #1
    This is NOT homework....

    If one takes the formula for work, which is force times distance, one can get the formula for power by taking the derivative of it with respect to time. But if we really do this correctly (by using calculus), we get:

    power=distance*(change of force/change in time)+force*(change in distance/change in time)
    power=distance*(change of force/change in time)+force*velocity

    Now we have two terms on the RHS that are not necessarily equal. However, because energy must be conserved, it would seem that that the two terms are somehow separable. In what ways is the first term "physical"? The question refers to the term "distance*(change of force/change in time)" It would suggest that power exists even when force is zero. There is something that does not make sense here....

    Imagine that we do the following:

    power=distance*(change of force/change in time)+force*velocity

    Multiply by distance and divide by velocity:

    power*distance/velocity=(distance^2/velocity)*(change of force/change in time)+force*distance

    (power*distance/velocity)-((distance^2/velocity)*(change of force/change in time))=force*distance

    Now taking the fact that force*distance=work=power*time ....

    (power*distance/velocity)-((distance^2/velocity)*(change of force/change in time))=power*time

    power-distance*(change of force/change in time)=power

    -distance*(change of force/change in time)=0

    There is something that really does not make sense here....




    Let's put it another way.

    Let's assume that power is EQUAL to force*(change in distance/change in time). But in order for this to occur, it cannot equal the time derivative of (force times distance) determined by applying product chain rule (as shown in the first equation of this post). Clearly, we end up trading something that does NOT make sense with another thing that does NOT make sense!

    If we integrate this with respect to time, we are left with the problem of choosing a constant.

    In light of the definition for power:

    time derivative of (force*distance)=distance*(change of force/change in time)+force*(change in distance/change in time)

    time derivative of (force*distance)=distance*(change of force/change in time)+time derivative of work

    time derivative of work=time derivative of (force*distance)-distance*(change of force/change in time)

    This is wacky. There must be something in here that I DON'T understand....
  2. jcsd
  3. Apr 19, 2009 #2

    Doc Al

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    Staff: Mentor

    I would use: dW = F dx, thus p = dW/dt = F dx/dt = Fv. (Where F and x are parallel.)
  4. Apr 19, 2009 #3
    To be honest, it's quite hard to follow because of all the equations have no symbols (velocity = v, etc).
  5. Apr 19, 2009 #4
    The formula for work is better expressed as dW = Fdx, in which F could be taken as a function of x. To evaluate the expression further:
    W = ∫Fdx ---- This is what you would derive. What you were deriving was W = Fx, which is valid for when the force is constant along the distance covered; for a constant force, the expression for power you gave would come down to F*dx/dt = F*v considering that dF/dt = 0.
    Last edited: Apr 19, 2009
  6. Apr 19, 2009 #5


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    Staff Emeritus
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    Homework Helper

    Just wanted to emphasize the following point:

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