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Homework Help: This guy is stealing my bases

  1. Oct 9, 2011 #1
    This guy is stealing my bases!!!

    nvm delete this please i got it
    Last edited: Oct 9, 2011
  2. jcsd
  3. Oct 9, 2011 #2
    Re: This guy is stealing my bases!!!

    I got the same ballpark (get it?) answer, -661J, using both common methods:

    [itex]W = \int^{x_{f}}_{x_{i}} F_{k} . ds = F_{k} . s = -Fs[/itex]

    where F (the force vector) is given by [itex]\frac{m\Delta v}{\Delta t}[/itex] where Δt is given by [itex]\Delta t = \frac{2\Delta x}{\Delta v}[/itex]

    as well as

    [itex]W_{k} = \Delta K = -\frac{m(\Delta v)^{2}}{2}[/itex]

    so I would say you can most likely enter -664J assuming you minded your p's and q's properly with your significant figures. (The smallest number of significant figures I saw was actually 2, in the 68kg, so that might be an issue.)

    But the work done by friction is by convention usually negative: the force of friction categorically acts in the opposite direction of the displacement, so the dot product of the friction-force function and the differential displacement will always be negative. (Because [itex]\theta = \pi[/itex].)

    EDIT: I'm glad that you were able to get the correct answer on your own. That said, threads where the problem has been solved are often useful to other people who might happen upon it via Google or other means, so that they can see how to derive the solution for a similar problem. Please don't make a habit of deleting your problems and attempted solutions. And when you come up with a solution, please post it and how you got it. For example, in this case where the difference appears to have simply been the sign of your answer, that fact can be very instructive; it's often helpful for first-year physics students to see that the work done by friction has a negative value, and it's often even more helpful for them to see why it has a negative value.
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