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Negative Work but Average Force positive in this situation?

  1. Jul 21, 2012 #1
    I just want to make sure I'm understanding negative work correctly. An example problem where there is a bullet with an initial velocity of x m/s and a final velocity of 0 m/s. It is stopped by a wood block and the question asks to determine the amount of work done on the bullet by the block. If I use the ΔKE = W formula (we are assuming here that the only work being done is by block so W is just the W of the block) I will get a negative number because KE final is 0 but KE initial is a positive # (using the bullets movement along the positive x axis). Therefore, the work done on the bullet by the block is negative. This makes sense to me in that I understand negative work means the work done is in the opposite direction of the displacement. So in this instance, the change in KE and the amount of work done are the same and are both negative (change in KE is negative not KE itself).

    To determine the average force the block exerts on the bullet I would use the Work equation: F (cos theta)(s). Because this is horizontal but the work being done is opposite the displacment, theta is 180 degrees, correct? Therefore, the average force exerted by the block will not be a negative number, correct? Example: Work= -X joules= Fcos180 x s (positive displacement #). Divide by s, then divide by cos 180 (-1) to get F. Negative divided by negative is positive so average force exerted is positive.

    Do I have this correct? Am I understanding negative work properly and how to correctly use 180 for theta in horizontal situations?
    Last edited: Jul 21, 2012
  2. jcsd
  3. Jul 21, 2012 #2
    The "F" and "s" in that formula are the magnitudes of the force and displacement.
    The magnitude of a vector is positive and this is so in all cases.
    The sign of the force itself depends on the direction that you take positive.
    If you consider positive in the direction of the velocity, then the force is negative (and displacement positive).
    The sign of the work depends on the direction of the force relative to the displacement so it is independent of the choice of positive axis.
  4. Jul 21, 2012 #3
    Okay, if in the W= F x cos theta x s equation, the F and s are the magnitudes of F and s then if I am solving for F using that equation the F I am solving for will just be the magnitude of F therefore positive or negative is not an issue if I am understanding that correctly. I understand that the magnitude of a vector is positive, but I wasn't sure if that equation involved magnitudes of the F and s. Now that I know it does it makes that part more clear. If I was looking for the sign of the force itself in the example I gave, then that would be negative because my sign convention is that positive is in the direction of the velocity as I noted in my first post.

    Now, for the sign of the work. If my equation of W= KE final- KE initial gets me a negative result, then isn't the work done (in this case where the only work being considered is that of the block on the bullet) negative also? When you say "The sign of the work depends on the direction of the force relative to the displacement so it is independent of the choice of positive axis." does this differ from my logic behind the KEf - KEi equation and work being negative in this case?
  5. Jul 21, 2012 #4
    Yes, if the kinetic energy decreases the work done by the net force is negative.
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