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Describing Work Done on an Object through KE

  1. Jul 2, 2011 #1
    1. The problem statement, all variables and given/known data
    The question asks about an electron moving with energy K1, then says a force acts on the particle so that its velocity is 1/4th of what it was originally, but in the opposite direction. Find the amount of work done on the particle.

    b) Then, decide if the direction of the particle has an effect on your answer.


    2. Relevant equations
    K1-K2=W
    K= 0.5mv^2


    3. The attempt at a solution
    I solved the solution correctly by substituting v2=(-1/4)v1 to find that the work equals
    (-15/16)mv^2. Working on part b), I found that whether I substitute (1/4)v1 or (-1/4)v1, you get the same answer because the quantity is squared, so the work would be the same, which agrees with the answer key. However, I am confused on how this is possible. If I understand this correctly, whether a force causes the electron to slow down to one fourth of its original velocity or whether it is strong enough to actually propel it in the other direction with the same magnitude, because the KE is the same, the work done on the particle is the same.

    This is what I am confused about: This seems to be saying that if a baseball is thrown at 100mph toward a batter, it doesn't matter if the batter whiffs or hits it back at 100 mph, because in both instances the batter works just as hard on the ball because in both cases the ball has the same kinetic energy, so K1-K2 equals zero. Is the above accurate? If so, I guess I am still having issues with the definition of work in the physics sense, but it doesn't seem right to me.
     
  2. jcsd
  3. Jul 2, 2011 #2

    ideasrule

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    Homework Helper

    Yes, that's right. The batter does negative work as he decelerates the ball from 100mph to 0, but he compensates for that by doing positive work while accelerating the ball from 0 to 100mph in the other direction. The total is 0, because the only energy the batter spent was in accelerating the bat to the right speed.

    If this seems counterintuitive, let's suppose that we replace the batter with a large brick wall. Neglecting loses due to the deformation of the ball, the ball reflects off the wall at exactly the same speed it had before, even though the wall does absolutely nothing.
     
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