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Thread: Equal and opposite

  1. #1 Equal and opposite 
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    If I had 2 bodies floating in space. Body 1 of Mass 10Kg and body 2 of mass 5Kg and then mass 1 exerted a force of 100N on body 2 both, assuming newtons 3rd law, should begin to accelerate away from one another.

    Using F = Ma then making a (acceleration) the subject of the equation we find that body 1 takes on an acceleration of 10m/s/s. Meanwhile in the other direction body 2 has an a of 20m/s/s.

    So far so good, a smaller mass suffers greater displacement under the same force...

    When I worked out the work done on each object in joules it seemed counter intuitive. J = Fd or work = force x distance. In this case the distance is either 10m in the first second or 20.

    Thusly for body 1. the work done = 100 x 10 or 1000J
    for body 2 the work done is 100 x 20 or 2000J

    In other words twice as much work has been done using the same force simply by moving a smaller object. Why doesnt the formula take into account the mass of the object being moved? I know this cant be right but ive looked up the formula and it seems set in stone.


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  3. #2  
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    In this case the distance is either 10m in the first second or 20.

    Body (1), which is accelerating uniformly at 10 ms<sup>-2</sup>, would travel 5m from rest in one second. Body (2), which is accelerating at 20 ms<sup>-2</sup>. would travel 10m from rest in one second. This is incidental.

    The masses are taken into account in calculating the accelerations and distances moved.

    In other words twice as much work has been done using the same force simply by moving a smaller object.

    If two bodies are subject to the same force, the one that moves the greater distance* will have had more work done on it. There is no requirement that the work done in one direction should be the same as the work done in the opposite direction.

    Work = Force(F) x distance moved(d)

    d = (1/2)at<sup>2</sup> in which a is acceleration

    a = F/m

    so, Work done = (1/2m)F<sup>2</sup>t<sup>2</sup>
    (m is taken into account)

    Putting your numbers into this will give the work done in one second as 500 J for body (1) and 1000 J for body (2).

    * the distances travelled should be measured from the centre of mass of the two bodies, as the position of this shouldn't be changed by a force exerted by one body on the other.


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  4. #3  
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    Think about when you shoot a gun. The bullet gets a lot more kinetic energy than the gun's recoil energy. If it didn't, the gun would put just as big a hurt on you as the bullet does on the target. Your calculation is correct.
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