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Equal and Opposite variable forces on a mass - does it accelerate?

  1. Oct 2, 2008 #1
    1. A mass is attached to a spring that is mounted on the end of an air track. The mass is pushed by your hand along the track, providing a force, F hand. The spring provides a force on the mass, F spring, equal to F hand but opposite in direction. Assume (frictionless?) air track is being used. The forces vary linearly (classic F vs x, work is area under the curve, net work equals zero on the spring/mass system.) How do I explain why the mass moves, if the forces on the mass are always equal and opposite? Is the mass accelerating, or moving at a constant velocity? I am confused!



    2. Relevant equations f(t) = m (dv/dt) ?



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 2, 2008 #2
    If there is a force, there is an acceleration. Think of [tex] F = ma [/tex]

    As for why it moves. You are correct, your hand pushing on the mass is the same force of the mass pushing back on your hand. However, this is looking at it as a closed system. More force is coming outside of your defined system that is 'biasing' the direction of the force (ie. from you).

    I just realised that biasing is a really bad word to use in this context. More force is added from outside the system to overcome the force that is pushing back on your hand. The equal and opposite force for this force is also located outside the system.

    Hope that makes sense.
     
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