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Pushing a ball through a viscous fluid

  1. Nov 11, 2007 #1
    We push a ball through a viscous fluid with constant external force. As the ball moves, it compresses a spring. The spring resists compression with an elastic force f=kd, where k is the spring constant. When this force balances the external force, the ball stops moving at d=f/k. Throughout the process, the applied force is fixed, so the work done is fd=f^2/k and energy stored in the spring is 1/2kd^2 or 1/2f^2/k.
    Suppose that we suddenly reduce the external force to a value of f1 that is smaller than the original external force.The ball moves in the opposite direction.
    a. How far does the ball move and how much work it does against the external force f1?
    b. For what constant value of f1 will the useful work be maximal? Show that the useful work output is half of what is stored in the spring =1/4f^2/k.
    c. How could we make this process more efficient?

    Please help with the solution...
    a. The elastic force is equal to the external force + friction force? How do we get the distance? I am confused.
    b. The work that is done on the ball by the spring is 1/4f^2/k. Do we need to include the friction force here when the ball is now moving in the opposite direction?
    c. Free energy transduction is most efficient when it proceeds by the incremental, controlled release of many small constraints. What steps do we need to take to make it more efficient? What are the constraints?
  2. jcsd
  3. Nov 13, 2007 #2


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    Ignore frictional forces. Without the viscous fluid, the ball wouldn't just stop at the equilibrium positions. It would oscillate. The purpose of the fluid is just to damp oscillations, and it does this by taking away the kinetic energy of the ball at an equilibrium point. And the question only asks what work is done against the force f1, not the work done against the fluid. Once you know the work done relative to spring and external forces, you can deduce the energy lost to the fluid using conservation. You don't have to calculate it directly. The point of c) is that there may be another way to change the force from f to f1 (other than a sudden shift) that would let the amount of work done against the external force to increase, and hence the amount of work absorbed by the fluid to decrease.
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