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This question makes no sense ?

  1. Nov 13, 2008 #1
    A spring with spring constant 800 N/m compressed 0.200 m is released and projects a 0.800 kg mass along a frictionless surface. The mass reaches a surface area where mk = 0.400 and comes to a stop.

    I don't need any answers here, I can calculate the force exterted by speed and kinetic energy and everything else related to this statement, but what exactly does "The mass reaches a surface area where mk = 0.400 and comes to a stop" mean? This is supposed to be a frictionless surface, so what would stop the mass from moving?
  2. jcsd
  3. Nov 13, 2008 #2


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    It means that for the first part of the problem, the mass is sliding on a frictionless surface. Then at some point, the ground changes from this magical frictionless surface into a surface that does have friction (and the coefficient of sliding friction between the new surface and mass is given to you as 0.400).
  4. Nov 13, 2008 #3
    ahhh ok.

    These questions that involve physical situations that could never, ever possibly happen are so frustrating. It's supposed to be physics. We're modeling the real world. If they just want people to plug stuff into formulas, they should be teaching math.

    Anyway, thank you.
  5. Nov 13, 2008 #4


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    Well, it's not completely unphysical: you could imagine a surface that has rough spots on it. Something sliding along that surface would see a transition from one coefficient of friction to another. Say it's a puck sliding on an air hockey table and someone kicks out the plug - the puck goes from sliding with very little friction (maybe for the purpose of modeling you could say no friction) to some friction. Or say you had a block sliding on ice that eventually gets to a portion of the ice that wasn't hit by the zamboni so it's all choppy...same deal. Maybe these aren't the best examples, but I hope you see my point: when you do model a real situation, you may have a discontinuity like this to deal with and it makes your model (and the math that goes with it) a little messy.
  6. Nov 13, 2008 #5


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    We can't model the real world! :biggrin:

    (except, of course, by actually building models! :rolleyes:)

    We have to model an approximation!

    Get used to it! :smile:
  7. Nov 13, 2008 #6
    Alright then, approximate this:

    Gilgabog, the Lord of Peptermeed Castle, threw a fireball at Queen Poptarded. Calculate the change in buoyant force on Gilgabog's bunteryup as n approaches infinity. Keep in mind that the force of static friction only applies on Tuesdays (for the purposes of this question only, of course).
  8. Nov 13, 2008 #7


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    I'm sorry, but I really can't let this slide. Mathematics is much more than just 'plugging stuff' into formulas, don't judge what you don't know. In fact 'plugging stuff' into formulas isn't really mathematics.

    Modelling physical phenomena is by no means a trivial exercise, some approximations must be made to arrive at an analytical solution, but that doesn't matter provided that the model produces predictions that agree with observations. For example, in electrostatics we assume that point charges exist because it greatly simplifies problems whilst making predictions that agree with experiment observations. Of course point charges do not exist, but consider the alternative: instead of modelling a current as the flow of point charges through a surface, imaging undertaking the same analysis but replacing the point charges with spheres of non-zero radius which carry some finite charge density.
  9. Nov 14, 2008 #8
    Point taken. I never thought of it as the output of our model agreeing with observations. My understanding of the limit of a function at infinity is the point at which we can no longer observe change, ie any solution we obtain will match what we see in real life, because we can't see beyond a certain precision.

    Good insight.
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