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Retarding Force proportional to the velocity falling particl

  1. Feb 12, 2016 #1
    Example problem from a Classical Mechanics book.
    Find the displacement and velocity of a particle undergoing vertical motion in a medium having a retarding force proportional to the velocity.

    F = m(dv/dt) = -mg-kmv

    dv/(kv + g) = -dt

    (1/k)ln(kv + g) = -t + c

    kv + g = e^(-kt+kc)

    v = (dz/dt) = (-g/k) + ((k(Vo) + g)/k)e^(-kt)

    I dont see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
    I know a property of exponents allows you to separate the exponents powers when they are being added, then each one becomes the power of an exponent which is where e^-kt comes from, but Im not sure about where the other half of the exponent goes.
  2. jcsd
  3. Feb 12, 2016 #2
    Hi Decypher:

    I think y ou have a sigh error. mg is the force of gravity pulling the mass towards rthe Earth, and kmv is a retarding force, slowing down the pull. The direction of these forces are opposite.

    Hope this is helpful.

  4. Feb 12, 2016 #3
    I spent a great amount of time on this part you are mentioning. The book does not provide a great description on this statement.
    "Where -kmv represents a positive upward force since we take z and v = z(dot) to be positive upward, and the motion is downward-that is, v<0, so that -kmv > 0."
    I think they are saying that because we are treating a positive velocity as one that is going higher on the z axis, whe have to treat -kmv as kmv because when a negative velocity is plugged in, kmv becomes an opposing force once again. It makes more sense to me to have it be g-kv just like a force opposing gravity should be.
  5. Feb 12, 2016 #4
    Hi Decypher:

    OK. I think the first quoted sentence would have been clearer if the word "upward" had preceded "vertical".

    The above quote is not what you want to do. I suggest you take the equation
    and fill in the missing step: subtract g from both sides, and divide both sides by k. Then make the substitution v = dz/dt.

    Then you should be able to make a substitution for c (which is an arbitrary constant of integration) which will give you what you want.

    Hope this helps.

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