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Velocity emparted to a mass by the decompression of a spring

  1. Mar 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A 1.5 inch spring is .75 inches at solid length when fully compressed. The spring force is rated at 400 lbs per inch. A 3 kilogram mass will be pushed by the spring. What is the velocity of the mass at the instant the spring has fully decompressed. It will take the spring 1/20 of a second to decompress

    2. Relevant equation

    An equation was offered to me by a tutor as follows

    V=(2F*t2)/3.14M

    Where F is the force of the spring in newtons, T2 is the time it takes for the spring to decompress , M is the mass being pushed

    3. The attempt at a solution
    I don't have any trouble solving this equation because I have all the variables. What troubles me is I dont think it makes intuitive sense because of the T2 term. The slower the spring decompresses the more the velocity goes up, which just doesnt seem logical

    I am seeking comments , and possibly an alternative formula.

    The formula above was derived by use of hooks law and the F=MA resolved into a differential equation, and I cant follow the derivation well enough to see if there was a mistake
     
  2. jcsd
  3. Mar 25, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    I don't recognize the given formula, and as you say it doesn't appear to behave properly. I'm thinking that it may be an effort to use a mass-spring oscillator analysis to determine a velocity at a specific time in the cycle. Otherwise I can't see why there appears to be a pi value in it.

    I would suggest that you approach this using conservation of energy and ignore the time.
     
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