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Vertically driven mass against spring

  1. Apr 23, 2010 #1
    Please help!
    Im not sure which is the correct theory to use for this problem.
    A mass 1kg is driven vertically upwards against a spring. The spring has no initial compression and a spring rate of 400N/m. The mass will reach a height of 0.180m.
    What is the input energy if ideal conditions are assumed here.
    I have been given the formula E=mgS+kS²,
    but do not understand why simply adding :
    E(gravity)=mgS and E(spring)=½kS² do not give me the same value.
    Further to this, what would be the effect of adding preload to the spring?
     
  2. jcsd
  3. Apr 23, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks like a typo to me.
    I suspect that's what they meant to do.
     
  4. Apr 23, 2010 #3
    Thanks,
    The derivation for E=mgS+kS² went like this:
    V²=U²+2aS, V=0
    therefore,
    U²=-2aS, where a=g(9.81)
    KE=½mU²
    Substituting gives:
    KE=½m*(2gS) = mgS

    adding in the spring, Fspring=kS
    the deceleration will now be (g+[kS/m])
    which will give:
    KE=[g+(kS/m)]*mS = gms+kS²

    working this through, I cant see any errors but maybe im missing something?
     
  5. Apr 23, 2010 #4

    Doc Al

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    Staff: Mentor

    Here only gravity acts. The acceleration is constant.

    The error here is that the acceleration is no longer constant. Hint: What's the average acceleration as the spring compresses from x = 0 to x = S?
     
  6. Apr 23, 2010 #5
    OK,
    The average acceleration will be ½kS, meaning that this now gives the work done to compress the spring from rest for the second half, and gravitational energy for the first half.
    Excellent, thats what I had expected but had missed the point of it being average. This now makes sense, MANY THANKS.
     
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