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Springs and Spring Energy

  1. Oct 18, 2006 #1
    Hello all, I am looking for some assistance with a physics problem that I have for my physics class. Any help would be greatly appreciated because I have no idea where to start. Thank you all ahead of time for your help.

    The problem is:
    A 10-g mass is attached to the end of an unstressed, light,
    vertical spring (k = 49 N/m) and then dropped. Answer the
    following questions by considering the potential energy due
    to the spring plus the potential energy due to gravity, i.e.
    measure distances from the equilibrium position of the spring
    with no mass attached. (a) What is the maximum speed of the
    falling mass? (b) How far does the mass drop before coming
    to rest momentarily? (c) Repeat (a) and (b), but answer the
    questions by considering the potential energy of the spring
    with the mass attached, i.e. measure distances from the
    equilibrium position of the spring with the mass attached.

    I don't even know how to start with the problem so any help would be most appreciated.

    -Rhaen-
     
  2. jcsd
  3. Oct 18, 2006 #2
    Ok so far all I have been thinking I could do with part a is this...

    (a)
    (1/2)mv^2 + mg(y_1-y_2) = (1/2)mv^2 + mg(y_1-y_2)

    This comes from the equation K_1 + U_1 = K_2 + U_2

    I don't know if I am missing something in that equation though. I know that there will be cancelations, I think the U_2 and K_1 possibly, but I'm not sure exactly how I would calculate the velocity out of that equation. Thank you for any light you can shine on this.

    -Rhaen-
     
  4. Oct 18, 2006 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Don't forget spring potential energy: 1/2 k y^2

    Now apply conservation of energy:
    KE_1 + GPE_1 + SPE_1 = KE_2 + GPE_2 + SPE_2

    Hint: Given the initial speed and your reference point for measure PE, all the terms on the left are zero.
     
  5. Oct 18, 2006 #4
    So then the equation qould go:

    0 = (1/2)mv^2 + mg(y_1-y_2) + (1/2)ky^2
    0 = (.1kg)v^2 + (.2kg)(9.8)(y_1-y_2) + (49)y^2

    If that is the case then how would I calculate the y distance so that I can have only the variable for velocity remaining? Thank you for your time.

    -Rhaen-
     
  6. Oct 18, 2006 #5

    Doc Al

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    Staff: Mentor

    I would write the equation like this:
    0 = (1/2)mv^2 + mgy + (1/2)ky^2

    You are measuring the potential energy from the y=0 (unstretched) point. Now isolate the KE to one side, giving KE as a function of y. The find what value of y maximizes the KE. (The answer will be some negative value for y.)
     
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