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Homework Help: Question on spring constant w/potential energy

  1. Jun 14, 2012 #1
    Hi all. I'm having trouble understanding a homework problem. Here it is...

    1. The problem statement, all variables and given/known data

    The question, from my book, is: "When a mass m sits at rest on a spring, the spring is compressed by a distance d from its undeformed length. Suppose instead tht the mass is released from rest when it barely touches the undeformed spring. Find the distance D that the spring is compressed before it is able to stop the mass. Does D = d?"

    In my own words: there is a spring attached to the ground, sticking up vertically. A person holds an object at the top of the spring, in contact with the spring but not compressing it at all (Position 1). The person lets go, and the mass drops, compresses the spring, until the spring stops the mass (Position 2). Then it goes back and forth until it settles (Position 3).

    2. Relevant equations / The attempt at a solution

    The chapter is "Conservation of Energy" so that's the formula I'm using. I start by finding K, Ugrav and Uel at each position:

    Position 1:
    K = 0 (velocity = 0)
    Ugrav = mg * 0 (we choose the undeformed length of the spring as x = 0)
    Uel = 1/2 * k * 02 (again, x = 0)

    Position 2:
    K = 0 (velocity = 0)
    Ugrav = -mgD (we choose the maximum compression before the spring stops the object as x = D. D is downward so we use a negative sign. Our goal is to find D in terms of d.)
    Uel = 1/2 * k * D2

    Position 3:
    K = 0 (velocity = 0)
    Ugrav = -mgd
    Uel = 1/2 * k * d2

    So far, so good, right? Now, the total mechanical energy at one position is equal to that at any other, right? I might have the terminology wrong, but what it means is:

    E1 = E2 = E3
    ∴ K1 + U1 = K2 + U2 = K3 + U3
    ∴ 0 = 1/2 * k * D2 - mgD = 1/2 * k * d2 - mgd

    Now, in order to solve for D in terms of d, I need to eliminate the other unknown: k. I do it like this:

    E1 = E3
    ∴ 0 = 1/2 * k * d2 - mgd
    ∴ k = 2mg / d


    I can also do this:

    Since F = ma, and F = kx
    then mg = kx
    ∴ k = mg / d

    I have two different possibilities for k.

    According to the book, the k = mg / d result is correct; plugging that into the previous formula using E1 = E2 results in

    0 = 1/2 * k * D2 - mgD
    mgD = mgD2 / 2d
    D = D2 / 2d
    2d = D

    and that is the answer.

    But, if you use k = 2mg / d, then d = D. That's what I'm missing: how do I know what the right way to solve k is? I'm pretty sure I have my formulas and my math right, so is there just a conceptual element that I'm not thinking of? What is it?

    Thanks in advance.
  2. jcsd
  3. Jun 14, 2012 #2
    "∴ 0 = 1/2 * k * D2 - mgD = 1/2 * k * d2 - mgd"

    You are using conservation of energy on two different problems. One is when the system oscillates. The other is when the system is at rest. During oscillation, the kinetic energy is not zero when it passes the neutral point.
  4. Jun 14, 2012 #3
    Okay, I think I get it... because when I say

    1/2 * k * D2 - mgD = 1/2 * k * d2 - mgd

    I'm not specifying when we're at "d"... and obviously it will have a velocity as soon as you let go of the object and it passes that point, whereas if you leave the system alone for an hour and let it... settle... or whatever the term is, then it won't have velocity... and this equation can't account for that.

    We haven't done anything with oscillation so this bit's foreign to me, but I think I see now.
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