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Compression in Accelerating Springs.

  1. Sep 26, 2010 #1
    Suppose I have two masses with mass [tex]M[/tex] each connected together by a spring with spring constant [tex]k[/tex] like this:
    [PLAIN]http://img827.imageshack.us/img827/4896/springs1.png [Broken]

    Then I apply a force of [tex]F[/tex] at both ends:
    [PLAIN]http://img833.imageshack.us/img833/2562/spring2.png [Broken]

    Then using Hooke's law [tex]F=kx[/tex], the compression of the spring [tex]x[/tex] would be given by:

    So now if I apply the same force [tex]F[/tex] just to the right end like this:
    [PLAIN]http://img810.imageshack.us/img810/4236/spring3.png [Broken]

    The acceleration of the system is given by Newton's 2nd Law [tex]F=ma[/tex]
    [tex]F=2Ma [/tex]

    And the acceleration of the mass on the left is the same as the entire system so the force is:
    [tex]F_{left mass}=Ma [/tex]
    [tex]F_{left mass}=M\frac{F}{2M}[/tex]
    [tex]F_{left mass}=\frac{F}{2}[/tex]

    This force must be supplied by the spring so using hookes law:

    This value is half as much as before.

    Here is where I'm confused can anyone explain the difference between this and the situation described in this post http://scienceblogs.com/dotphysics/2008/10/fake-vs-real-forces/" [Broken] where it is said the compression is "the compression is EXACTLY the same before"

    What mistakes have I made?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 27, 2010 #2
  4. Sep 27, 2010 #3
    I'm not worried about that statement.

    After the spring has compressed, the acceleration of the mass on the left is the same as the entire system
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