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Hanging Mass on a Spring- Fg = 1/2 Fx

  1. Dec 3, 2011 #1
    Hi everyone! So I'm kind of new to physics forums but I've been having a major problem with this scenario and I can't figure it out. Its not exactly a direction question but more of a situation that doesn't make sense to me...

    Say we have a spring that's hanging vertically. And we say we attach a mass at the end of the spring and the spring stretches to maximum extension. According to Hooke's Law, the force exerted by the spring (Fx) is proportional to the extension of the spring. So Fx=kx. In addition, the force exerted by the spring is equal (but in the opposite direction) to the force applied on the spring.

    Following this situation, we would assume the force exerted on the spring is the Fg of the hanging mass, which would be equal to mg. Therefore, the restoring force of the spring should be equal to the Fg of the mass (mg).

    However, when I work out the problem in terms of energy equations, it appears that the restoring force is actually double the Fg exerted by the object. However,

    My question is this.... how can we prove that the restoring force is double the force exerted by the hanging? Or, am I just doing something incorrect in my energy calculations.

    Thanks!
     
  2. jcsd
  3. Dec 3, 2011 #2

    PeterO

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    Homework Helper

    In this scenario, if you carefully lowered the mass, and stabilised everything, the mass would be hanging on an extended spring. At that point, the force exerted by the spring would equal the weight of the mass.

    If however you are attaching the mass and letting it fall until the spring finally stops the mass, the string will be stretched much more. The mass will also bounce up and down - and in the ideal situation it will even return to its original height before again plunging down.

    A quick analysis of that situation will show that at maximum extension the spring is stretched twice as much as the "carefully lowered to equilibrium" scenario.
     
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