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    kother's Avatar
    kother Posts: 3, Reputation: 1
    New Member
     
    #1

    Jun 27, 2015, 05:51 AM
    SHM
    The upper end of vertical spring of natural length 250 mm is attached to a fixed point. When a small object mass 0.15 kg is attached to the lower end of the spring , the spring stretches to an equilibrium length of 320 mm. calculate the extension of the spring at equilibrium
    Calculate the spring constant
    Curlyben's Avatar
    Curlyben Posts: 18,514, Reputation: 1860
    BossMan
     
    #2

    Jun 27, 2015, 08:25 AM
    What formula do you need for this task ?
    ma0641's Avatar
    ma0641 Posts: 15,675, Reputation: 1012
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    #3

    Jun 27, 2015, 11:34 AM
    What did you calculate? Try Hooke's law. This is a very basic physics experiment that you should know. We will help, not do!
    kother's Avatar
    kother Posts: 3, Reputation: 1
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    #4

    Jun 27, 2015, 11:45 AM
    I don't think it is hooke's law. Can you just give me the formula that I need?

    The answer should be 70mm but I am not sure which formula to use to get to the answer
    ma0641's Avatar
    ma0641 Posts: 15,675, Reputation: 1012
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    #5

    Jun 27, 2015, 11:55 AM
    Consider a simple helical spring that has one end attached to some fixed object, while the free end is being pulled by a force whose magnitude is . Suppose that the spring has reached a state of equilibrium, where its length is not changing anymore. Let be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched). Hooke's law states that
    or, equivalently,

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