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Thread: Hooke's Law

  1. #1 Hooke's Law 
    Forum Sophomore
    Join Date
    Sep 2005
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    I have some questions about Hooke's law. I have posted my thoughts, but getting stuck at the end. Can someone please help me? I would appreciate!

    1) An archer pulls her bow string back 0.400m by exerting a force that increases uniformly from 0 to 230N. How much work is done in pulling the bow?

    F=kx, so the equivalent spring constant is 575N/m. The elastic potential energy is (1/2)kx^2=(1/2)(575)(0.400)^2=46J. So is the work done by the archer 46J? I am not sure about that because I was told that WORK DONE and ENERGY are not entirely the same thing.....

    2)

    Total spring constant=65.0x4=260N/m
    Amount of stretch from equilibrium position=0.365m
    F=kx
    =(260)(0.365)
    =94.9N
    Now do I have to divide this by 2 to get 47.4N? I think this is getting really TRICKY. Is the man stretching and applying a force on both sides? Or is he applying a force of 94.9N on each side?


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  3. #2 Re: Hooke's Law 
    Forum Freshman Captain_Anubis's Avatar
    Join Date
    May 2006
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    Quote Originally Posted by kingwinner
    I have some questions about Hooke's law. I have posted my thoughts, but getting stuck at the end. Can someone please help me? I would appreciate!

    1) An archer pulls her bow string back 0.400m by exerting a force that increases uniformly from 0 to 230N. How much work is done in pulling the bow?

    F=kx, so the equivalent spring constant is 575N/m. The elastic potential energy is (1/2)kx^2=(1/2)(575)(0.400)^2=46J. So is the work done by the archer 46J? I am not sure about that because I was told that WORK DONE and ENERGY are not entirely the same thing.....

    2)

    Total spring constant=65.0x4=260N/m
    Amount of stretch from equilibrium position=0.365m
    F=kx
    =(260)(0.365)
    =94.9N
    Now do I have to divide this by 2 to get 47.4N? I think this is getting really TRICKY. Is the man stretching and applying a force on both sides? Or is he applying a force of 94.9N on each side?
    Umm... for the first one there I think that it should be the force constant there, times the distance one to get the force done and then once again to get work (since as far as I remember work = force x distance) making the answer 92J I think... But I wouldn't trust me unless I get some reassurance.. I'm kinda crazy..

    For the second one there I don't think it has to be divided by 2, the 94.9 you calculated would be the equivalent of tension if he were pulling on two ends of a rope... since tension is equal to the force applied to the rope on only one end I'm pretty sure the same counts here, making 94.9 your final answer.... but once again, I may be wrong...


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