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Thread: work done by spring force

  1. #1 work done by spring force 
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    Suppose we have a spring with one end fixed and block of mass m is attached with the other free end. This position of the free end is x=0. The spring constant k=50 N/m.
    Now if we apply a constant force of 100N on the block to displace it rightward, the spring stretches and applies leftward force on the block. By Hooks statement we can find the displacement of the block from x=0 and that is x=2 (lets treat x=2 as meters, so 2 meter displacement). After two meter displacement the block stops (because forces are balanced now).

    Now since the block is stationary before and after the displacement so we can apply the work energy theorem, By work energy theorem we get

    Work done by Applied force= 200 J
    Work done by Spring force = -200 J

    Now if we apply the integral formula as integral of kxdx from 0 to 2 we different result that is

    Work done by spring force = -100 J

    Why this is so? Can we not apply the integral formula here? Or whats the problem?

    Suppose we have a spring with one end fixed and block of mass m is attached with the other free end. This position of the free end is x=0. The spring constant k=50 N/m.
    Now if we apply a constant force of 100N on the block to displace it rightward, the spring stretches and applies leftward force on the block. By Hooks statement we can find the displacement of the block from x=0 and that is x=2 (lets treat x=2 as meters, so 2 meter displacement). After two meter displacement the block stops (because forces are balanced now).

    Now since the block is stationary before and after the displacement so we can apply the work energy theorem, By work energy theorem we get

    Work done by Applied force= 200 J
    Work done by Spring force = -200 J

    Now if we apply the integral formula as integral of kxdx from 0 to 2 we different result that is

    Work done by spring force = -100 J

    Why this is so? Can we not apply the integral formula here? Or whats the problem?


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  3. #2  
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    .... To much work done to read....

    Work = Force * Distance...

    I don't see that anywhere..

    Example,

    50N/m over 2 meter is 100N work done.


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  4. #3  
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    Quote Originally Posted by ziaharipur View Post
    Suppose we have a spring with one end fixed and block of mass m is attached with the other free end. This position of the free end is x=0. The spring constant k=50 N/m.
    Now if we apply a constant force of 100N on the block to displace it rightward, the spring stretches and applies leftward force on the block. By Hook’s statement we can find the displacement of the block from x=0 and that is x=2 (lets treat x=2 as meters, so 2 meter displacement). After two meter displacement the block stops (because forces are balanced now).

    Now since the block is stationary before and after the displacement so we can apply the work energy theorem, By work energy theorem we get

    Work done by Applied force= 200 J
    Work done by Spring force = -200 J

    Now if we apply the integral formula as integral of –kxdx from 0 to 2 we different result that is

    Work done by spring force = -100 J

    Why this is so? Can we not apply the integral formula here? Or what’s the problem?
    The problem is that the force you apply will not be a constant 100N. It will vary between 0 at the beginning of travel and 100 at the end of travel, and the average is 50. Or, if you did apply a constant 100N and the spring is providing no opposing force, there would be an acceleration in accordance with Newton's law F=ma. Then when the spring is finally compressed to the point where it is exerting a force or 100N, the mass now has kinetic energy, and will overshoot the 2 meter mark.
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