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Work done of a spring.

  1. Jul 25, 2011 #1
    I have been scratching my head on this problem and hope that someone might solve my problem.
    The question is about work done by a variable force on a spring. When a spring is stretched x distance from its initial position by an external force.Why the work done by the external force is not equal to F(external) * x?
     
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  3. Jul 25, 2011 #2

    ZapperZ

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    Because the force is a variable and varies as a function of x.

    Zz.
     
  4. Jul 25, 2011 #3
    suppose, the natural position (where no force is acting on the spring) is at x=0 (stable). It's obvious that it will not stretch unless some external force is applied. Let us say, the action of any external force will lead to instability. Now, the spring exerts a force against the external force to attain stability. So, more the instability, more the force the spring is going to exert...
     
  5. Jul 25, 2011 #4
    What u meant by function of x?
     
  6. Jul 25, 2011 #5
    Is it because as the spring is stretched more, it takes more force to stretch it?
     
  7. Jul 25, 2011 #6
    as the spring is stretched more, more force is exerted by the spring on the body applying the external force.

    if you apply a continuous and unchanging force F on an object attached spring whose other end is fixed to a wall, the object will first accelerate, till the acceleration becomes 0 (due to the backward force exerted by the spring). When the acceleration becomes 0, the object is not at rest but at a certain velocity. Now, it will start decelerating till the object come to rest.
     

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  8. Jul 25, 2011 #7

    rcgldr

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  9. Jul 25, 2011 #8

    ZapperZ

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    Are you familiar with Hooke's law? I'm a bit puzzled that you are asked to solve a problem using springs, but this appears to be a surprise to you.

    Zz.
     
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