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Spring and distance?

  1. Jan 6, 2007 #1
    Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate to stiffer springs. An object is attached to the lower end of a 100-coil spring that is hanging from the ceiling. The spring stretches by 0.170 m. The spring is then cut into two identical springs of 50 coils each. As the drawing shows, each spring is attached between the ceiling and the object. By how much does each spring stretch?

    I know the equation F=-Kx applies...I know that it should be 2k and 1/2x from there confused?? * i tryed dividing 0.170 by 2 and times it by 2 ?? what do ??
     
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  3. Jan 6, 2007 #2

    HallsofIvy

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    Your statement "I know that it should be 2k and 1/2x" makes no sense. WHAT should be 2k and 1/2x? And where did you get that "1/2x" from?

    You are told that the spring constant is inversely proportional to the number of coils of the spring of the spring. You are also told that attaching an object of weight w (unknown) it stretches by 0.170 m. That means its initial spring constant is k= w/0.170. Since the number of coils is cut in half, and k is inversely proportional to that, the new k is 2w/0.170. The same weight will be stretched a distance w= (2w/0.170)x so that x= 0.170: half the previous stretch. Prehaps that's where you got that 1/2x but that's the result, not another "multiplier".
     
  4. Jan 6, 2007 #3
    2k and 1/2x ...F= 2k 1/2x in comparison to the 100 coils so that would be for the 50 coils....I am still confused on what to do with the distance
     
  5. Jan 6, 2007 #4

    Doc Al

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    Do it step by step. As Halls' showed, first find the k of the original spring (express it in term of the unknown weight w). Then find the new spring constant of the two shorter springs. Then figure out how much each smaller spring stretches when they support the original weight. (Note that both springs attach to the same weight, so how much of that weight must each spring support?)
     
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