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anonymous

  • one year ago

Can someone help me with this problem!!!! A spring has a spring constant of 50N/m. If you stretch the spring 0.2m past its natural length, what force does the spring apply ? A.250n B.49.8n C.10n D.25n

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  1. anonymous
    • one year ago
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    Hooke's law f= - k x http://en.wikipedia.org/wiki/Hooke%27s_law

  2. anonymous
    • one year ago
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    F=-kx F=-50x 0.2 =-10 would 10 be the answer ?

  3. anonymous
    • one year ago
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    since you are stretching the spring the displacement is negative: \[F=-k\Delta x = -50\frac{N}{m}\times (-0.2m) = 10 N\] you just need to correct that sign!

  4. anonymous
    • one year ago
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    Thank you !!!!

  5. anonymous
    • one year ago
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    @Greg_D could you help me with another problem ?

  6. anonymous
    • one year ago
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    sure,, but make another post if it is a different problem...

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