anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Hooke's law f= - k x http://en.wikipedia.org/wiki/Hooke%27s_law
anonymous
  • anonymous
F=-kx F=-50x 0.2 =-10 would 10 be the answer ?
anonymous
  • anonymous
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!

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anonymous
  • anonymous
Thank you !!!!
anonymous
  • anonymous
@Greg_D could you help me with another problem ?
anonymous
  • anonymous
sure,, but make another post if it is a different problem...

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