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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hooke's law f=  k x http://en.wikipedia.org/wiki/Hooke%27s_law

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0F=kx F=50x 0.2 =10 would 10 be the answer ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since 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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Greg_D could you help me with another problem ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sure,, but make another post if it is a different problem...
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