Question

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

Answer 1

Hooke's law
f= - k x
http://en.wikipedia.org/wiki/Hooke%27s_law

Answer 2

F=-kx
F=-50x 0.2 =-10 would 10 be the answer ?

Answer 3

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!

Answer 4

Thank you !!!!

Answer 5

@Greg_D could you help me with another problem ?

Answer 6

sure,, but make another post if it is a different problem...