What is the spring constant if a 0.25 kg rock stretches a spring by 0.25 m?

AI Thread Summary
To find the spring constant when a 0.25 kg rock stretches a spring by 0.25 m, Hooke's Law (F = -kx) is applicable. The force exerted by the rock due to gravity is calculated using F = mg, where m is the mass and g is the acceleration due to gravity. Substituting the values, the force is 0.25 kg * 9.81 m/s², resulting in 2.4525 N. The spring constant k can then be determined by rearranging Hooke's Law to k = F/x, where x is the stretch of the spring. The final calculation yields a spring constant of approximately 9.81 N/m.
iamtrojan3
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Homework Statement


Hang a 0.25 kg rock from the end of a spring. If the spring stretches 0.25 m, what is the spring constant?


Homework Equations


i'd assume
F/A=Y(Change in L/Lo)


The Attempt at a Solution


I can do it if there was Area, but there isn't. there's probably another equation for this thta i missed in class.
Thanks to anyone tat help:smile:
 
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Looks more like a Hooke's law problem to me??
 
like as in f=-kx. what orce is present besides the springs?
 
yea it is hooks law, thanks again
 
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