**Help with Using hookes law?**

if you have your initial length of the spring…then you add different lengths to the spring E.G the spring is 2.10 cm and with 100g the top of the spring is 11.85 bottom is 17.2 therefore new length is 5.35cm…carring on till 1000g.What is the extension but more so how do you calculate the constant or what is the constant…Sorry if the question is confusin not sure how to explain it well

http://en.wikipedia.org/wiki/Hooke%27s_law

This should help. In general, whatever extension (or compression) occurs when a force x is applied is proportional to the force applied. So twice the force generates twice the extension. Three times the force generates three times the extension.

The following did not copy and paste correctly but if you go to the link above you can read it and will likely understand how to determine the spring constant.

“For systems that obey Hooke’s law, the extension produced is directly proportional to the load:

mathbf{F}=-kmathbf{x}

where:

mathbf{x} is the distance that the spring has been stretched or compressed away from the equilibrium position, which is the position where the spring would naturally come to rest (meters),

mathbf{F} is the restoring force exerted by the material (Newtons), and

mathbf{k} is the force constant (or spring constant). The constant has units of force per unit length (newtons per meter).

When this holds, we say that the behavior is linear. If shown on a graph, the line should show a direct variation. There is a negative sign on the right hand side of the equation because the restoring force always acts in the opposite direction of the x displacement (when a spring is stretched to the left, it pulls back to the right).”

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