Applying Hooke's Law to real life problems

When an elastic object - such as a spring - is stretched, the increased length is called its extension. The extension of an elastic object is directly proportional to the force applied to it
F = -kx

F   = Force (N)
-k  = Spring constant N/m
x   = Extension (m)

Hooke's Law can be explained using a small experiment I carried out at home. Force can also be calculated using the formula F=ma where (m) is mass and (a) is acceleration. A mattress contains springs inside and I wanted to find out the spring constant applied to the mattress.

Acceleration is calculated to be 9.81 m / s²  
I used my sister for this experiment. Her mass is 50kg.
Therefore F = 50 x 9.81 = 490.5 N

For the first formula, I have got the force and the extension, which is the depth of the mattress/length of springs. Therefore to find the spring constant, I will have to rearrange the formula to create the following:

k = F/x

Using the data I have got from the second formula, I now need to apply that to the first formula, which has been rearranged, to find the spring constant.

k = 490.5(N)/0.23(m)
   = 2132.608696 (N/m)
Therefore, a force of 2132.608696 (N/m) would be needed to extend the spring per 0.23m