Hi Shaista,

For the MCAT, you are expected to know
F = kdx for the force exerted by a spring, and
PE = 1/2kx^2 for the potential energy in a spring

One way to remember these two is that the potential energy equation is the integral of the force equation.

For the passage, it is stated in the first paragraph that E is the proportionality constant, and e (eta) is the strain on the bone.

They then state that hooke's law can be expressed as:

sigma = Ee, which is analogous to F = kdx.

You can then integrate this to get the PE of the spring, where

PE = 1/2Ee^2, which is analogous to PE = 1/2kx^2.

Please let me know if this is unclear!

Katt