Derive elastic strain energy?

How do you derive E=1/2kx^2 for the elastic strain energy in a deformed material sample obeying Hooke's law?

1 Answer
May 21, 2017

From Hooke's law,

vecF = -kvecd,

where vecF is the restoring force (the force that brings the spring back to equilibrium, and is thus negative), k is the force constant, and vecd is the positive horizontal displacement from the equilibrium position (defined as zero).

Force is given by units of "N", and energy is given by units of "N"cdot"m", or "N"cdot"m"^2"/m".

Imagine compressing the spring by an infinitesimally small displacement dvecd so that its elastic potential energy increases.

This switches the sign of vecF to be positive, because the direction of compression is the opposite of the direction of the restoring force.

If we integrate the force over a certain compression displacement vecd = vecx_f - vecx_0, where vecx_f is the new length of the compressed spring, and vecx_0 is the equilibrium length of the uncompressed spring, the "elastic strain energy" or elastic potential energy is:

E = int_(vecx_0)^(vecx_f) vecF dvecd

= +int_(vecx_0)^(vecx_f) kvecddvecd = +k/2|[vecd^2]|_(vecx_0)^(vecx_f)

= k/2 (vecx_f^2 - vecx_0^2)

Since we defined vecx_f - vecx_0 as the compression displacement, vecx_f > vecx_0.

If we define E relative to the equilibrium length vecx_0 being our zero, then we redefine the displacement so that:

d^2 = vecx_f^2 - cancel(vecx_0^2)^(0) = (vecx_f - cancel(vecx_0)^(0))^2

E = k/2(vecx_f^2)

=> color(blue)(E = 1/2kd^2)

In terms of x instead of d, the notation is the only thing that changes when we write:

E = 1/2kx^2