Question

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?

Answer 1

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`