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#