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Mathematics for Physics

99 Elastic Potential Energy Formula Derivation

A derivation of the elastic potential energy formula that is used to obtain \(U_{\mathrm{Sp}} = \dfrac{1}{2}k(\Delta s)^{2}\) from the spring force.

definition [d] (Elastic Potential Energy Formula Derivation) From Logan: Hooke’s law is \(F(x) = -kx\), and potential energy is the negative integral of the force,

  • \(V(x) = -\displaystyle\int (-kx)\, dx = \dfrac{1}{2}kx^{2}\) .

where

  • \(F(x)\) is the spring restoring force.
  • \(k\) is the spring constant.
  • \(x\) is the displacement from equilibrium.
  • \(dx\) is the displacement differential.
  • \(V(x)\) is the elastic potential energy.
  • \(\displaystyle\int (-kx)\, dx\) is the integral of the spring force.

definition [d] (Elastic Potential Energy Formula Derivation) From Knight: Hooke’s law is \((F_{\mathrm{Sp}})_{s} = -k\Delta s\). Integrating from \(s_{i}\) to \(s_{f}\) gives

  • \(W = \displaystyle\int_{s_{i}}^{s_{f}}\bigl[-k(s - s_{\mathrm{eq}})\bigr]\, ds = -\left[\dfrac{1}{2}k(\Delta s_{f})^{2} - \dfrac{1}{2}k(\Delta s_{i})^{2}\right]\) .

With \(\Delta U_{\mathrm{Sp}} = -W\),

  • \(U_{\mathrm{Sp}} = \dfrac{1}{2}k(\Delta s)^{2}\) .

where

  • \((F_{\mathrm{Sp}})_{s}\) is the spring force along the stretch coordinate.
  • \(k\) is the spring constant.
  • \(\Delta s\) is the displacement from equilibrium.
  • \(s\) is the position along the spring coordinate.
  • \(s_{\mathrm{eq}}\) is the equilibrium length coordinate.
  • \(s_{i}\) and \(s_{f}\) are the initial and final positions.
  • \(ds\) is the spring-coordinate differential.
  • \(\Delta s_{i}\) and \(\Delta s_{f}\) are the initial and final displacements from equilibrium.
  • \(W\) is the work done by the spring force.
  • \(\Delta U_{\mathrm{Sp}}\) is the change in elastic potential energy.
  • \(U_{\mathrm{Sp}}\) is the elastic potential energy.
  • \(\displaystyle\int_{s_{i}}^{s_{f}}\bigl[-k(s - s_{\mathrm{eq}})\bigr]\, ds\) is the work integral of the spring force.

99.1 Elementary Example

99.1.1 Simple

For \(k = 200\,\mathrm{N/m}\) and stretch \(\Delta s = 0.10\,\mathrm{m}\),

\[ U_{\mathrm{Sp}} = \dfrac{1}{2}k(\Delta s)^{2} = 1\,\mathrm{J} \]

where

  • \(k\) is the spring constant.
  • \(\Delta s\) is the displacement from equilibrium.
  • \(U_{\mathrm{Sp}}\) is the elastic potential energy.

99.1.2 General

Stretching from \(\Delta s_{i} = 0\) to \(\Delta s_{f} = 0.20\,\mathrm{m}\) with the same \(k\) gives

\[ \Delta U_{\mathrm{Sp}} = \dfrac{1}{2}k(\Delta s_{f})^{2} - \dfrac{1}{2}k(\Delta s_{i})^{2} = 4\,\mathrm{J} \]

where

  • \(k\) is the spring constant.
  • \(\Delta s_{i}\) is the initial displacement from equilibrium.
  • \(\Delta s_{f}\) is the final displacement from equilibrium.
  • \(\Delta U_{\mathrm{Sp}}\) is the change in elastic potential energy.