For a system of particles and for rigid bodies the work \( W \) done by external forces on the system is equal to the change in the kinetic energy \( \Delta K \) of the system: $$ \begin{equation} \tag{4.64} W = \Delta K\qquad \text{for rigid bodies} \end{equation} $$ The rigid body work-energy equation (4.64) is a special result of the more generic mechanical energy balance to be derived in the following. The special rigid body form follow as there are no internal deformations for a rigid body exposed to external forces.
Consider a body of volume \( V \) and surface area \( A \), which is subjected to body forces \( \mathbf{b} \) and surface forces \( \mathbf{t} \). These forces exert work on the body, which per unit time is expressed by: $$ \begin{equation} \tag{4.65} P = \int_V \boldsymbol{b \cdot v} \, \rho \, dV + \int_A \boldsymbol{t \cdot v} \, dA \end{equation} $$
Further, we define the kinetic energy for the body in a natural manner as: $$ \begin{equation} \tag{4.66} K = \int_V \frac{1}{2} \rho \mathbf{v} \cdot \mathbf{v} \, dV \end{equation} $$ and from equation (2.34) we get that the rate of change of the kinetic energy for the body is given by: $$ \begin{equation} \tag{4.67} \dot{K} = \int_V \dot{\boldsymbol{v}} \cdot \boldsymbol{v} \rho \, dV \end{equation} $$ Now, from the integral form of Cauchy's stress theorem (see equation (2.65) one may express: $$ \begin{equation} \tag{4.68} \int_A \mathbf{v\cdot t} \, dA = \int_A \mathbf{v \cdot T \cdot n } \, dA = \int_A (\mathbf{v \cdot T}) \cdot \mathbf{n} \, dA = \int_V \text{div} (\mathbf{v \cdot T}) \, dV \end{equation} $$ The latter expression of equation (4.68) need some consideration: $$ \begin{equation} \tag{4.69} \text{div} (\mathbf{v \cdot T}) = \left (v_i T_{ik} \right)_{,k} = v_{i,k} T_{ik} + v_{i} T_{ik,k} = T_{ik} D_{ik} + v_{i} T_{ik,k} \end{equation} $$ where we have introduced the strain rate tensor \( \mathbf{D} \) as defined in Eqs. (3.78) and (3.79). The expressions in equation (4.69) have the coordinate invariant representation: $$ \begin{equation} \tag{4.70} \text{div} (\mathbf{v \cdot T}) = \mathbf{T:D} + \mathbf{v} \cdot \text{div} \mathbf{T} = \mathbf{T:D} + \mathbf{v} \cdot \left ( \rho \dot{\mathbf{v}} - \rho \mathbf{b} \right) \end{equation} $$ where the latter equality follow from Cauchy's equations of motion (2.89). Further, if we define the stress power \( P_d \) as: $$ \begin{equation} \tag{4.71} P_d = \int_V \mathbf{T:D} \, dV \end{equation} $$
the equations (4.71), (4.70), and (4.68) may be substutited into equation (4.65) to give: The mechanical energy balance equation: $$ \begin{equation} P = \dot{K} + P_d \tag{4.72} \end{equation} $$
One may show (see e.g. p.185 [2] ) that the stress power \( P_d \) represents the deformation work done in the body per unit time. The stress power results in a change in the internal energy, which in part may be recovarable elastic energy, and partly may be represented by an increase in the temperature of the body, and in heat conducted to the surroundings of the body.