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===Derivation for a particle in constrained movement=== In particle dynamics, a formula equating work applied to a system to its change in kinetic energy is obtained as a first integral of [[Newton's laws of motion|Newton's second law of motion]]. It is useful to notice that the resultant force used in Newton's laws can be separated into forces that are applied to the particle and forces imposed by constraints on the movement of the particle. Remarkably, the work of a constraint force is zero, therefore only the work of the applied forces need be considered in the work–energy principle. To see this, consider a particle P that follows the trajectory {{math|'''X'''(''t'')}} with a force {{math|'''F'''}} acting on it. Isolate the particle from its environment to expose constraint forces {{math|'''R'''}}, then Newton's Law takes the form <math display="block"> \mathbf{F} + \mathbf{R} = m \ddot{\mathbf{X}}, </math> where {{mvar|m}} is the mass of the particle. ====Vector formulation==== Note that n dots above a vector indicates its nth [[time derivative]]. The [[scalar product]] of each side of Newton's law with the velocity vector yields <math display="block"> \mathbf{F}\cdot\dot{\mathbf{X}} = m\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}},</math> because the constraint forces are perpendicular to the particle velocity. Integrate this equation along its trajectory from the point {{math|'''X'''(''t''<sub>1</sub>)}} to the point {{math|'''X'''(''t''<sub>2</sub>)}} to obtain <math display="block"> \int_{t_1}^{t_2} \mathbf{F} \cdot \dot{\mathbf{X}} dt = m \int_{t_1}^{t_2} \ddot{\mathbf{X}} \cdot \dot{\mathbf{X}} dt. </math> The left side of this equation is the work of the applied force as it acts on the particle along the trajectory from time {{math|''t''<sub>1</sub>}} to time {{math|''t''<sub>2</sub>}}. This can also be written as <math display="block"> W = \int_{t_1}^{t_2} \mathbf{F}\cdot\dot{\mathbf{X}} dt = \int_{\mathbf{X}(t_1)}^{\mathbf{X}(t_2)} \mathbf{F}\cdot d\mathbf{X}. </math> This integral is computed along the trajectory {{math|'''X'''(''t'')}} of the particle and is therefore path dependent. The right side of the first integral of Newton's equations can be simplified using the following identity <math display="block"> \frac{1}{2}\frac{d}{dt}(\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}) = \ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}, </math> (see [[product rule]] for derivation). Now it is integrated explicitly to obtain the change in kinetic energy, <math display="block">\Delta K = m\int_{t_1}^{t_2}\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}dt = \frac{m}{2}\int_{t_1}^{t_2}\frac{d}{dt} (\dot{\mathbf{X}} \cdot \dot{\mathbf{X}}) dt = \frac{m}{2} \dot{\mathbf{X}}\cdot \dot{\mathbf{X}}(t_2) - \frac{m}{2} \dot{\mathbf{X}}\cdot \dot{\mathbf{X}} (t_1) = \frac{1}{2}m \Delta \mathbf{v}^2 , </math> where the kinetic energy of the particle is defined by the scalar quantity, <math display="block"> K = \frac{m}{2} \dot{\mathbf{X}} \cdot \dot{\mathbf{X}} =\frac{1}{2} m {\mathbf{v}^2}</math> ====Tangential and normal components==== It is useful to resolve the velocity and acceleration vectors into tangential and normal components along the trajectory {{math|'''X'''(''t'')}}, such that <math display="block"> \dot{\mathbf{X}}=v \mathbf{T}\quad\text{and}\quad \ddot{\mathbf{X}}=\dot{v}\mathbf{T} + v^2\kappa \mathbf{N},</math> where <math display="block"> v=|\dot{\mathbf{X}}|=\sqrt{\dot{\mathbf{X}}\cdot\dot{\mathbf{X}}}.</math> Then, the [[scalar product]] of velocity with acceleration in Newton's second law takes the form <math display="block"> \Delta K = m\int_{t_1}^{t_2}\dot{v}v \, dt = \frac{m}{2} \int_{t_1}^{t_2} \frac{d}{dt}v^2 \, dt = \frac{m}{2} v^2(t_2) - \frac{m}{2} v^2(t_1),</math> where the kinetic energy of the particle is defined by the scalar quantity, <math display="block"> K = \frac{m}{2} v^2 = \frac{m}{2} \dot{\mathbf{X}} \cdot \dot{\mathbf{X}}. </math> The result is the work–energy principle for particle dynamics, <math display="block"> W = \Delta K. </math> This derivation can be generalized to arbitrary rigid body systems.
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