User:Prof McCarthy/work

One dimensional case
For the case in which a body of mass m moves along a straight line, we have Newton's Law, F=ma, and we can compute the power at any instant to be
 * $$Fv = mav.$$

Integrating this power from time t1 to time t2 we obtain the work
 * $$W=\int_{t_1}^{t_2}Fvdt = \int_{t_1}^{t_2}mavdt.$$

The right side of this equation simplifies to be the change in kinetic energy
 * $$\int_{t_1}^{t_2}mavdt = \int_{t_1}^{t_2}m\frac{dv^2}{dt} dt = \frac{mv^2(t_2)}{2} - \frac{mv^2(t_1)}{2} .$$

If the force F is constant, then the work is computed to be
 * $$W=\int_{t_1}^{t_2}Fvdt =\int_{t_1}^{t_2}F\frac{dr}{dt}dt =F(r(t_2)-r(t_1)).$$

or
 * $$W = F\Delta r = \Delta KE.$$

Work and Kinetic Energy
In rigid body dynamics, a formula equating work and the change in kinetic energy of the system is obtained as a first integral of Newton's second law of motion.

To see this, consider a particle P that follows the trajectory X(t) with a force F acting on it. Newton's second law provides a relationship between the force and the acceleration of the particle as
 * $$ \mathbf{F}=m\ddot{\mathbf{X}}, $$

where m is the mass of the particle.

The scalar product of each side of Newton's law with the velocity vector yields
 * $$ \mathbf{F}\cdot\dot{\mathbf{X}} = m\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}},$$

which is integrated from the point X(t1) to the point X(t2) to obtain
 * $$ \int_{t_1}^{t_2} \mathbf{F}\cdot\dot{\mathbf{X}} dt = m\int_{t_1}^{t_2}\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}dt. $$

The left side of this equation is the work of the force as it acts on the particle along the trajectory from time t1 to time t2. This can also be written as
 * $$ W = \int_{t_1}^{t_2} \mathbf{F}\cdot\dot{\mathbf{X}} dt =   \int_{\mathbf{X}(t_1)}^{\mathbf{X}(t_2)} \mathbf{F}\cdot d\mathbf{X}.  $$

This integral is computed along the trajectory 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 identity
 * $$ \frac{1}{2}\frac{d}{dt}(\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}) = \ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}, $$

which can be integrated explicitly to obtain the change in kinetic energy,
 * $$\Delta K = m\int_{t_1}^{t_2}\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}dt = \frac{m}{2}\int_{t_1}^{t_2}\frac{d}{dt}(\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}) dt = \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}(t_2) - \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}} (t_1), $$

where the kinetic energy of the particle is defined by the scalar quantity,
 * $$ K = \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}. $$

The result is the work-energy principle for rigid body dynamics,
 * $$ W = \Delta K. \!$$

This derivation can be generalized to arbitrary rigid body systems.

Using a wedge to hold in place
A wedge is often inserted under a block that cannot rise in order to lock the block in place. Deformation of the wedge provides a vertical force against the block. This vertical force has components normal and tangential to the sloped face of the wedge.