Work

Suppose that a particle subject to a force ${\bf f}$ undergoes an infinitesimal displacement $d{\bf r}$. The net work that the force does on the particle (i.e., the net energy transferred to the body by the force) is

$$dW = {\bf f}\cdot d{\bf r}.$$ (1.28)

(See Section A.6.) In other words, the work is the product of the displacement and the component of the force parallel to the displacement. It follows from Equation (1.22) that

$$dW = m\,\frac{d{\bf v}}{dt}\,\cdot d{\bf r}.$$ (1.29)

However, $d{\bf r} = {\bf v}\,dt$, so we obtain

$$dW = m\,\frac{d{\bf v}}{dt}\,\cdot{\bf v}\,dt = m\,{\bf v}\cdot d{\bf v}.$$ (1.30)

Furthermore,

$${\bf v}\cdot d{\bf v} = v_x\,dv_x + v_y\,dv_y + v_z\,dv_z = \frac... ...) + \frac{1}{2}\,d(v_y^{\,2})+ \frac{1}{2}\,d(v_z^{\,2}) = \frac{1}{2}\,d(v^2),$$ (1.31)

where $v=\vert{\bf v}\vert = (v_x^{\,2}+v_y^{\,2}+v_z^{\,2})^{1/2}$ is the particle's speed. It follows from the previous two equations that

$$dW = dK,$$ (1.32)

where

$$K = \frac{1}{2}\,m\,v^{\,2}.$$ (1.33)

Here, $K$ is known as kinetic energy, and is the energy that the particle possesses by virtue of its motion. Equation (1.32) can be integrated to give the work-energy theorem,

$$W= {\mit\Delta} K.$$ (1.34)

According to this theorem, the net work done by the force acting on the particle in a given time interval is equal to the change in the particle's kinetic energy during the same time interval.

Suppose that the force is a function of the particle's displacement, ${\bf r}$. If the particle moves from point $A$ to point $B$ along any path then Equations (1.28) and (1.34) imply that

$$W = \int_{{\bf r}_A}^{{\bf r}_B}{\bf f}\cdot d{\bf r} = K_B-K_A,$$ (1.35)

where ${\bf r}_A$ denotes the displacement of point $A$, et cetera, $K_A$ is the kinetic energy at point $A$, et cetera, and $d{\bf r}$ is an element of the path. (See Section A.14.)