Power

Suppose that an object moves in a general force-field ${\bf f}({\bf r})$. We now know how to calculate how much energy flows from the force-field to the object as it moves along a given path between two points. Let us now consider the rate at which this energy flows. If $dW$ is the amount of work that the force-field performs on the mass in a time interval $dt$ then the rate of working is given by

$$P = \frac{dW}{dt}.$$ (176)

In other words, the rate of working--which is usually referred to as the power--is simply the time derivative of the work performed. Incidentally, the mks unit of power is called the watt (symbol W). In fact, 1 watt equals 1 kilogram meter-squared per second-cubed, or 1 joule per second.

Suppose that the object displaces by $d{\bf r}$ in the time interval $dt$. By definition, the amount of work done on the object during this time interval is given by

$$d W = {\bf f}\!\cdot\!d{\bf r}.$$ (177)

It follows from Eq. (176) that

$$P = {\bf f}\!\cdot\!{\bf v},$$ (178)

where ${\bf v} = d{\bf r}/dt$ is the object's instantaneous velocity. Note that power can be positive or negative, depending on the relative directions of the vectors ${\bf f}$ and ${\bf v}$. If these two vectors are mutually perpendicular then the power is zero. For the case of 1-dimensional motion, the above expression reduces to

$$P = f v.$$ (179)

In other words, in 1-dimension, power simply equals force times velocity.