Work

When a force is applied to an object and that object moves, work is performed. The work performed from moving an object from position $\vec{r}_1$ to position $\vec{r}_2$ is,

$\large W = \int \limits_{\vec{r}_1}^{\vec{r}_2} \vec{F}\cdot d\vec{r}$ [J].

The units of work are Joules, the same as for energy. If a machine has to perform a certain task, the work associated with that task can be calculated to determine how much energy will be needed to perform that task. Power measured in Watts is the amount of Joules of energy needed per second. By calculating the work associated with a certain task and dividing this by the time needed to complete the task, the power requirement can be determined.

If the force is constant, it can be taken out of the integral,

$\large W = \vec{F}\cdot\int \limits_{\vec{r}_1}^{\vec{r}_2} d\vec{r}=\vec{F}\cdot(\vec{r}_2-\vec{r}_1)$ [J].

In this case, the work is just the scalar product of the force and the distance that the object has moved. If the force is a function of position, the integral has to be evaluated. The scalar product in the definition of work can be written out as the sum of three integrals,

$\large W = \int \limits_{\vec{r}_1}^{\vec{r}_2} \vec{F}\cdot d\vec{r}= \int \limits_{x_1}^{x_2} F_xdx + \int \limits_{y_1}^{y_2} F_ydy + \int \limits_{z_1}^{z_2} F_zdz$ [J].

Each of these integrals must be evaluated. There is an app for evaluating such integrals numerically.

The work is positive if the object moves in the direction that the force is pushing. It is negative if the object moves in the opposite direction that the force is pushing.