Work and energy

Work is performed by applying a force $\vec{F}$ to push a particle from position $\vec{r}_1$ to position $\vec{r}_2$. If the force is constant, the work is,

$$W = \vec{F}\cdot\left(\vec{r}_2-\vec{r}_1\right)=\vec{F}\cdot\Delta\vec{r}.$$

The units of work are Joules, J = Nm. Theses are the units of energy.

Often the force changes while the particle is being moved. Then the work is written as an integral,

\[ \begin{equation} W = \int\limits_{\vec{r}_1}^{\vec{r}_2}\vec{F}\cdot d\vec{r}\hspace{1cm}\text{[J]}. \end{equation} \]

The integral can be thought of as being the sum of many small displacements $d\vec{r}$. For each small displacement, the force is about constant so the work performed is $dW = \vec{F}\cdot d\vec{r}= F_xdx + F_ydy + F_zdz$. The expression for work can be written out in terms of its the three components.

\[ \begin{equation} W = \int\limits_{x_1}^{x_2}F_x dx + \int\limits_{y_1}^{y_2}F_y dy+\int\limits_{z_1}^{z_2}F_z dz\hspace{1cm}\text{[J]}. \end{equation} \]

Example 1

Consider a linear spring. The force needed to compress a linear spring depends on how far it has been compressed $F = kx$. The work required to compress the spring from $x=0$ to $x=x_1$ is,

$$W = \int\limits_0^{x_1} kxdx= \frac{kx_1^2}{2}.$$

The motion is only in the $x$-direction so the integrals in $y$ and $z$ are zero.

Example 2

The force needed to compress a certain nonlinear spring depends is $F = kx^3$. The work required to compress the spring from $x=0$ to $x=x_1$ is,

$$W = \int\limits_{0}^{x_1} kx^3 dx= \frac{kx_1^4}{4}.$$

Example 3

The force needed to move a particle with charge $q$ in an electric field $\vec{E}$ is $\vec{F} = -q\vec{E}$. The work required to move this particle from $x=0$ to $x=x_1$ is,

$$W = \int\limits_{0}^{x_1} -qE_x dx= -qE_xx_1.$$

The form below will integrate a force in one dimension. This program would have to be used three times to find the work due to moving in the $x$, $y$, and $z$ directions.

$F_x(x)=$  [N]
in the range from $x_1=$  [m] to $x_2=$  [m].

 $x$ [m]    $F_x$ [N]

  

$F_x$ [N]

$x$ [m]

Work is the integral of the force,

$$W = \int\limits_{x_1}^{x_2}F_x dx +W(x_1).$$

This app calculates the work needed to move from $x_1$ to $x_2$. If some work was performed to get to $x_1$ it can be added as the integration constant.

$W(x_1)=$

The integral is calculated numerically using a method called Simpson's rule.

 $x$ [m]   $W$ [J]

  

$W$ [J]

$x$ [m]

Questions