3.2 Work

As force is applied to an object, energy will be added or taken away. We call this energy work. Work is defined by the force applied multiplied by the distance the force was applied over. For a constant force, this can be written mathematically as $W=\boldsymbol{F}\cdot\boldsymbol{s}=\lvert\boldsymbol{F}\rvert\lvert% \boldsymbol{s}\rvert\cos\theta$ (check that this has units of energy!). We can think of work as the mechanism for transferring energy.

If the force is changing continuously over time, we need to be more careful about how we define the work. We break up the path of the object into small sections over which the force is approximately constant and integrate over all these small sections to get the total work.

Definition 3.2.

The work done by a force over an infinitesimal distance $\operatorname{\mathrm{d}}\!\boldsymbol{r}$ is

\operatorname{\mathrm{d}}\!W=\boldsymbol{F}\cdot\operatorname{\mathrm{d}}\!% \boldsymbol{r}.

(3.25)

The total work is then defined as

W=\int\operatorname{\mathrm{d}}\!W=\int_{\boldsymbol{r}_{1}}^{\boldsymbol{r}_{% 2}}\boldsymbol{F}\cdot\operatorname{\mathrm{d}}\!\boldsymbol{r},

(3.26)

where $\boldsymbol{r}_{1}$ and $\boldsymbol{r}_{2}$ are the start and end points of the path being integrated over.

This is a line integral, so evaluating it will involve choosing a coordinate system and splitting the path up into components along each axis. Let’s do a couple of examples now.

3.2.1 Calculating Work Done

In cartesian coordinates, the force vector is given by $\boldsymbol{F}=F_{x}\hat{\boldsymbol{\imath}}+F_{y}\hat{\boldsymbol{\jmath}}+F% _{z}\hat{\boldsymbol{k}}$ and the infinitesimal displacement is $\operatorname{\mathrm{d}}\!\boldsymbol{r}=\operatorname{\mathrm{d}}\!x\hat{% \boldsymbol{\imath}}+\operatorname{\mathrm{d}}\!y\hat{\boldsymbol{\jmath}}+% \operatorname{\mathrm{d}}\!z\hat{\boldsymbol{k}}$ . The dot product $\boldsymbol{F}\cdot\operatorname{\mathrm{d}}\!\boldsymbol{r}$ is $F_{x}\operatorname{\mathrm{d}}\!x+F_{y}\operatorname{\mathrm{d}}\!y+F_{z}% \operatorname{\mathrm{d}}\!z$ . Thus, the integral in equation 3.26 becomes

W=\int_{x_{1}}^{x_{2}}F_{x}\operatorname{\mathrm{d}}\!x+\int_{y_{1}}^{y_{2}}F_% {y}\operatorname{\mathrm{d}}\!y+\int_{z_{1}}^{z_{2}}F_{z}\operatorname{\mathrm% {d}}\!z,

(3.27)

where the limits of the integrals are the components of the start and end points of the path.

Example 3.3.

Consider a force $\boldsymbol{F}=xy\hat{\boldsymbol{\imath}}+x^{2}y\hat{\boldsymbol{\jmath}}$ . What is the work done on an object that moves along a straight line from $(0,0)$ to $(4,3)$ while under the influence of this force.

Note that the components of this force $F_{x}$ and $F_{y}$ both depend on $x$ and $y$ , so to evaluate the integrals we will have to eliminate the other variable by parameterising the path. Luckily this is easy for a straight line, the path can be written as $y=\frac{3}{4}x$ . Then the work is

$\displaystyle W$	$\displaystyle=\int_{0}^{4}xy\operatorname{\mathrm{d}}\!x+\int_{0}^{3}x^{2}y% \operatorname{\mathrm{d}}\!y$	(3.28)
	$\displaystyle=\int_{0}^{4}x\left(\frac{3}{4}x\right)\operatorname{\mathrm{d}}% \!x+\int_{0}^{3}\left(\frac{4}{3}y\right)^{2}y\operatorname{\mathrm{d}}\!y$	(3.29)
	$\displaystyle=\frac{3}{4}\int_{0}^{4}x^{2}\operatorname{\mathrm{d}}\!x+\frac{1% 6}{9}y^{3}\operatorname{\mathrm{d}}\!y$	(3.30)
	$\displaystyle=\left.\frac{1}{4}x^{3}\right\|_{0}^{4}+\left.\frac{4}{9}y^{4}% \right\|_{0}^{3}$	(3.31)
	$\displaystyle=$52\text{\,}\mathrm{J}$.$	(3.32)

Because work is defined by a line integral, its value may change depending on the path taken, even if the start and end points are the same. This is quite often the case. Let’s do the above example again, but have the object follow a different path.

Example 3.4.

Calculate the work done on an object under the influence of the force from example 3.3, but this time following a path from $(0,0)$ to $(4,0)$ , then to $(4,3)$ .

We can break this path up into two separate straight lines to integrate over. The total work done is the sum of the work over both paths. Over the first part, $y=0$ , so $F_{x}=xy=0$ , so no work is done over this section of the path. For the second part, $x=4$ and the work done is

W=\int_{0}^{3}16y\operatorname{\mathrm{d}}\!y=\left.8y^{2}\right|_{0}^{3}=$72% \text{\,}\mathrm{J}$.

(3.33)

Now let’s look at how to calculate work done in polar coordinates, in 2D to begin with. Force in polar coordinates is given by $\boldsymbol{F}=F_{r}\hat{\boldsymbol{r}}+F_{\theta}\hat{\boldsymbol{\theta}}$ , and the infinitesimal displacement is $\operatorname{\mathrm{d}}\!\boldsymbol{r}=\operatorname{\mathrm{d}}\!r\hat{% \boldsymbol{r}}+r\operatorname{\mathrm{d}}\!\theta\hat{\boldsymbol{\theta}}$ . Thus the work done is given by

	$\displaystyle W$	$\displaystyle=\int(F_{r}\hat{\boldsymbol{r}}+F_{\theta}\hat{\boldsymbol{\theta% }})\cdot(\operatorname{\mathrm{d}}\!r\hat{\boldsymbol{r}}+r\operatorname{% \mathrm{d}}\!\theta\hat{\boldsymbol{\theta}})$		(3.34)
		$\displaystyle=\int_{r_{1}}^{r_{2}}F_{r}\operatorname{\mathrm{d}}\!r+\int_{% \theta_{1}}^{\theta_{2}}F_{\theta}r\operatorname{\mathrm{d}}\!\theta.$		(3.35)

3.2 Work

Definition 3.2.

3.2.1 Calculating Work Done

Example 3.3.

Example 3.4.

Example 3.5.