15.4 An Application of Line Integrals: Work

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An important application of a line integral is to work. Recall that work is the energy transferred to or from an object by a force acting on the object.

Let's review what we already know about work:

• Constant force: The work $W$ done by a constant force $F$ in moving an object a distance $s$ along a straight line in the direction of $F$ is defined to be $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { W=Fs }}$$ When the force $F$ acts in the same direction as the motion, the work done is positive; if the force $F$ acts in a direction opposite to the motion, the work is negative.
• Variable force: In Chapter 6, the definition of work was generalized to the case where the force $F$ is variable. The work $W$ done by a variable force $F=F(x)$ acting in the direction of the motion of an object as that object moves along a straight line from $x=a$ to $x=b$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { W=\int_{a}^{b}F(x)dx }}$$
• Constant vector force: In Chapter 10, we investigated the work $W$ done by a constant force vector $\mathbf{F}$ acting on an object as that object moves in the direction of a vector $\mathbf{r}$ for a distance $\Vert \mathbf{r}\Vert$ and found that $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { W=\mathbf{F}\,{\cdot}\, \mathbf{r} }}$$
• Variable vector force: Now we investigate the work done by a vector field $\mathbf{F}$ (called a force field) acting on an object as that object moves along a smooth curve $C$ from the point $A$ on $C$ to the point $B$ on $C$.

Suppose $$\begin{equation*} \mathbf{F}=\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j} \end{equation*}$$

is a force field acting on an object at the point $(x,y)$ in some open, connected set $R$. We assume the functions $P$ and $Q$ have continuous, first-order partial derivatives at each point in $R$ and that $C$ is a smooth curve lying entirely in $R.$ Finally, suppose $C$ is defined by the vector function $$\begin{equation*} \mathbf{r}=\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\qquad a\leq t\leq b \end{equation*}$$

Partition the closed interval $[a,b]$ into $n$ subintervals $$\begin{equation*} \lbrack a,t_{1}],\quad \lbrack t_{1},t_{2}],\quad \ldots ,\quad \lbrack t_{i-1},t_{i}],\quad \ldots ,\quad \lbrack t_{n-1},b] \end{equation*}$$

and denote the length of each subinterval by $\Delta t_{1},$ $\Delta t_{2},$ $\ldots ,$ $\Delta t_{n}$. Corresponding to each number $a=t_{0},$ $t_{1}, t_{2}, \ldots , t_{n}=b$ of the partition, there is a sequence of points $A=p_{0},$ $p_{1}, \ldots , p_{n}=B$ on the curve $C$. These points subdivide the curve into $n$ subarcs of lengths $\Delta s_{1}, \ \Delta s_{2}, \ldots , \ \Delta s_{n}$, as shown in Figure 29.

The norm $\Vert \Delta \Vert$ of the partition is the largest length of the subarcs $\Delta s_{i},$ $i=1, 2, \ldots , n$. Let $p_{i}^{\ast}=(x_{i}^{\ast },y_{i}^{\ast })$ be a point on the $i$th subarc $p_{i-1}p_{i}$ of $C$. If the norm $\Vert \Delta \Vert$ is small enough, then the work $W_{i}$ done by the force field $\mathbf{F}$ in moving an object along the arc $p{_{i-1}p_{i}}$ can be approximated by the work done by the constant force $$\begin{equation*} \mathbf{F}_{i}^{\ast }=\mathbf{F}(x_{i}^{\ast },y_{i}^{\ast }) \end{equation*}$$

in moving the object along the directed line segment $\Delta \mathbf{r}_{i}=\overrightarrow{p_{i-1}p_{i}}.$ Then $$\begin{equation*} W_{i}\approx \mathbf{F}_{i}^{\ast }\,{\cdot}\, \Delta \mathbf{r}_{i} \end{equation*}$$

The total work $W$ done by $\mathbf{F}$ along $C$ from the point $A$: $\mathbf{r}=\mathbf{r}(a)$ to the point $B$: $\mathbf{r}=\mathbf{r}(b)$ is $W=\lim\limits_{\left\Vert \Delta \right\Vert \rightarrow 0}\sum\limits_{i=1}^{n}W_{i}=\lim\limits_{\left\Vert \Delta \right\Vert \rightarrow 0}\sum\limits_{i=1}^{n}\mathbf{F}_{i}^{\ast }\,{\cdot}\, \Delta \mathbf{r}_{i}.$ This leads to the following definition:

1 Compute Work

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See Figure 30 for the force field $\mathbf{F}(x,y)=-y\mathbf{i}+x\mathbf{j}$ and the curve $C$.

Work and Kinetic Energy

The work $W$ done by a force field $\mathbf{F}$ in moving an object along a piecewise-smooth curve $C$ traced out by $\mathbf{r}=\mathbf{r}(t)$, $a\leq t\leq b$, is $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { W=\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{a}^{b}\mathbf{F} \,{\cdot}\, \mathbf{r}' (t)\,dt }}$$

Using Newton's Second Law of Motion, $\mathbf{F}=m\mathbf{r}'' (t)$, where $\mathbf{r}'' (t)$ is the acceleration of the object, we find that work $W$ can be expressed as $$\begin{eqnarray*} W=\int_{a}^{b}\mathbf{F}\,{\cdot}\, \mathbf{r}' (t)\,dt&\underset{\underset{\color{#0066A7}{\mathbf{F}=m\mathbf{r}'' (t)}}{\color{#0066A7}{\uparrow }}}{=}&\int_{a}^{b}m \mathbf{r}'' (t)\,{\cdot}\, \mathbf{r}' (t)\,dt\underset{\underset{\color{#0066A7} {\dfrac{d}{dt}[\mathbf{r}' (t)\,{\cdot}\, \mathbf{r}' (t)]=2 [\mathbf{r}'' (t)\,{\cdot}\, \mathbf{r}' (t) ]}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{m}{2}\int_{a}^{b}\dfrac{d}{dt}[ \mathbf{r}^{\prime }(t)\,{\cdot}\, \mathbf{r}^{\prime }(t)]\,dt \nonumber\\ &\underset{\underset{\color{#0066A7}{\mathbf{r}^{\prime }\!\cdot\!\mathbf{r}^{\prime } = \left\Vert \mathbf{r}^{\prime }\right\Vert ^{2}}}{\color{#0066A7}{\uparrow }}}{=}&\dfrac{m}{2}\int_{a}^{b}\dfrac{d}{dt}\left\Vert \mathbf{r}' (t) \right\Vert ^{2}dt=\left. \dfrac{m}{2}\left\Vert \mathbf{r' (t)}\right\Vert ^{2}\right] _{a}^{b}\tag{1} \end{eqnarray*}$$

Kinetic energy $K$ is the energy of motion. The kinetic energy $K$ of an object of mass $m$ that moves along a curve with speed $\left\Vert \mathbf{v} \right\Vert$ is given by $$\begin{eqnarray*}\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] {{ K=\dfrac{1}{2}m\Vert \mathbf{v}\Vert ^{2}\underset{\underset{\color{#0066A7}{\hbox{\(\mathbf{v}=\mathbf{r}' (t)\)}}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{m}{2}\left\Vert \mathbf{r' (t)}\right\Vert ^{2} \\[-6.3pt] }}} \tag{2} \end{eqnarray*}$$

By combining (1) and (2), we obtain $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { W=K(b)-K(a) }}$$

which relates work $W$ to kinetic energy $K$. In other words, $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \left[ \begin{eqnarray*} \hbox{Work \(W\) done by a force field \(\mathbf{F}\)} \\ \hbox{in moving an object from \(a\) to \(b\)} \end{eqnarray*}\right] = [ \hbox{Change in the kinetic energy \(K\) from \(a\) to \(b\)}] }}$$

If the curve $C$ is closed and the force field $\mathbf{F}$ is conservative, then it follows that $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \hbox{Work} =\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=0 }}$$

The work done by a conservative force field $\mathbf{F}$ is independent of the path. Physicists paraphrase this by saying that, in a conservative force field, work is a function of position and not path.

Gravitational force is an example of a conservative force field. But not all forces $\mathbf{F}$ are conservative, as we show in Example 2.

Work and Potential Energy

Suppose an object of mass $m$ moves along a smooth curve $C$ traced out by $\mathbf{r}=\mathbf{r}(t)$, $a\leq t\leq b,$ in a conservative force field $\mathbf{F}=\mathbf{F}(x,y)$. The potential energy $U=U(x,y)$ of the object due to $\mathbf{F}$ is the function $U$ for which $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \nabla U=-\mathbf{F} }}$$

That is, $U$ is the negative of the potential function for $\mathbf{F}$. The work $W$ done by $\mathbf{F}$ in moving an object along $C$ from $a$ to $b$ is $$\begin{eqnarray*} W=\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=-\int_{C}{\nabla }U\,{\cdot}\, d\mathbf{r}=- \big[U(x,y)\big] _{a}^{b}=- [U ( \mathbf{r} ( b) ) -U ( \mathbf{r} ( a) ) ] =U ( \mathbf{r} ( a) ) -U ( \mathbf{r}(b)) \end{eqnarray*}$$

This equation relates work $W$ to potential energy $U,$ as stated below. $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \left[ \begin{eqnarray*} \hbox{The work \(W\) done by a force field \(\mathbf{F}\)} \\ \hbox{in moving an object from \(a\) to \(b\)} \end{eqnarray*}\right] =[\hbox{The change in potential energy \(U\) from \(b\) to \(a\)}] }}$$

Now we can state and prove the Law of Conservation of Energy.