Work and Path-Dependent Forces

1. Generalizing the Definition of Work

You know from introductory physics that when a constant force $\vec{F}$ acts on an object that undergoes a straight-line displacement $\vec{d}$, the work done is given by the dot product:

This simple formula breaks down if the force is not constant or if the path is not a straight line. For instance, consider a block sliding along a curved track with friction, or a charged particle moving through a non-uniform electric field.

To handle these situations, we use the core strategy of calculus: we break the complex path into a series of infinitesimally small, essentially straight displacements, which we can call $d\vec{r}$. Over each tiny displacement, the force vector $\vec{F}$ can be considered constant. The small amount of work, $dW$, done over this segment is:

The total work done in moving the object along the entire path $C$ is the sum of all these infinitesimal contributions. This sum is precisely what a line integral represents:

This is the general definition of work done by a vector force field $\vec{F}$ along a path $C$. The integral measures the cumulative effect of the component of the force that acts along the direction of motion at every point on the path.

The following video provides an excellent conceptual bridge from the simple idea of work to this more powerful line integral formulation.

2. The Mechanics of Calculating the Line Integral

The expression $W={\int }_C\vec{F}\cdot d\vec{r}$ is the formal definition, but to actually compute it, we need a systematic procedure. This involves converting the line integral into a standard definite integral with respect to a single variable. The key is to parameterize the path.

Here is the step-by-step method:

1. Parameterize the Path, $\vec{r}(t)$: Describe the curve $C$ as a vector function of a single parameter, typically $t$. For a path in 3D space, this looks like $\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}$, where $t$ ranges from a starting value $a$ to an ending value $b$.

2. Find the Differential Displacement, $d\vec{r}$: Differentiate the parameterized path with respect to the parameter $t$ to get ${\vec{r}}^{\prime }(t)=\frac{d\vec{r}}{dt}$. Then, the differential displacement vector is $d\vec{r}={\vec{r}}^{\prime }(t)dt$.

3. Express the Force Field on the Path, $\vec{F}(\vec{r}(t))$: The force field $\vec{F}(x,y,z)$ is defined throughout space. You must evaluate it specifically along the path by substituting the parameterized components $x(t),y(t),z(t)$ into the expression for $\vec{F}$.

4. Set Up and Solve the Integral: Substitute these components into the line integral definition. The integral transforms into a standard definite integral in the parameter $t$:

   $$W={\int }_C\vec{F}\cdot d\vec{r}={\int }_a^b\vec{F}(\vec{r}(t))\cdot {\vec{r}}^{\prime }(t)dt$$

   This is now a single-variable integral that you can solve using standard calculus techniques.

The following resource provides a concise, text-based explanation of this computational process.

3. Example: Work in a Rotational Field

Let's work through a classic example. Consider a particle moving in a force field given by $\vec{F}(x,y)=y\hat{i}-x\hat{j}$. This is a "vortex" field that tends to push objects in circles around the origin. Let's calculate the work done by this force as a particle moves counter-clockwise along the upper half of a circle of radius 1, from point (1,0) to (-1,0).

1. Parameterize the Path: The unit circle can be parameterized using $t$ as the angle. Since we only want the upper semi-circle from (1,0) to (-1,0), our parameterization is:
   $\vec{r}(t)=\cos (t)\hat{i}+\sin (t)\hat{j}$, for $0\le t\le \pi$.

2. Find $d\vec{r}$:
   ${\vec{r}}^{\prime }(t)=\frac{d\vec{r}}{dt}=-\sin (t)\hat{i}+\cos (t)\hat{j}$.
   So, $d\vec{r}=(-\sin (t)\hat{i}+\cos (t)\hat{j})dt$.

3. Express $\vec{F}$ on the Path: Substitute $x=\cos (t)$ and $y=\sin (t)$ into the force field equation:
   $\vec{F}(\vec{r}(t))=\sin (t)\hat{i}-\cos (t)\hat{j}$.

4. Set Up and Solve: Now we compute the dot product and integrate.

   $$\vec{F}(\vec{r}(t))\cdot {\vec{r}}^{\prime }(t)=(\sin (t))(-\sin (t))+(-\cos (t))(\cos (t))=-{\sin }^2(t)-{\cos }^2(t)=-({\sin }^2(t)+{\cos }^2(t))=-1$$

   The work integral is:

   $$W={\int }_0^{\pi }(-1)dt=[-t{]}_0^{\pi }=-\pi$$

   The work done is $-\pi$. Note that if we moved the particle around the full circle (from $t=0$ to $t=2\pi$), the work would be $-2\pi$. The work done over a closed path is non-zero. This is the defining characteristic of a non-conservative force.

4. Path Dependence: Conservative vs. Non-Conservative Fields

The result of our example leads to a crucial distinction in physics.

• Conservative Force: A force is conservative if the work it does on an object moving between two points is independent of the path taken. Equivalently, the work done by a conservative force over any closed loop is zero. Gravity and ideal spring forces are classic examples. For these forces, we can define a potential energy function, where $W=-\Delta U$.
• Non-Conservative Force: A force is non-conservative if the work it does depends on the path taken. The work done over a closed loop is generally non-zero. Friction and the rotational field from our example are non-conservative. You cannot define a unique potential energy for such forces.

The following video offers a fantastic visual intuition for why some fields are conservative and others are not, by examining their structure.

This concept of path dependence is not just a mathematical curiosity. In the problems you'll be solving, the work done by friction critically depends on the length and shape of the path an object travels. Calculating this work correctly, using a line integral, is often the key to solving the problem.

Conclusion

In this lesson, you have generalized your understanding of work to encompass the most general cases found in mechanics. You've connected the physical concept of work to the mathematical tool of a line integral, which you first studied in a more abstract context.

Here are the key takeaways:

• The work done by a variable force $\vec{F}$ along a curved path $C$ is given by the line integral $W={\int }_C\vec{F}\cdot d\vec{r}$.
• To compute this integral, you must parameterize the path ($\vec{r}(t)$) and convert the line integral into a standard definite integral: $W={\int }_a^b\vec{F}(\vec{r}(t))\cdot {\vec{r}}^{\prime }(t)dt$.
• Force fields are classified based on path dependence. For conservative fields (like gravity), work is path-independent and zero over a closed loop. For non-conservative fields (like friction), work is path-dependent.

You are now equipped to analyze the energetics of motion in any force field, a skill that is fundamental to advanced dynamics.

In our next lesson, we will shift gears to another important topic in advanced mechanics: variable mass systems. We will learn how to apply Newton's second law in situations where the mass of the object is changing, such as in rocket propulsion or a falling chain. This will require setting up and solving new kinds of differential equations.