Animated Solution for Physics - Work, Energy, and Power: Consider a force $\mathbf{F}=-x\hat{\mathbf{i}}+y\hat{\mathbf{j}}$. The work done by this force in moving a particle from point $A(1,0)$ to $B(0,1)$ along the line segment is (all quantities are in SI units)

Imagine a particle moving on a flat two-dimensional plane. It starts its journey at point $A(1,0)$ on the x-axis and travels in a perfectly straight line to point $B(0,1)$ on the y-axis.

As it moves, it experiences a variable force given by $\mathbf{F}=-x\hat{\mathbf{i}}+y\hat{\mathbf{j}}$. Our goal is to find the total work done by this force during the journey.

To find the work done by a variable force, we must calculate the line integral of the force along the particle's path. The fundamental equation is:

Here, $d\mathbf{r}$ represents an infinitesimally small displacement vector along the path, which in two dimensions is written as $d\mathbf{r}=dx\hat{\mathbf{i}}+dy\hat{\mathbf{j}}$.

Let's substitute our specific force and displacement vectors into the dot product:

Taking the dot product is beautifully straightforward. The $\hat{\mathbf{i}}$ components multiply together, and the $\hat{\mathbf{j}}$ components multiply together:

Notice something incredible here? The $x$ and $y$ terms are completely separated! This means we don't even need to find the equation of the line segment $AB$. We can integrate them independently.

Since the terms are separated, we set up our definite integrals using the initial and final coordinates of the particle. The particle moves from $x=1$ to $x=0$, and from $y=0$ to $y=1$.

Now, we perform the basic polynomial integration:

Let's carefully plug in our upper and lower limits. Watch out for the minus signs!

Evaluating the expression gives us:

Conservative Forces

Did you notice that the work done only depended on the initial and final coordinates, and not on the actual straight-line path we took?

This is the hallmark of a conservative force. Because the force components depended only on their respective coordinates, the line integral was path-independent. If you were to calculate the work done by moving the particle along the x-axis to the origin, and then up the y-axis to point B, you would get the exact same answer!