Animated Solution for Physics - Work, Energy, and Power: A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force $\vec{F}=(\alpha y\hat{i}+2\alpha x\hat{j})\text{N}$, where x and y are in meter and $\alpha =-1{\text{ N/m}}^{-1}$. The work done on the particle by this force $\vec{F}$ will be ____ Joule.

Imagine you are tracking a particle moving through a two-dimensional force field. The particle doesn't just move randomly; it follows a very strict, staircase-like closed path: $A\to B\to C\to D\to E\to F\to A$.

The force acting on this particle is not constant. It changes depending on where the particle is in the xy-plane. The force is given by the vector equation:

We are given that the constant $\alpha =-1$. Substituting this right away, our force field simplifies to:

Our mission is to find the total work done by this force as the particle completes one full lap around this path.

Because the force varies with position, we cannot use the simple $W=F\times d$ formula. We must bring in the heavy artillery: the line integral. The work done by a variable force along a path is the integral of the dot product of the force vector and the infinitesimal displacement vector $d\vec{r}$.

We know that $d\vec{r}=dx\hat{i}+dy\hat{j}$. Taking the dot product with our force vector, we get the master expression for work:

To find the total work, we need to evaluate this integral for each straight-line segment of the path and add them all up.

Let's tackle this journey one segment at a time. The beauty of this staircase path is that along every segment, either $x$ or $y$ is constant, which makes one of the differentials ($dx$ or $dy$) exactly zero!

1. Path AB (Horizontal) Here, the particle moves from $x=0$ to $x=1.0$. The $y$-coordinate is locked at $y=1.0$, which means $dy=0$. Substituting $y=1.0$ and $dy=0$ into our master expression:

2. Path BC (Vertical) Now the particle drops down. The $x$-coordinate is constant at $x=1.0$, so $dx=0$. The $y$-coordinate goes from $1.0$ down to $0.5$. Substituting $x=1.0$ and $dx=0$:

3. Path CD (Horizontal) Moving left, $y$ is constant at $0.5$, so $dy=0$. The $x$-coordinate changes from $1.0$ to $0.5$.

4. Path DE (Vertical) Another drop. $x$ is constant at $0.5$, so $dx=0$. The $y$-coordinate goes from $0.5$ to $0$.

5. Paths EF and FA (The Easy Ones) For path EF, the particle moves along the x-axis where $y=0$ and $dy=0$. Looking at our expression $-ydx-2xdy$, substituting $y=0$ and $dy=0$ makes the whole thing zero! So, $W_{EF}=0$.

Similarly, for path FA along the y-axis, $x=0$ and $dx=0$. This makes the expression zero again. So, $W_{FA}=0$.

We have the work done for every piece of the puzzle. Now, we just sum them up to find the total work done over the closed loop.

The $-1$ and $+1$ beautifully cancel each other out, leaving us with:

Take a step back and look at what we just found. We calculated the work done around a completely closed loop, and the answer was not zero.

This is a profound result. It mathematically proves that the force field $\vec{F}=-y\hat{i}-2x\hat{j}$ is a non-conservative force. If it were a conservative force (like gravity or an ideal spring force), the total work done in returning to the starting point would always be exactly zero, regardless of the path taken. Because our result is $0.75\text{ J}$, this force field behaves more like friction or a swirling fluid, actively adding or removing mechanical energy from the system depending on the trajectory.