Work Done by Variable Vector Force along Circular Arc

To find the total work done on the particle by the variable force field, we first analyze the given force vector:

$\vec{F} = K \left[ \frac{x}{\left(x^2+y^2\right)^{3/2}} \hat{i} + \frac{y}{\left(x^2+y^2\right)^{3/2}} \hat{j} \right]$

We know that the position vector $\vec{r}$ from the origin in the $x-y$ plane is given by: $\vec{r} = x\hat{i} + y\hat{j}$

The distance $r$ of the particle from the origin is: $r = |\vec{r}| = \sqrt{x^2 + y^2}$

Substituting $\vec{r}$ and $r$ into the expression for $\vec{F}$: $\vec{F} = \frac{K}{r^3} (x\hat{i} + y\hat{j}) = \frac{K}{r^3} \vec{r} = \frac{K}{r^2} \hat{r}$

where $\hat{r} = \frac{\vec{r}}{r}$ is the radial unit vector pointing away from the origin.

Method 1: Using the definition of work done along the path

The infinitesimal work done $dW$ during a displacement $d\vec{r}$ is given by: $dW = \vec{F} \cdot d\vec{r}$

For a particle moving along a circular path of radius $a$ centered at the origin, the direction of motion at any point is along the tangent to the circle ($d\vec{r}$ is tangential). Since the radial unit vector $\hat{r}$ is perpendicular to the tangential displacement $d\vec{r}$ everywhere along the circular arc:

$\hat{r} \cdot d\vec{r} = 0$

Therefore, the work done along the path is: $W = \int_{(a,0)}^{(0,a)} \vec{F} \cdot d\vec{r} = \int_{(a,0)}^{(0,a)} \left(\frac{K}{r^2} \hat{r}\right) \cdot d\vec{r} = 0$

---

Method 2: Using the conservative nature of central forces

Since $\vec{F} = \frac{K}{r^2} \hat{r}$ is a central force, it is conservative. The force can be written as the negative gradient of a scalar potential energy function $U(r)$: $\vec{F} = -\nabla U(r)$

Integrating to find $U(r)$: $U(r) = \frac{K}{r}$

The total work done by a conservative force is equal to the negative change in potential energy: $W = -\Delta U = -\left[ U(r_f) - U(r_i) \right]$

The initial point is $(a,0)$, where $r_i = \sqrt{a^2 + 0^2} = a$. The final point is $(0,a)$, where $r_f = \sqrt{0^2 + a^2} = a$.

Thus: $U(r_i) = \frac{K}{a}$ $U(r_f) = \frac{K}{a}$

Substituting these values into the work equation: $W = -\left( \frac{K}{a} - \frac{K}{a} \right) = 0$

Hence, the total work done is $0$.

Correct Option: D