A particle moves in the $x-y$ plane under the action of a force, $\mathbf{F}=K\left[\frac{x}{\left(x^2+y^2\right)^{\frac{3}{2}}} \hat{\mathbf{i}}+\frac{y}{\left(x^2+y^2\right)^{\frac{3}{2}}} \hat{\mathbf{j}}\right]$ where, $K$ is a constant. Work done by the force when the particle moves from $(0,…

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A particle moves in the $x-y$ plane under the action of a force, $\mathbf{F}=K\left[\frac{x}{\left(x^2+y^2\right)^{\frac{3}{2}}} \hat{\mathbf{i}}+\frac{y}{\left(x^2+y^2\right)^{\frac{3}{2}}} \hat{\mathbf{j}}\right]$ where, $K$ is a constant. Work done by the force when the particle moves from $(0, a)$ to $(a, 0)$ along a circular path of radius $a$ about the origin is

Aicharya · Accepted Answer

According to the question,

Force on a particle moves in the $x$ - $y$ plane, $ \mathbf{F}=k\left[\frac{x}{\left(x^2+y^2\right)^{3 / 2}} \hat{\mathbf{i}}+\frac{y}{\left(x^2+y^2\right)^{3 / 2}} \hat{\mathbf{j}}\right] $ Now, Putting, $x=r \cos \theta, y=r \sin \theta[\because$ From figure, $]$ $ \begin{gathered} \mathbf{F}=k\left[\frac{r \cos \theta}{\left(r^2 \cos ^2 \theta+r^2 \sin ^2 \theta\right)^{3 / 2}} \hat{\mathbf{i}}\right. \\ \left.\quad+\frac{r \sin \theta}{\left(r^2 \cos ^2 \theta+r^2 \sin ^2 \theta\right)^{3 / 2}} \hat{\mathbf{j}}\right] \\ \because \quad \sin ^2 \theta+\cos ^2 \theta=1 \\ \quad \mathbf{F}=k\left[\frac{r}{r^3}(\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}})\right] \\ \quad \mathbf{F}=\frac{k}{r^2}[\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}}] \end{gathered} $ $\therefore$ Work done, $W=\mathbf{F} \times s(\because s=0$, for circle $)$ Now, we can say that the direction of force is along the radius of circle. Hence, the work done by this force will be zero

A particle moves in the $x-y$ plane under the action of a force,…

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