The work done on a particle of mass m by a fo | vedclass

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(D) The given force is $\vec{F} = K \left[ \frac{x}{(x^2+y^2)^{3/2}} \hat{i} + \frac{y}{(x^2+y^2)^{3/2}} \hat{j} \right]$.Using polar coordinates,$x =

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(D) The given force is $\vec{F} = K \left[ \frac{x}{(x^2+y^2)^{3/2}} \hat{i} + \frac{y}{(x^2+y^2)^{3/2}} \hat{j} ight]$.Using polar coordinates,$x = r \cos 	heta$ and $y = r \sin 	heta$,where $r^2 = x^2 + y^2$.Substituting these into the force expression: $\vec{F} = K \left[ \frac{r \cos 	heta}{r^3} \hat{i} + \frac{r \sin 	heta}{r^3} \hat{j} ight] = \frac{K}{r^2} (\cos 	heta \hat{i} + \sin 	heta \hat{j})$.Since $\cos 	heta \hat{i} + \sin 	heta \hat{j}$ is the unit radial vector $\hat{r}$,the force is $\vec{F} = \frac{K}{r^2} \hat{r}$.This is a central force acting in the radial direction.For a circular path of radius $a$ centered at the origin,the displacement vector $d\vec{l}$ is always perpendicular to the radial vector $\hat{r}$ (tangential to the circle).Since the force is purely radial and the displacement is purely tangential,the work done $W = \int \vec{F} \cdot d\vec{l} = 0$.

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