Question

The work done on a particle of mass m by a force,
$\sf \: \vec{F} = k \bigg\{ \dfrac{x}{(x {}^{2} + y {}^{2} ) {}^{ \frac{3}{2} } } \hat{i} +\dfrac{y}{(x {}^{2} + y {}^{2} ) {}^{ \frac{3}{2} } } \hat{j} \bigg \}$
when the particle is taken from the point (a,0) to the point (0,a) along a circular path of radius 'a'
about the origin in the xy plane is : (k is a constant of appropriate dimensions)
a) 2kπ/a
b) kπ/2a
c) zero
d) kπ/a ​

Answer 1

The force acting on the particle is,

$\small\text{ \longrightarrow\vec{\sf{F}}=\sf{\dfrac{kx}{(x^2+y^2)^{\frac{3}{2}}}\,\hat i+\dfrac{ky}{(x^2+y^2)^{\frac{3}{2}}}\,\hat j} }$

The particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the xy plane.

So we can say the particle is taken along the path $\small\sf{x^2+y^2=a^2}$ which is the equation of the circle of radius a with origin as center.

Let the displacement vector of the particle during a small time dt be $\small\text{ \vec{\sf{dr}}=\sf{dx\,\hat i+dy\,\hat j}. }$ The work done on the particle at this time is,

$\small\text{ \longrightarrow\textsf{dW}=\vec{\sf{F}}\cdot\vec{\sf{dr}} }$

$\small\text{ \longrightarrow\sf{dW}=\left(\sf{\dfrac{kx}{(x^2+y^2)^{\frac{3}{2}}}\,\hat i+\dfrac{ky}{(x^2+y^2)^{\frac{3}{2}}}\,\hat j}\right)\cdot\left(\sf{dx\,\hat i+dy\,\hat j}\right) }$

$\small\text{ \longrightarrow\sf{dW=\dfrac{kx\ dx}{(x^2+y^2)^{\frac{3}{2}}}+\dfrac{ky\ dy}{(x^2+y^2)^{\frac{3}{2}}}} }$

$\small\text{ \longrightarrow\sf{dW=\dfrac{k(x\ dx+y\ dy)}{(x^2+y^2)^{\frac{3}{2}}}} }$

Since the path equation $\small\sf{x^2+y^2=a^2}$ holds true,

$\small\text{ \longrightarrow\sf{dW=\dfrac{k(x\ dx+y\ dy)}{(a^2)^{\frac{3}{2}}}} }$

$\small\text{ \longrightarrow\sf{dW=\dfrac{k(x\ dx+y\ dy)}{a^3}} }$

Now the total work done on taking the particle from (a, 0) to (0, a) is,

$\small\text{ \displaystyle\longrightarrow\sf{W=\int\limits_{(a,\ 0)}^{(0,\ a)}\dfrac{k(x\ dx+y\ dy)}{a^3}} }$

$\small\text{ \displaystyle\longrightarrow\sf{W=\dfrac{k}{a^3}\int\limits_{(a,\ 0)}^{(0,\ a)}(x\ dx+y\ dy)} }$

$\small\text{ \displaystyle\longrightarrow\sf{W=\dfrac{k}{a^3}\left[\int\limits_a^0x\ dx+\int\limits_0^ay\ dy\right]} }$

$\small\text{ \displaystyle\longrightarrow\sf{W=\dfrac{k}{a^3}\left[\dfrac{1}{2}\left[x^2\right]_a^0+\dfrac{1}{2}\left[y^2\right]_0^a\right]} }$

$\small\text{ \displaystyle\longrightarrow\sf{W=\dfrac{k}{a^3}\left[-\dfrac{a^2}{2}+\dfrac{a^2}{2}\right]} }$

$\small\text{ \displaystyle\longrightarrow\sf{\underline{\underline{W=0}}} }$

Hence (c) is the answer.