A particle moves in the x-y plane under the action of a force, F =K [ x (x^2+y^2 )^ 3 2 i + y (x^2+y^2 )^ 3 2 j ] 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

Step-by-step solution

According to the question, Force on a particle moves in the x - y plane, F =k [ x (x^2+y^2 )^ 3 / 2 i + y (x^2+y^2 )^ 3 / 2 j ] Now, Putting, x=r , y=r [ From figure, ] gathered F =k [ r (r^2 ^2 +r^2 ^2 )^ 3 / 2 i . . + r (r^2 ^2 +r^2 ^2 )^ 3 / 2 j ] ^2 + ^2 =1 F =k [ r r^3 ( i + j ) ] F = k r^2 [ i + j ] gathered Work done, W= F s( 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.