Given : \(\vec{F} = (xy^2)\hat{i} + (x^2 y)\hat{j}\) N.
The work done by \(\vec{F}\) when a particle is taken along the semicircular path OAB where the coordinates of B are (4, 0) is :
Answer & explanation
Correct answer: option 1
\(W = \int F_x.dx + \int F_y.dy\)
\(= \int xy^2.dx + \int x^2ydy\)
\(= \frac{1}{2} \int d(x^2y^2)\)
\(= [\frac{x^2 y^2}{2}]^{(4, 0)}_{(0, 0)} = 0\)