Given ⃗F=(xy2)^i+(x2y)^jN. The work done by ⃗F when a particle is taken alo

Discussion Forum : Work Power And Energy

Question -

Given $⃗ F = (x y^{2})^i + (x^{2} y)^j N$ . The work done by $⃗ F$ when a particle is taken along the semicircular path OAB where the coordinates of B are (4,0) is

Options:

A .

\frac{65}{3} J

B .

\frac{75}{2} J

C .

\frac{73}{4} J

D . Zero

Answer: Option D
:
D

Given $⃗ F = (x y^{2})^i + (x^{2} y)^j$

W = $\int F_{x} d x + \int F_{y} d y$

= $\int x y^{2} d x + \int x^{2} y d y$

= $\frac{1}{2} \int d (x^{2} y^{2}) = {[\frac{x^{2} y^{2}}{2}]}_{(0, 0)}^{(4, 0)} = 0$

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