The work done on a particle of mass m by a force K[x(x² + y²)³/²^i + y(x² + y²)³/²^j], where K is a constant of appropriate dimensions, 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 x-y plane is:
2K xa
K xa
0
K x 2a
Solution:
The correct option is 0 dW = 𝐅 ⋅ d𝐫 = 𝐅 ⋅ (dxi^ + dyj^) W = K ∫ₐ₀ xdx/(x² + y²)³/² + ∫₀ₐ ydy/(x² + y²)³/² x² + y² = a² W = K/a³ ∫ₐ₀ xdx + ∫₀ₐ ydy = K/a³ (-a²/2 + a²/2) = 0