A particle of mass 500 g is moving in a straight line with velocity $$v = bx^{\frac{5}{2}}$$. The work done by the net force during its displacement from $$x = 0$$ to $$x = 4$$ m is (Take b = 0.25 m$$^{\frac{-3}{2}}$$s$$^{-1}$$).

A mass $$m = 500$$ g $$= 0.5$$ kg moves with velocity $$v = bx^{5/2}$$ where $$b = 0.25$$ m$$^{-3/2}$$s$$^{-1}$$.

Using the work-energy theorem, the work done equals the change in kinetic energy: $$\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$.

Since at $$x = 0$$ the velocity is $$v_i = b(0)^{5/2} = 0$$, and at $$x = 4$$ m the velocity is $$v_f = b(4)^{5/2} = 0.25 \times 4^{5/2}$$.

Noting that $$4^{5/2} = (4^{1/2})^5 = 2^5 = 32$$, it follows that $$v_f = 0.25 \times 32 = 8 \text{ m/s}$$.

Substituting into the expression for work gives $$W = \frac{1}{2}(0.5)(8)^2 - 0 = \frac{1}{2}(0.5)(64) = 16 \text{ J}$$.

The correct answer is Option D: 16 J.