A body of mass $$0.5 \text{ kg}$$ travels on a straight line path with velocity $$v = (3x^2 + 4) \text{ m s}^{-1}$$. The net work done by the force during its displacement from $$x = 0$$ to $$x = 2 \text{ m}$$ is

We are given a body of mass $$m = 0.5 \text{ kg}$$ with velocity $$v = (3x^2 + 4) \text{ m/s}$$, and we need to find the net work done as it moves from $$x = 0$$ to $$x = 2 \text{ m}$$.

By the Work-Energy Theorem, the net work done equals the change in kinetic energy, so we have $$W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)$$.

At the initial position $$x = 0$$ the velocity is $$v_i = 3(0)^2 + 4 = 4 \text{ m/s}$$, and at the final position $$x = 2$$ it becomes $$v_f = 3(2)^2 + 4 = 12 + 4 = 16 \text{ m/s}$$.

Substituting these values into the work expression yields $$W = \frac{1}{2} \times 0.5 \times (16^2 - 4^2)$$, which simplifies to $$W = 0.25 \times (256 - 16)$$ and then to $$W = 0.25 \times 240$$, giving $$W = 60 \text{ J}$$.

Hence, the correct answer is Option B.