1. Given data

• Mass of particle: $m = 500\text{ gm} = 0.5\text{ kg}$
• Velocity as a function of position: $v = b x^{5/2}$
• Constant: $b = 0.25\, \text{m}^{-3/2}\text{s}^{-1}$
• Displacement: from $x=0$ to $x=4\text{ m}$

We need the work done by the net force.

2. Use work-energy theorem

The net work done is equal to the change in kinetic energy: $W_{\text{net}} = \Delta K = \frac12 m v_f^2 - \frac12 m v_i^2$

So we first find the velocities at $x=0$ and $x=4$.

3. Initial velocity at $x=0$

$v_i = b(0)^{5/2} = 0$

4. Final velocity at $x=4$

$v_f = b(4)^{5/2}$

Now, $4^{5/2} = (\sqrt{4})^5 = 2^5 = 32$

Hence, $v_f = 0.25 \times 32 = 8\text{ m/s}$

5. Compute change in kinetic energy

$W_{\text{net}} = \frac12 (0.5)(8^2) - 0$

$W_{\text{net}} = 0.25 \times 64 = 16\text{ J}$

6. Check options

• A: $2\text{ J}$
• B: $4\text{ J}$
• C: $8\text{ J}$
• D: $16\text{ J}$

So the correct option is D.

Final Answer

$\boxed{16\text{ J}}$