The kinetic energy (KE) of an object moving with velocity $V$ and mass $m$ is given by the formula:

$KE = \frac{1}{2} m V^2$

However, in the context of this problem, the object is moving with a constant velocity inclined at $30^\circ$ to the horizontal. The velocity $V$ is the horizontal component of the object's total velocity $v_{\text{total}}$. Therefore, we first need to relate $V$ to the total velocity $v_{\text{total}}$.

The horizontal velocity component can be expressed as:

$V = v_{\text{total}} \cos(30^\circ) = \frac{\sqrt{3}}{2} v_{\text{total}}$

So, the total velocity $v_{\text{total}}$ is:

$v_{\text{total}} = \frac{2V}{\sqrt{3}}$

Now we substitute this into the kinetic energy formula:

$KE = \frac{1}{2} m v_{\text{total}}^2$

Substituting $v_{\text{total}} = \frac{2V}{\sqrt{3}}$:

$KE = \frac{1}{2} m \left( \frac{2V}{\sqrt{3}} \right)^2 = \frac{1}{2} m \cdot \frac{4V^2}{3} = \frac{2mV^2}{3}$

Thus, the correct answer is:

3. $\frac{2mV^2}{3}$

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Would you like more details or have any further questions? Here are 5 related questions to expand on this concept:

1. How does the kinetic energy change if the angle of inclination increases?
2. What happens to the kinetic energy if the mass of the object doubles?
3. How is the vertical velocity component related to the kinetic energy?
4. If the object were moving on a frictionless surface, how would the forces affect its energy?
5. What is the total velocity if the horizontal component is $V$ and the angle is $45^\circ$?

Tip: Always break down inclined motion into components to simplify the analysis of forces and energies.