The minimum force required to pull the block up a smooth inclined plane is equal to the component of its weight along the incline, $F = mg \sin \theta$ . Given: Weight $W = mg = 2 \text{ kN} = 2000 \text{ N}$, length of incline $d = 10 \text{ m}$, and $\theta = 15^\circ$. The required force is $F = 2000 \times \sin 15^\circ = 2000 \times 0.2588 = 517.6 \text{ N}$. The minimum work done is given by $W = F \times d$ . $W = 517.6 \text{ N} \times 10 \text{ m} = 5176 \text{ J} = 5.176 \text{ kJ} \approx 5.17 \text{ kJ}$. Alternatively, work done equals the change in potential energy $= mgh = mg(d \sin \theta) = 2000 \times 10 \times 0.2588 = 5176 \text{ J} = 5.176 \text{ kJ}$.