1. Given data

• Mass falling per second: $\dot m = 0.5\ \text{kg s}^{-1}$
• Speed of conveyor belt: $v = 5\ \text{m s}^{-1}$
• Sand is dropped from a stationary dropper, so its initial horizontal velocity is $u=0$

2. What happens physically?

When sand lands on the moving belt, it must be accelerated horizontally from $0$ to $v=5\,\text{m s}^{-1}$.

The belt provides this horizontal momentum to the sand. Therefore, an external agent must supply power to the belt to maintain constant speed.

3. Force required on the belt

Rate of change of momentum of sand: $F = \dot m (v-u)$

Substitute values: $F = 0.5(5-0) = 2.5\ \text{N}$

So the resisting force on the belt due to sand is $2.5\,\text{N}$.

4. Power needed to keep belt moving

Power is $P = Fv$

Thus, $P = 2.5 \times 5 = 12.5\ \text{W}$

5. Check using energy method

Each second, $0.5\,$kg of sand is given kinetic energy: $\frac{1}{2}\dot m v^2 = \frac{1}{2}(0.5)(25)=6.25\ \text{W}$

But this is only the increase in kinetic energy of sand. The actual power supplied by the motor is $P = Fv = \dot m v^2 = 12.5\ \text{W}$

because during slipping, some power is dissipated as heat due to friction. Hence the required motor power is not $6.25\,$W.

6. Option check

• A: $1.25\,$W — incorrect
• B: $2.5\,$W — incorrect
• C: $6.25\,$W — incorrect
• D: $12.5\,$W — correct

Therefore, the correct answer is: $\boxed{12.5\ \text{W}}$