A conveyor belt is moving with constant velocity (V). Sand is being dropped on the belt at the rate of M kg / s . The force necessary to keep the belt moving with a constant velocity Vm / s will be

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According to Newton's second law of motion, the force applied is equal to the rate of change of momentum.

$F = \frac{d p}{d t} = \frac{d ( m v )}{d t}$

Since the velocity of the conveyor belt is constant, $v = V$ and $\frac{d v}{d t} = 0$ .

$F = v \frac{d m}{d t} + m \frac{d v}{d t} = V \frac{d m}{d t}$

Given that sand is dropped at the rate of $M kg/s$ , we have $\frac{d m}{d t} = M$ .

$F = V \times M = M V N$

Answer: MV N

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