Gravels are dropped on a conveyor belt at the rate of $0.5 k g / s e c$ . The extra force required in $n e w t o n$ to keep the belt moving at $2 m / s e c$ . is:

1 N

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2 N

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4 N

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0.5 N

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Solution

The correct option is B $1 N$
The net force given on the system is given as:

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

$= u \frac{d m}{d t}$ + $m \frac{d v}{d t}$

Here, v is constant as mentioned in the question itself

Hence, $\frac{d v}{d t} = 0$

Thus, $F = u \frac{d m}{d t} = 2 \times 0.5 = 1 N$

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