When a belt moves horizontally at a constant speed of $1.5 {m s}^{- 1}$ , gravel is falling on it at $5 {k g s}^{- 1}$ . Then the extra power needed to drive the belt is :

11.25 W

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37.5 W

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75 W

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0.75 W

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Solution

The correct option is A $11.25 W$
Velocity of belt $v = 1.5$ $m s^{- 1}$
Rate at which the mass of gravel falls $\frac{d m}{d t} = 5$ $k g s^{- 1}$

$∴$ Force acting on the gravel $F = \frac{d p}{d t} = \frac{d m}{d t} v = 5 \times 1.5 = 7.5$ $N$
Power needed $P = F v = 7.5 \times 1.5 = 11.25$ $W$

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