Work Energy Power

JEE Main level, single correct, medium

155 of 183

Question 155 of 183

Sand is being dropped from a stationary dropper at a rate of 0.5kgs−10.5kg{s}^{-1} on a conveyor belt moving with a velocity of 5ms−15m{s}^{-1}. The power needed to keep belt moving with the same velocity will be

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Answer: (D) 12.5W12.5W

When the sand is dropped on the conveyor belt its velocity will become equal to that of the conveyor. Therefore, v=5ms−1v=5m{s}^{-1}.
The force required to keep the conveyor belt moving will be equal to the gain in the momentum by the sand per second.
⇒F=dmdtv⇒F=0.5×5=2.5N\Rightarrow F=\dfrac{dm}{dt}v \Rightarrow F=0.5\times 5=2.5N
Now the power required to keep the conveyor belt moving will be,
P=Fv⇒P=2.5×5=12.5W.P=Fv \Rightarrow P=2.5\times 5=12.5W.

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