Sand is being dropped from a stationary dropper at a rate of 0.5 kgs-1 on a conveyor belt moving with a velocity of 5ms-1. The power needed to keep the belt moving with the same velocity will be:

<!- [if gte mso 9]><xml> </xml><! [endif]-> P = F → . v → = v d P d t = v 2 ( d m d t ) = 0 . 5 × 2 5 = 1 2 . 5 W

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Physics Laws of Motion Physics Class 11th Laws of Motion

Sand is being dropped from a stationary dropper at a rate of 0.5 kgs^-1 on a conveyor belt moving with a velocity of 5ms^-1. The power needed to keep the belt moving with the same velocity will be:

Option 1 -

1.25 W

Option 2 -

2.5 W

Option 3 -

6.25 W

Option 4 -

12.5 W

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$P = \vec{F} . \vec{v} = v \frac{d P}{d t} = v^{2} (\frac{d m}{d t}) = 0.5 \times 25 = 12.5 W$

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