Sand particles drop vertically at the rate of 2 $k g s^{- 1}$ on a conveyor belt moving horizontally with a velocity of 0.2 $m s^{- 1}$ .
The time rate of change of kinetic energy of sand particles is

0.4

J s^{- 1}

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0.8

J s^{- 1}

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0.04

J s^{- 1}

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0.2

J s^{- 1}

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Solution

The correct option is C 0.04 $J s^{- 1}$
Time rate of change of KE
$= \frac{1}{2} \frac{d m}{d t} v^{2}$ $\frac{d m}{d t} = 2. k g / s$
$= \frac{1}{2} \times 2 \times {0.2}^{2}$ $v = 0.2 m / s$
$= 0.04 J / s .$
Hence, the answer is $0.04 J / s o r 0.04 J s^{- 1}$

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Sand particles drop vertically at the rate of 2kgs−1 on a conveyor belt moving horizontally with a velocity of 0.2ms−1.The time rate of change of kinetic energy of sand particles is

The correct option is C 0.04Js−1Time rate of change of KE =12dmdtv2 dmdt=2.kg/s=12×2×0.22 v=0.2m/s=0.04J/s.Hence, the answer is 0.04J/sor0.04Js−1

Sand particles drop vertically at the rate of 2 $k g s^{- 1}$ on a conveyor belt moving horizontally with a velocity of 0.2 $m s^{- 1}$ .
The time rate of change of kinetic energy of sand particles is

The correct option is C 0.04 $J s^{- 1}$
Time rate of change of KE
$= \frac{1}{2} \frac{d m}{d t} v^{2}$ $\frac{d m}{d t} = 2. k g / s$
$= \frac{1}{2} \times 2 \times {0.2}^{2}$ $v = 0.2 m / s$
$= 0.04 J / s .$
Hence, the answer is $0.04 J / s o r 0.04 J s^{- 1}$