Sand is being dropped from a stationary dropper at a rate of $0.5 \,\mathrm{kgs}^{-1}$ on a conveyor belt moving with a velocity of $5 \mathrm{~ms}^{-1}$. The power needed to keep the belt moving with the same velocity will be :
Solution
<p>${{dm} \over {dt}} = 0.5$ kg/s</p>
<p>$v = 5$ m/s</p>
<p>$F = {{vdm} \over {dt}} = 2.5$ kg m/s<sup>2</sup></p>
<p>$P = \overline F \,.\,\overline v = (2.5)(5)$ W</p>
<p>$= 12.5$ W</p>
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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