Question Number 49966 by Necxx last updated on 12/Dec/18
$${A}\:{conveyor}\:{belt}\:{is}\:{driven}\:{at}\:{velocity} \\ $$$${v}\:{by}\:{a}\:{motor}.{Sand}\:{drops}\:{vertically} \\ $$$${on}\:{to}\:{the}\:{belt}\:{at}\:{a}\:{rate}\:{of}\:\boldsymbol{{m}}{kg}/{s}. \\ $$$${What}\:{is}\:{the}\:{additional}\:{power} \\ $$$${needed}\:{to}\:{keep}\:{the}\:{conveyor}\:{belt} \\ $$$${moving}\:{at}\:{a}\:{steady}\:{speed}\:{when} \\ $$$${the}\:{sand}\:{starts}\:{to}\:{fall}\:{on}\:{it}? \\ $$$$\left.{a}\left.\right)\left.\frac{\mathrm{1}}{\mathrm{2}}\left.{mv}\:{b}\right){mv}\:{c}\right)\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \:{d}\right){mv}^{\mathrm{2}} \\ $$
Commented by Necxx last updated on 12/Dec/18
$${please}\:{help}\:{sirs} \\ $$
Answered by mr W last updated on 12/Dec/18
$${in}\:{time}\:\Delta{t}\:{sand}\:{of}\:{mass}\:{m}\Delta{t}\:{drops}\:{on} \\ $$$${to}\:{the}\:{belt}\:{and}\:{will}\:{be}\:{moved}\:{in} \\ $$$${horizontal}\:{direction}\:{from}\:{zero}\:{speed} \\ $$$${to}\:{speed}\:{v}.\:{the}\:{work}\:{done}\:{to}\:{the}\:{sand}\:{is} \\ $$$${W}=\frac{\mathrm{1}}{\mathrm{2}}\left({m}\Delta{t}\right){v}^{\mathrm{2}} \\ $$$${the}\:{power}\:{needed}\:{is}\:{P}=\frac{{W}}{\Delta{t}}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \\ $$
Commented by Necxx last updated on 13/Dec/18
$${thanks}\:{boss} \\ $$