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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 mkg/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?  a)(1/2)mv b)mv c)(1/2)mv^2  d)mv^2

$${A}\:{conveyor}\:{belt}\:{is}\:{driven}\:{at}\:{velocity} \\ $$$${v}\:{by}\:{a}\:{motor}.{Sand}\:{drops}\:{vertically} \\ $$$${on}\:{to}\:{the}\:{belt}\:{at}\:{a}\:{rate}\:{of}\:\boldsymbol{{m}}{kg}/{s}. \\ $$$${What}\:{is}\:{the}\:{additional}\:{power} \\ $$$${needed}\:{to}\:{keep}\:{the}\:{conveyor}\:{belt} \\ $$$${moving}\:{at}\:{a}\:{steady}\:{speed}\:{when} \\ $$$${the}\:{sand}\:{starts}\:{to}\:{fall}\:{on}\:{it}? \\ $$$$\left.{a}\left.\right)\left.\frac{\mathrm{1}}{\mathrm{2}}\left.{mv}\:{b}\right){mv}\:{c}\right)\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \:{d}\right){mv}^{\mathrm{2}} \\ $$

Commented by Necxx last updated on 12/Dec/18

please help sirs

$${please}\:{help}\:{sirs} \\ $$

Answered by mr W last updated on 12/Dec/18

in time Δt sand of mass mΔt 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=(1/2)(mΔt)v^2   the power needed is P=(W/(Δt))=(1/2)mv^2

$${in}\:{time}\:\Delta{t}\:{sand}\:{of}\:{mass}\:{m}\Delta{t}\:{drops}\:{on} \\ $$$${to}\:{the}\:{belt}\:{and}\:{will}\:{be}\:{moved}\:{in} \\ $$$${horizontal}\:{direction}\:{from}\:{zero}\:{speed} \\ $$$${to}\:{speed}\:{v}.\:{the}\:{work}\:{done}\:{to}\:{the}\:{sand}\:{is} \\ $$$${W}=\frac{\mathrm{1}}{\mathrm{2}}\left({m}\Delta{t}\right){v}^{\mathrm{2}} \\ $$$${the}\:{power}\:{needed}\:{is}\:{P}=\frac{{W}}{\Delta{t}}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \\ $$

Commented by Necxx last updated on 13/Dec/18

thanks boss

$${thanks}\:{boss} \\ $$

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