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An-object-of-mass-24kg-is-accelerated-up-a-frictionless-plane-inclined-at-an-angle-of-37-Starting-at-the-bottom-from-the-rest-it-covers-a-distance-of-18m-in-3secs-a-what-is-the-average-power-requir




Question Number 28646 by NECx last updated on 28/Jan/18
An object of mass 24kg is  accelerated up a frictionless plane  inclined at an angle of 37°.  Starting at the bottom from the  rest,it covers a distance of 18m  in 3secs.  a)what is the average power  required to accomplish the  process?  b)what is the instantaneous power  required at the end of the 3second  interval?
$${An}\:{object}\:{of}\:{mass}\:\mathrm{24}{kg}\:{is} \\ $$$${accelerated}\:{up}\:{a}\:{frictionless}\:{plane} \\ $$$${inclined}\:{at}\:{an}\:{angle}\:{of}\:\mathrm{37}°. \\ $$$${Starting}\:{at}\:{the}\:{bottom}\:{from}\:{the} \\ $$$${rest},{it}\:{covers}\:{a}\:{distance}\:{of}\:\mathrm{18}{m} \\ $$$${in}\:\mathrm{3}{secs}. \\ $$$$\left.{a}\right){what}\:{is}\:{the}\:{average}\:{power} \\ $$$${required}\:{to}\:{accomplish}\:{the} \\ $$$${process}? \\ $$$$\left.{b}\right){what}\:{is}\:{the}\:{instantaneous}\:{power} \\ $$$${required}\:{at}\:{the}\:{end}\:{of}\:{the}\:\mathrm{3}{second} \\ $$$${interval}? \\ $$
Answered by ajfour last updated on 28/Jan/18
s=(1/2)at^2   ⇒   a=((2s)/t^2 ) =((36)/9)=4m/s^2   v=u+at =0+4×3=12m/s  F−mgsin 37°=ma  ⇒ F=240×(3/5)+24×4            =144+96 =240N  P_(avg) =(W_F /(△t))=((240×18)/3)=1440watt  P_(inst) =Fv =240×4×3 =2880watt .
$${s}=\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\Rightarrow\:\:\:{a}=\frac{\mathrm{2}{s}}{{t}^{\mathrm{2}} }\:=\frac{\mathrm{36}}{\mathrm{9}}=\mathrm{4}{m}/{s}^{\mathrm{2}} \\ $$$${v}={u}+{at}\:=\mathrm{0}+\mathrm{4}×\mathrm{3}=\mathrm{12}{m}/{s} \\ $$$${F}−{mg}\mathrm{sin}\:\mathrm{37}°={ma} \\ $$$$\Rightarrow\:{F}=\mathrm{240}×\frac{\mathrm{3}}{\mathrm{5}}+\mathrm{24}×\mathrm{4}\: \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{144}+\mathrm{96}\:=\mathrm{240}{N} \\ $$$${P}_{{avg}} =\frac{{W}_{{F}} }{\bigtriangleup{t}}=\frac{\mathrm{240}×\mathrm{18}}{\mathrm{3}}=\mathrm{1440}{watt} \\ $$$${P}_{{inst}} ={Fv}\:=\mathrm{240}×\mathrm{4}×\mathrm{3}\:=\mathrm{2880}{watt}\:. \\ $$

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