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Derive an expression for power and velocity. |
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Answer» Solution :The work done by a force `vecF` for a displacement `dvecr` is `W=intvecF.dvecr""...(1)` Left hand side of the equation (1) can be written as `W=intdW=int(dW)/(dt)dt""...(2)` (MULTIPLIED and divided by dt) Since, velocity is `vecv=(dvecr)/(dt),dvecr=vecvdt`. Right hand side of the equation (1) can be written as `intvecF.dvecr=int(vecF(dvecr)/(dt))dt` `=int(vecF.vecv)dt[vecv=(dvecr)/(dt)]""...(3)` Substituting equation (2) and equation (3) in equation (1), we get `int((dW)/(dt)-vecF.vecv)dt=0` This relation is true for any arbitrary value of dt. It is implied that the term within the bracket must be equal to zero, i.e., `(dW)/(dt)-vecF.vecv=0` or `(dW)/(dt)=vecF.vecv` Eg: A vehicle of mass 1250 kg is DRIVEN with an ACCELERATION `0.2ms^(-2)` along a straight level road against an external resistive force 500N. Calculate the power delivered by the vehicle's engine if the velocity of the vehicle is `30MS^(-1)`. The vehicle's engine has to do work against resistive force and make vehicle to move with an acceleration. THEREFORE, power delivered by the vehicle engine is P = (resistive force + mass `xx` acceleration) (velocity) `P=vecF_(-"tot")vecv=(F_("resistive")+F)vecv` `=(F_("resistive")+ma)vecv` `=(500N+(1250kg)xx(0.2ms^(-2)))` Power = 22.5 kW |
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