Derive an expression for power and velocity.

Explanation

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