Question

Arrive at an expression for power and velocity. Give some examples for the same.

Answer 1

The work done by a force `vecF` for a displacement `dvecr` is W = `intvecF.vecdr` ......................(i)

Left-hand side of the equation (i) can be written as

W = `int`dW = `int(dW)/dt dt` (multiplied and divided by dt) ......................(ii)

Since, velocity is `vecv = (dvecr)/dt; dvecr = vecv  dt.`

Right hand side of the equation (i) can be written as dt

`intvecF.dvecr = int(vecF.(dvecr)/dt) dt = int(vecf. vecv) dt [vecv = (dvecr)/dt]` ..............(iii)

Substituting equation (ii) and equation (iii) in equation (i), we get

`int(dW)/dt dt = int(vecF.vecv)dt`

`int((dW)/dt - vecF.vecv)dt = 0`

This relation is true for any arbitrary value of dt. This implies that the term within the bracket must be equal to zero, i.e.,

`(dW)/dt - vecF.vecv = 0` or `(dW)/dt = vecF.vecv`

Hence power P = `vecF. vecv`

Example:

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 work against resistive force and make the vehicle move with acceleration. Therefore power delivered by the vehicle engine is

P = (resistive force + mass × acceleration) (velocity)

`P = vec F_"-tot" vecv = (F _(resistive)+ F)vecv`

`= (F_(resistive) + ma)vecv`

`= (500N + (1250kg) xx (0.2ms^-2))`

Power = 22.5 kW