A car of mass 500 kg is moving up along an inclined surface of slope `1/(25)` at a constant speed of `72 km*h^(-1)`. If the coefficient of friction between the road and the car wheel is 0.1 , find the power of the car engine `(g= 9.8 m*s^(-2)`).

Explanation

Frictional force, `f=mu mg cos theta ,v=72 km*h^(-1) `
`=20 m*s^(1-) , m=500 kg, g=9.8m*s^(-2), mu =0.1 and tan theta =1/(25) ` [Fig.1.28].

`sin theta =1/sqrt(626)`
`and cos theta =(25)/(sqrt(626))`
F= force opposing the motion of the car
`=f+mg sin theta =mg(mu cos theta +sin theta)`
`therefore` Power of the car engine ,
`P=Fv=mg(mucos theta +sintheta)v`
`=500xx9.8xx((0.1xx25)/sqrt(626)+1/sqrt(626))xx20= 13709 J*s^(-1)=13709 W.`