When an automobile moving with a speed of `36km//h` reaches an upward inclined road of angle `30^(@)`, its engine is switched off. If the coefficient of friction is 0.1, how much distance will the automobile move before coming to rest ? Take `g=10ms^(-2)`.

Explanation

Here, initial speed,
`u=36km//h=(36xx1000)/(60xx60)ms^(-1)=10ms^(-1)`
`theta =30^(@), mu=0.1, s=?`
Final velocity, `v=0`
Work done is moving up the inclined road
`=` K.E. of the vehicle , figure,
`(mg sintheta +F)xxs=(1)/(2)m u^(2)`
`(mg sintheta+muR)xxs=(1)/(2)m u^(2)`
`(mg sintheta +mu mg cos theta)s=(1)/(2)m u^(2)`
`s=((1)/(2)mu^(2))/(mg(sintheta+mu costheta))=(u^(2))/(2g(sintheta +mucostheta))`
`=(10xx10)/(2xx10(sin30^(@)+0.1cos30^(@)))=8.53m`