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)`.

Options

A. 12.53m
B. 21.42m
C. 15.43m
D. 8.53m

Answer

D

Explanation

Given, initial speed, u=36km/h
`=(36 xx 1000)/(60xx 60)=10 ms^(-1)`

`theta= 30^(@), mu= 0.1s`= ?
Here, work done in moving up the inclined road = KE of the vehicle
`(mg sin theta + F)s= (1)/(2) m u^(2)`
`(mg sin theta+ muR)xx s= (1)/(2) m u^(2)`
`(mg sin theta + mu mg cos theta) xx s=(1)/(2) m u^(2)`
`s=((1)/(2) mu^(2))/(mg (sin theta + mu cos theta)) = (u^(2))/(2g (sin theta + mu cos theta))`
`=(10 xx 10)/(2 xx 10 xx (sin 30^(@) + 0.1 cos 30^(@)))= 8.53m`