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

Here, initial speed,

u = 36 km/h

= (36 × 1000) / (60 × 60) = 10 ms⁻¹

θ = 30°, μ = 0.1, s = ?

Final velocity, v = 0

Work done in moving up the inclined road = K.E. of the vehicle,

(mg sin θ + F) × s = 1/2 m u²

(mg sin θ + μR) × s = 1/2 m u²

(mg sin θ + μmg cos θ) s = 1/2 m u²

s = (1/2 m u²) / [m g (sin θ + μ cos θ)]

s = u² / [2 g (sin θ + μ cos θ)]

s = (10 × 10) / [2 × 10 (sin 30° + 0.1 cos 30°)]

s = 8.53 m