Example 64. When an automobile moving with a speed of36 km/h reaches an upward inclined road of angle 30, itsengine is switched off. If the coefficient of friction involved is0.1, how much distance will the automobile move beforecoming to rest ? Take g = 10 ms^-2

Example 64. When an automobile moving with a speed of36 km/h reaches a

1.	Example 64. When an automobile moving with a speed of36 km/h reaches an upward inclined road of angle 30, itsengine is switched off. If the coefficient of friction involved is0.1, how much distance will the automobile move beforecoming to rest ? Take g = 10 ms^-2
Answer» <p>F = normal force by the incline f = kinetic frictional force </p><p>perpendicular to incline, force equation is given as F = mg Cos30 eq-1μ = coefficient of friction = 0.1kinetic frictional force is given as f = μFusing eq-1f = μmg Cos30 eq-2parallel to incline force equation is given as</p><p>mg Sin30 + f = ma </p><p>using eq-2</p><p>mg Sin30 + μmg Cos30 = ma </p><p>a = g (Sin30 + μCos30)</p><p>a = (9.8) (Sin30 + (0.1)Cos30)</p><p>a = 5.75 m/s²</p><p>consider the upward parallel to incline direction as positive </p><p>v₀ = initial velocity = 36 km/h = 10 m/s </p><p>v = final velocity when it comes to stop = 0 m/s </p><p>a = acceleration = - 5.75 m/s²</p><p>d = distance travelled before stopping </p><p>using the equation </p><p>v² = v²₀ + 2 a d </p><p>0² = 10² + 2 (- 5.75) d</p><p>d = 8.7 m </p>