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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 |
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Answer» F = normal force by the incline f = kinetic frictional force 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 mg Sin30 + f = ma using eq-2 mg Sin30 + μmg Cos30 = ma a = g (Sin30 + μCos30) a = (9.8) (Sin30 + (0.1)Cos30) a = 5.75 m/s² consider the upward parallel to incline direction as positive v₀ = initial velocity = 36 km/h = 10 m/s v = final velocity when it comes to stop = 0 m/s a = acceleration = - 5.75 m/s² d = distance travelled before stopping using the equation v² = v²₀ + 2 a d 0² = 10² + 2 (- 5.75) d d = 8.7 m |
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