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NEET 1992 (AIPMT) · Friction Basics

Friction Basics question from NEET 1992 (AIPMT)

Q38NEET 1992 · AIPMTEasy

Friction Basics

Consider a car moving along a straight horizontal road with a speed of 72 km/h72\text{ km/h}. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (taking g=10 m/s2g = 10\text{ m/s}^{2})

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Answer: (b)

  1. v=72 km/h=20 m/sv = 72\text{ km/h} = 20\text{ m/s}; maximum retardation a=μg=5 m/s2a = \mu g = 5\text{ m/s}^2.
  2. s=v22a=40010s = \dfrac{v^2}{2a} = \dfrac{400}{10}
  3. s=40s = 40 m, so the answer is (b).
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