Friction: Practice Problem & Solution
Consider a car moving along a straight horizontal road with a speed of \(72\text{ km/h}\). 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 \(g = 10\text{ m/s}^2)\): (1992)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Friction. Understanding the underlying formula is key to arriving at the correct answer below:
Convert speed to m/s: \(v = 72 \times 5/18 = 20\text{ m/s}\). Deceleration due to friction is \(a = \mu_s g = 0.5 \times 10 = 5\text{ m/s}^2\). Using \(v^2 = u^2 + 2as\), \(0 = (20)^2 - 2(5)s\). So, \(10s = 400\) and \(s = 40\text{ m}\).
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