Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (g = 10 m/s²):

30 m

40 m

72 m

20 m

Step-by-Step Solution

Unit Conversion: First, convert the initial speed from km/h to m/s. $u = 72 \text{ km/h} = 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}$
Deceleration due to Friction: The retarding force is the kinetic friction, $f_k = \mu_k N = \mu_k mg$ . According to Newton's Second Law, the deceleration $a$ is: $a = \frac{f_k}{m} = \frac{\mu_k mg}{m} = \mu_k g$ $a = 0.5 \times 10 \text{ m/s}^2 = 5 \text{ m/s}^2$ [Source 85, 87].
Stopping Distance: Using the kinematic equation $v^2 = u^2 - 2as$ , where final velocity $v = 0$ : $0 = (20)^2 - 2(5)s$ $400 = 10s$ $s = 40 \text{ m}$ Alternatively, using the work-energy theorem: Work done by friction = Change in Kinetic Energy ( $\mu_k mg s = \frac{1}{2}mu^2$ ), which yields the same result [Source 50, 98].

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