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NEET PHYSICSLAWS OF MOTIONEasy

Question

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):

A

30 m

B

40 m

C

72 m

D

20 m

Step-by-Step Solution

  1. Unit Conversion: First, convert the initial speed from km/h to m/s. u=72 km/h=72×518 m/s=20 m/su = 72 \text{ km/h} = 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}
  2. Deceleration due to Friction: The retarding force is the kinetic friction, fk=μkN=μkmgf_k = \mu_k N = \mu_k mg. According to Newton's Second Law, the deceleration aa is: a=fkm=μkmgm=μkga = \frac{f_k}{m} = \frac{\mu_k mg}{m} = \mu_k g a=0.5×10 m/s2=5 m/s2a = 0.5 \times 10 \text{ m/s}^2 = 5 \text{ m/s}^2 [Source 85, 87].
  3. Stopping Distance: Using the kinematic equation v2=u2−2asv^2 = u^2 - 2as, where final velocity v=0v = 0: 0=(20)2−2(5)s0 = (20)^2 - 2(5)s 400=10s400 = 10s s=40 ms = 40 \text{ m} Alternatively, using the work-energy theorem: Work done by friction = Change in Kinetic Energy (μkmgs=12mu2\mu_k mg s = \frac{1}{2}mu^2), which yields the same result [Source 50, 98].

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONconsidermovingstraighthorizontalcoefficient

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