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NEET PHYSICSLAWS OF MOTIONMedium

Question

A block of mass mm is in contact with the cart CC as shown in the figure. The coefficient of static friction between the block and the cart is μ\mu. The acceleration aa of the cart that will prevent the block from falling satisfies:

A

a>mgμa > \frac{mg}{\mu}

B

a>gμma > \frac{g}{\mu m}

C

a≥gμa \ge \frac{g}{\mu}

D

a<gμa < \frac{g}{\mu}

Step-by-Step Solution

  1. Forces Analysis: Consider the block of mass mm against the vertical surface of the accelerating cart.
  • Horizontal Direction: The cart accelerates with acceleration aa. The normal reaction NN provided by the vertical surface of the cart on the block causes this acceleration. According to Newton's Second Law: N=maN = ma.
  • Vertical Direction: The weight mgmg acts downwards. To prevent the block from falling, the static friction force fsf_s acting upwards must balance the weight. Thus, fs=mgf_s = mg.
  1. Friction Condition: For the block not to slip, the required static friction must be less than or equal to the limiting value (fs≤fs,maxf_s \le f_{s,max}). The limiting friction is given by fs,max=μNf_{s,max} = \mu N [NCERT Class 11, Physics Part I, Section 5.9 Friction].
  2. Derivation: fs≤μNf_s \le \mu N Substitute fs=mgf_s = mg and N=maN = ma: mg≤μ(ma)mg \le \mu (ma) Cancel the mass mm (since m>0m > 0): g≤μag \le \mu a a≥gμa \ge \frac{g}{\mu} Thus, the minimum acceleration required is g/μg/\mu.

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 MOTIONcontactfigurecoefficientstaticfriction

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