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NEET PHYSICSLAWS OF MOTIONMedium

Question

A body of mass mm is kept on a rough horizontal surface (coefficient of friction = μ\mu). A horizontal force is applied to the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by F\vec{F}, where:

A

F=mg+μmg|\vec{F}| = mg + \mu mg

B

F=μmg|\vec{F}| = \mu mg

C

Fmg1+μ2|\vec{F}| \le mg\sqrt{1+\mu^2}

D

F=mg|\vec{F}| = mg

Step-by-Step Solution

  1. Identify Forces: The forces exerted by the surface on the body are the Normal Reaction (NN) and the Static Friction (fsf_s).
  • Since the body is on a horizontal surface and there is no vertical acceleration, N=mgN = mg [NCERT Class 11, Physics Part I, Chapter 5, Section 5.9].
  • The static friction fsf_s balances the applied horizontal force to keep the body at rest. Its magnitude adjusts such that 0fsfs,max0 \le f_s \le f_{s,max}, where the limiting friction fs,max=μN=μmgf_{s,max} = \mu N = \mu mg [NCERT Class 11, Physics Part I, Eq. 4.14].
  1. Resultant Force (Contact Force): The resultant force F\vec{F} (often called the total contact force) is the vector sum of N\vec{N} and fs\vec{f_s}. Since N\vec{N} is vertical and fs\vec{f_s} is horizontal, they are perpendicular. F=N2+fs2|\vec{F}| = \sqrt{N^2 + f_s^2}
  2. Apply Limits: Substituting N=mgN = mg and the inequality for friction:
  • Minimum F|\vec{F}| occurs when fs=0f_s = 0 (no applied force): Fmin=(mg)2+0=mg|\vec{F}|_{min} = \sqrt{(mg)^2 + 0} = mg.
  • Maximum F|\vec{F}| occurs when friction is limiting (fs=μmgf_s = \mu mg): Fmax=(mg)2+(μmg)2=mg1+μ2|\vec{F}|_{max} = \sqrt{(mg)^2 + (\mu mg)^2} = mg\sqrt{1+\mu^2}.
  1. Conclusion: The magnitude of the resultant force satisfies the condition: Fmg1+μ2|\vec{F}| \le mg\sqrt{1+\mu^2}

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 MOTIONhorizontalsurfacecoefficientfrictionhorizontal

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