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NEET 2019
Exam
NEET
Year
2019
Subject
Physics
Chapter
Laws of Motion

NEET 2019 · Physics · Laws of Motion

A body of mass m is kept on a rough horizontal surface (coefficient...

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

Options

  1. A∣F∣=mg+μmg|\mathrm{F}|=m g+\mu m g
  2. B∣F∣=μmg|\mathrm{F}|=\mu m g
  3. C∣F∣≤mg1+μ2|\mathrm{F}| \leq m g \sqrt{1+\mu^2}
  4. D∣F∣=mg|\mathrm{F}|=m g

Correct answer

C. ∣F∣≤mg1+μ2|\mathrm{F}| \leq m g \sqrt{1+\mu^2}

Step-by-step solution

\text{ The situation can be drawn as }A body of mass m is kept on a rough horizontal surface (coefficient of friction =μ ). Horizontal force is applied on \\ The frictional force, f=μN=μmg[∵\mathrm{f}=\mu \mathrm{N}=\mu \mathrm{mg} \quad[\because\mathrm{N}=\mathrm{mg}]] From Free body diagram (FBD), the resultant force is ∣F∣=N2+f2=(mg)2+(μmg)2=mg1+μ2\begin{aligned} |F| & =\sqrt{\mathrm{N}^2+\mathrm{f}^2}\\ & =\sqrt{(\mathrm{mg})^2+(\mu \mathrm{mg})^2}=\mathrm{mg} \sqrt{1+\mu^2} \end{aligned}This is the minimumforcerequiredtomovetheobject.Butasthebodyisnotmoving∴∣minimum force required to move the object. But as the body is not moving \therefore \quad|\mathrm{F}| \leq \mathrm{mg} \sqrt{1-\mu^2}$
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