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JEE Mains · Physics · STD 11 - Friction
  • JEE Mains 2021
  • Medium
  • Numerical

The coefficient of static friction between a wooden block of mass \(0.5\, kg\) and a vertical rough wall is \(0.2\) The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be \(N\) \(\left[ g =10\, ms ^{-2}\right]\)

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Answer & Solution

Correct Answer

(A) \(25\)

Step-by-step Solution

Detailed explanation

\(F.B.D.\) of the block is shown in the diagram

Since block is at rest therefore

\(fr - mg =0\) \(........(1)\)

\(F- N =0\)  \(..........(2)\)

\(fr \leq \mu N\)

In limiting case

\(fr =\mu N =\mu F\)  \(........(3)\)

Using eq. \((1)\) and \((3)\)

\(\therefore \mu F = mg\)

\(\Rightarrow F=\frac{0.5 \times 10}{0.2}=25 N\)

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