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 \(N\) \(\left[ g =10\, ms ^{-2}\right]\)
Pick an option to check your answer.
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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