Laws of Motion

JEE Main level, numerical answer, medium

168 of 210

Question 168 of 210

The coefficient of static friction between a wooden block of mass 0.5kg0.5kg and a vertical rough wall is 0.2.0.2. The magnitude of the horizontal force that should be applied on the block to keep it adhere to the wall will be______NN. ⌈g=10ms−2]\left⌈g=10m{s}^{-2}\right]

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Answer: 25

F.B.D.F.B.D. of the block is shown in the diagram

Since block is at rest therefore

fr−mg=0...(1)fr-mg=0...\left(1\right)
F−N=0...(2)F-N=0...\left(2\right)
fr≤μNfr\leq \mu N
In limiting case

fr=μN=μF…(3)fr=\mu N=\mu F\ldots \left(3\right)
Using eq.
(1)\left(1\right) and (3)\left(3\right)
∴μF=mg∴\mu F=mg
⇒F=0.5×100.2=25N\Rightarrow F=\dfrac{0.5\times 10}{0.2}=25N

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