A small block starts slipping down from a point B on an inclined plane AB, which is making an angle $θ$ with the horizontal. Section BC is smooth and the remaining section CA is rough with a coefficient of friction $μ$ . It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC=2AC, the coefficient of friction is given by $μ = k \tan θ$ . The value of k is_____

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Work done by gravity in moving from B to A = mg sing $θ$ (BA)
Work done by friction in moving from C to A $= μ m g \cos θ (C A)$
Total work = $Δ K E = 0$
$\Rightarrow m g \sin θ . (B A) - μ m g \cos θ C A = 0$
$\Rightarrow μ = \tan θ (\frac{B A}{C A}) \Rightarrow μ = \tan θ (\frac{B C + C A}{C A}) \Rightarrow μ = \tan θ (\frac{2 C A + C A}{C A}) = 3 \tan θ$