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

Correct answer: 0 Let AC = ℓ ∴ BC = 2ℓ ∴ AB = 3ℓ Apply work – Energy theorem W _ t + W _ mg = mg ( 3ℓ ) sinθ - μ mg cosθ ( ℓ ) = 0 + 0 μ mgcosθℓ = 3mg / sinθ μ = 3tanθ = ktanθ k = 3

JEE Main 20202 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 2 September
Shift: Morning Shift
Subject: Physics
Chapter: Laws of Motion

JEE Main 2020 · 2 September · Morning Shift · Physics · Laws of Motion

A small block starts slipping down from a point B on an inclined...

A small block starts slipping down from a point $B$ on an inclined plane $AB$ , which is making an angle $\theta$ with the horizontal section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$ . 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 $\mu = ktan \theta$ . The value of $k$ is .......