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 = ktan θ. The value of k is _________. Numerical answer type. Work it out, then open the answer.

Answer: 3. Let AC = l; then BC = 2l and AB = 3l. Starts and ends at rest: mg(3l)sin θ = μ mgcos θ · l μ = 3tan θ So k = 3.

A small block starts slipping down from a point B on an inclined… | JEE Main 2020 (2 Sep, Shift 1)

Q28JEE Main 2020 · 2 SEP · SHIFT 1Medium

Conservation of Energy

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 _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Let $AC = l$ ; then $BC = 2l$ and $AB = 3l$ .
Starts and ends at rest: $mg(3l)\sin\theta = \mu mg\cos\theta \cdot l$
$\mu = 3\tan\theta$
So $k = 3$ .

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Conservation of Energy question from JEE Main 2020 (2 Sep, Shift 1)

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