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Frictional force on the box f = μmg. Acceleration in the box, a = f/m = μmg/m.

1. v² = u² + 2as
2. s = 2/5 w.r.t. belt
3. distance = 0.4 m
4. μ = 2 sin θ / cos θ

Correct answer: μ = 2 sin θ / cos θ

Solution

The equation μ = 2 sin θ / cos θ is a valid expression for the coefficient of friction in terms of the angle of inclination θ. The other options are unrelated to the given context.

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