A conveyor belt is moving at a constant speed of 2 ms ⁻ 1. A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box will move relativce to belt before coming to rest on it, taking g = 10 m s ⁻ 2,, is (a) 0.4 m (b) 1.2 m (c) 0.6 m (d) zero

Answer: (a). Friction accelerates the box: a = μ g = 5 m/s². Relative to the belt, the box starts at 2 m/s backwards and decelerates at 5 m/s². srel = v²/2a = 4/10 srel = 0.4 m, so the answer is (a).

A conveyor belt is moving at a constant speed of 2 ms ⁻ 1. A box is… | NEET 2011 (AIPMT, Mains)

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Friction Basics

A conveyor belt is moving at a constant speed of 2 ms ${}^{-}$ 1. A box is gently dropped on it. The coefficient of friction between them is $\mu$ = 0.5. The distance that the box will move relativce to belt before coming to rest on it, taking g = 10 m s ${}^{-}$ 2,, is

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Answer: (a)

Friction accelerates the box: $a = \mu g = 5\text{ m/s}^2$ .
Relative to the belt, the box starts at $2\text{ m/s}$ backwards and decelerates at $5\text{ m/s}^2$ .
$s_{rel} = \dfrac{v^2}{2a} = \dfrac{4}{10}$
$s_{rel} = 0.4$ m, so the answer is (a).

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