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A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box will move relative to belt before coming to rest on it taking g = 10 m/s², is

1. 1.2 m
2. 0.6 m
3. zero
4. 0.4 m

Correct answer: 0.6 m

Solution

The box will experience a frictional force that decelerates it relative to the belt. The deceleration is a = μg = 0.5 × 10 = 5 m/s². Using the equation v² = u² - 2as, where v = 0, u = 2 m/s, and a = 5 m/s², we get s = u² / (2a) = (2²) / (2 × 5) = 0.4 m.

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