Friction: Practice Problem & Solution
A conveyor belt is moving at a constant speed of \( 2 \text{ m/s} \). A box is gently dropped on it. The coefficient of friction between them is \( \mu = 0.5 \). The distance that the box will move relative to belt before coming to rest on it, taking \( g = 10 \text{ ms}^{-2} \). (2011 Mains)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Friction. Understanding the underlying formula is key to arriving at the correct answer below:
Initial relative speed of box \( u = 2 \text{ m/s} \). Friction force \( f = \mu mg \). Acceleration \( a = \frac{f}{m} = \mu g = 0.5 \times 10 = 5 \text{ m/s}^2 \). Using \( v^2 = u^2 - 2as \) (since friction causes deceleration). With \( v = 0 \), distance \( s = \frac{u^2}{2a} = \frac{(2)^2}{2 \times 5} = \frac{4}{10} = 0.4 \text{ m} \).
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