A bag is gently dropped on a conveyer on a conveyor belt moving at a speed of $2 m / s$ . the coefficient of friction between the conveyor bell and bag is $0.4$ . Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is: [Take $g = 10 m / s^{- 2}$ ]

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$2 m$

$0.5 m$

$3.2 m$

$0.8 m$

answer is B.

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Detailed Solution

Retardation due to friction $= \frac{μ M g}{M} = μ g$
Now, $s = \frac{v^{2} - u^{2}}{2 a} = \frac{0^{2} - 2^{2}}{2 x - μ g} = \frac{2}{0.4 \times 10} = 0.5 m$

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