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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 $m s^{- 2}$ is:

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0.6 m

zero

1.2 m

0.4 m

answer is D.

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Detailed Solution

Force, F = $μ m g$

Retardation of the block on the belt

a = $\frac{F}{m}$ = $\frac{μ m g}{m}$ = $μ g$

From, $v^{2} = u^{2} + 2 a s$

0 = ${(2)}^{2} - 2 (μ g) s$

s = $\frac{4}{2 \times 0.5 \times 10}$ = 0.4 m

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