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The conveyor belt is designed to transport packages of various weights. Each $10 kg$ package has a coefficient of kinetic friction $μ_{k} = 0.15$ . If the speed of the conveyor belt is $5 m / s$ , and then it suddenly stops, determine the distance the package will slide before coming to rest. (Take, $g = 9.8 m / s^{2}$ )

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6.5m

7.5m

8.5m

9.5m

answer is C.

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Detailed Solution

Retardation $a = μ_{1} g = 0.15 \times 9.8 = 1.47 m / s^{2}$ Distance travelled before sliding stops is,

$s = \frac{v^{2}}{2 a} = \frac{(5)^{2}}{2 \times 1.47} = 8.5 m$

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