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A block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is (g = 10m/s²)

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0.02

0.03

0.06

0.01

answer is C.

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Detailed Solution

$u = 6 m / s, v = 0, t = 10 s a = \frac{v - u}{t} = \frac{- 3}{5} m / s^{2}$
"-ve" means retardation
Here retardation is due to friction
f = ma
$\begin{array}{l} μ mg = ma \\ μ = \frac{a}{g} = \frac{3}{50} = 0 .06 \end{array}$

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