A block of mass $10\, kg$ starts sliding on a surface with an initial velocity of $9.8\, ms ^{-1}$. The coefficient of friction between the surface and bock is $0.5$. The distance covered by the block before coming to rest is: [use $g =9.8\, ms ^{-2}$ ].........$m$
JEE MAIN 2022, Medium
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$a =-\mu g =-0.5 \times 9.8=-4.9$ $m / s ^{2}$
$d =\frac{ v ^{2}}{2 a }=\frac{9.8 \times 9.8}{2(4.9)}$
$=9.8\, m$
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