A block of mass 10, kg starts sliding on a surface with an initial velocity of 9.8, ms ^{-1} . coeff

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4-2.Friction

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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 $g =9.8\, ms ^{-2}$ ].........$m$

A

$4.9$

B

$9.8$

C

$12.5$

D

$19.6$

(JEE MAIN-2022)

Solution

$a =-\mu g =-0.5 \times 9.8=-4.9$ $m / s ^{2}$

$d =\frac{ v ^{2}}{2 a }=\frac{9.8 \times 9.8}{2(4.9)}$

$=9.8\, m$

Standard 11

Physics

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