Q 11-04-130JEE MainJEE Main 2022 (24 Jun, Shift 1)Easy
A block of mass $10\ \text{kg}$ starts sliding on a surface with an initial velocity of $9.8\ \text{m s}^{-1}$. The coefficient of friction between the surface and block is $0.5$. The distance covered by the block before coming to rest is: [use $g = 9.8\ \text{m s}^{-2}$]
Answer: (A) $9.8\ \text{m}$
Retardation due to friction: $a = \mu g = 0.5\times 9.8 = 4.9\ \text{m s}^{-2}$.
$$s = \frac{u^2}{2a} = \frac{(9.8)^2}{2\times 4.9} = 9.8\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics