Q 11-04-110JEE MainJEE Main 2023 (31 Jan, Shift 2)Easy
A body of mass $10\ \text{kg}$ is moving with an initial speed of $20\ \text{m s}^{-1}$. The body stops after $5\ \text{s}$ due to friction between body and the floor. The value of the coefficient of friction is (Take $g=10\ \text{m s}^{-2}$)
Answer: (D) 0.4
Deceleration $a=\dfrac{20}{5}=4\ \text{m s}^{-2}=\mu g\Rightarrow\mu=0.4$.
Solution by Sreeraj P, M.Sc Physics