Q 11-04-126JEE MainJEE Main 2023 (1 Feb, Shift 1)Easy
A block of mass $5$ kg is placed at rest on a table of rough surface. Now, if a force of $30$ N is applied in the direction parallel to surface of the table, the block slides through a distance of $50$ m in an interval of time $10$ s. Coefficient of kinetic friction is (given $g=10\ \text{m s}^{-2}$)
Answer: (C) 0.50
$50=\tfrac12a(10)^2\Rightarrow a=1\ \text{m s}^{-2}$. $30-\mu\times50=5\times1\Rightarrow\mu=0.5$.
Solution by Sreeraj P, M.Sc Physics