Q 11-05-206JEE MainJEE Main 2025 (3 Apr, Shift 2)Easy
A block of mass $1\ \text{kg}$, moving along $x$ with speed $v_i = 10\ \text{m/s}$, enters a rough region ranging from $x = 0.1\ \text{m}$ to $x = 1.9\ \text{m}$. The retarding force acting on the block in this range is $F_r = -kx\ \text{N}$, with $k = 10\ \text{N/m}$. Then the final speed of the block as it crosses the rough region is:
Answer: (D) $8\ \text{m/s}$
Work done by the retarding force:
$$W = -\int_{0.1}^{1.9}10x\,dx = -5\left(1.9^2 - 0.1^2\right) = -5(3.61 - 0.01) = -18\ \text{J}$$
Work–energy theorem: $\tfrac12(1)v^2 = \tfrac12(1)(10)^2 - 18 = 32 \Rightarrow v = 8\ \text{m/s}$.
Solution by Sreeraj P, M.Sc Physics