Q75 / 75 · JEE Main 2025 (3 Apr Shift 2) · Physics
A block of mass 1 kg , moving along x with speed $\mathrm{v}_{\mathrm{i}}=10 \mathrm{~m} / \mathrm{s}$ enters a rough region ranging from $\mathrm{x}=0.1 \mathrm{~m}$ to $\mathrm{x}=1.9 \mathrm{~m}$. The retarding force acting on the block in this range is $\mathrm{F}_{\mathrm{r}}=-\mathrm{kx} \mathrm{N}$, with $\mathrm{k}=10 \mathrm{~N} / \mathrm{m}$. Then the final speed of the block as it crosses rough region is
- A$10 \mathrm{~m} / \mathrm{s}$
- B$4 \mathrm{~m} / \mathrm{s}$
- C$6 \mathrm{~m} / \mathrm{s}$
- Correct answerD$8 \mathrm{~m} / \mathrm{s}$
Solution
$\begin{aligned} & a=\frac{F}{m}=-10 x \\ & v \frac{d v}{d x}=-10 x\end{aligned}$
$\int_{10}^{\mathrm{v}} \mathrm{vdv}=-10 \int_{0.1}^{1.9} \mathrm{xdx}$
$\begin{aligned} & \frac{v^2-100}{2}=-10\left(\frac{1.9^2-0.1}{2}\right)^2 \\ & v=8 \mathrm{~m} / \mathrm{s}\end{aligned}$
$\int_{10}^{\mathrm{v}} \mathrm{vdv}=-10 \int_{0.1}^{1.9} \mathrm{xdx}$
$\begin{aligned} & \frac{v^2-100}{2}=-10\left(\frac{1.9^2-0.1}{2}\right)^2 \\ & v=8 \mathrm{~m} / \mathrm{s}\end{aligned}$
Hence, the correct answer is option: 'D'.