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 ${\mathrm{F}}_{\mathrm{r}}=-\mathrm{kxN}$, with $\mathrm{k}=10\mathrm{N}/\mathrm{m}$. Then the final speed of the block as it crosses rough region is [JEE Main 2025, 3 Apr (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: d) $8\mathrm{m}/\mathrm{s}$. \begin{equation} \begin{aligned} & a=\frac{F}{m}=-10 x \ & v \frac{d v}{d x}=-10 x \ & \int_{10}^v v d v=-10 \int_{0.1}^{1.9} x d x \ & \frac{v^2-100}{2}=-10\left(\frac{1.9^2-0.1}{2}\right)^2 \ & v=8 \mathrm{~m} / \mathrm{s} \end{aligned} \end{equation}

A block of mass \(1\) kg , moving along x with speed \({\mathrm{v}}_{\mathrm{i}}=10\mathrm{m}/\mathrm{s}\) enters a roug…

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