JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2025
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(8 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & a=\frac{F}{m}=-10 x \\ & v \frac{d v}{d x}=-10 x\end{aligned}\)
\(\int_{10}^{\mathrm{v}} \mathrm{vdv}=-10 \int_{0.1}^{1.9} \mathrm{xdx}\)
\(\begin{aligned} & \frac{v^2-100}{2}=-10\left(\frac{1.9^2-0.1}{2}\right)^2 \\ & v=8 \mathrm{~m} / \mathrm{s}\end{aligned}\)
\(\int_{10}^{\mathrm{v}} \mathrm{vdv}=-10 \int_{0.1}^{1.9} \mathrm{xdx}\)
\(\begin{aligned} & \frac{v^2-100}{2}=-10\left(\frac{1.9^2-0.1}{2}\right)^2 \\ & v=8 \mathrm{~m} / \mathrm{s}\end{aligned}\)
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