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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 3 April · Evening Shift · Physics · Work, Power and Energy

A block of mass 1 kg , moving along x with speed v_ i =10 ~m / s...

A block of mass 1 kg , moving along x with speed vi=10 m/s\mathrm{v}_{\mathrm{i}}=10 \mathrm{~m} / \mathrm{s} enters a rough region ranging from x=0.1 m\mathrm{x}=0.1 \mathrm{~m} to x=1.9 m\mathrm{x}=1.9 \mathrm{~m}. The retarding force acting on the block in this range is Fr=−kxN\mathrm{F}_{\mathrm{r}}=-\mathrm{kx} \mathrm{N}, with k=10 N/m\mathrm{k}=10 \mathrm{~N} / \mathrm{m}. Then the final speed of the block as it crosses rough region is

Options

  1. A10 m/s10 \mathrm{~m} / \mathrm{s}
  2. B4 m/s4 \mathrm{~m} / \mathrm{s}
  3. C6 m/s6 \mathrm{~m} / \mathrm{s}
  4. D8 m/s8 \mathrm{~m} / \mathrm{s}

Correct answer

D. 8 m/s8 \mathrm{~m} / \mathrm{s}

Step-by-step solution

a=Fm=−10xvdvdx=−10x\begin{aligned} & a=\frac{F}{m}=-10 x \\ & v \frac{d v}{d x}=-10 x\end{aligned}
∫10vvdv=−10∫0.11.9xdx\int_{10}^{\mathrm{v}} \mathrm{vdv}=-10 \int_{0.1}^{1.9} \mathrm{xdx}
v2−1002=−10(1.92−0.12)2v=8 m/s\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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