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

Physics · Work, Power & Energy

JEE Main 2025 — 3 April, Evening Shift — Question 51

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

Option A:
$6 \mathrm{~m} / \mathrm{s}$
Option B:
$10 \mathrm{~m} / \mathrm{s}$
Option C:
$4 \mathrm{~m} / \mathrm{s}$
Option D:
$8 \mathrm{~m} / \mathrm{s}$
Correct

Answer: D

Step-by-step solution

$W=\Delta k$ so $\left|\mathrm{W}_{f}\right|=$ loss in KE

$\frac{1}{2} k\left((1.9)^{2}-(0.1)^{2}\right)=\frac{1}{2} \times 1\left\{10^{2}-v^{2}\right\}$

$10 \times 1.8 \times 2=100-v^{2}$

$v^{2}=64$ $v=8 \mathrm{~m} / \mathrm{s}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Work Done by a Constant and Variable Force

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Answer: D

Step-by-step solution

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