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JEE Main 2025 (3 Apr, Shift 2) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2025 (3 Apr, Shift 2)

Q9JEE Main 2025 · 3 APR · SHIFT 2Medium

Inertia, First & Second Laws

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

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Answer: (c)

  1. Work by retarding force: W=−∫0.11.910x dx=−5(1.92−0.12)=−5(3.6)=−18 JW = -\int_{0.1}^{1.9} 10x\,dx = -5(1.9^2 - 0.1^2) = -5(3.6) = -18\text{ J}
  2. Initial KE =12(1)(10)2=50 J= \frac{1}{2}(1)(10)^2 = 50\text{ J}, so final KE =50−18=32 J= 50 - 18 = 32\text{ J}
  3. 12v2=32⇒v=8 m/s\frac{1}{2}v^2 = 32 \Rightarrow v = 8\text{ m/s}. So the answer is (c) 8 m/s8\text{ m/s}.
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