a=mF=−10xvdxdv=−10x
∫10vvdv=−10∫0.11.9xdx
2v2−100=−10(21.92−0.1)2v=8 m/s
JEE Main 2025 — Physics Mechanics
A block of mass 1 kg , moving along x with speed vi=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 Fr=−kxN, with k=10 N/m. Then the final speed of the block as it crosses rough region is
Held on 3 Apr 2025 · Verified 6 Jul 2026.
10 m/s
4 m/s
6 m/s
8 m/s
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