Acceleration of the block a=mf=−212x=−6x (Here the block is retarding)
Now, using, a=vdxdv
⇒vdv=adx
On integrating, we get
∫4vvdv=−6∫0.51.5xdx
⇒2v2−42=−6(2(1.5)2−(0.5)2)
⇒v2=16−12⇒v=2ms−1
JEE Main 2022 — Physics Mechanics
A block of mass 2kg moving on a horizontal surface with speed of 4ms−1 enters a rough surface ranging from x=0.5m to x=1.5m. The retarding force in this range of rough surface is related to distance by F=−kx where k=12Nm−1. The speed of the block as it just crosses the rough surface will be
Held on 28 Jun 2022 · Verified 6 Jul 2026.
2ms−1
2.5ms−1
1.5ms−1
zero
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