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Question: Answered & Verified by Expert
A block of mass m=1 kg moving on a horizontal surface with speed u=2 m s-1 enters a rough horizontal patch ranging from x = 0.10 m to x = 2.00 m. If the retarding force fr on the block in this range is inversely proportional to x over this range i.e.
fr=-kx for 0.10<x<2.00
fr=0 for x<0.10 and x>2.00
If k=0.5 J then, the speed of this block ( in m s-1 ) as it crosses the patch is ( use ln   20=3 )
PhysicsWork Power EnergyJEE Main
Solution:
1677 Upvotes Verified Answer
The correct answer is: 1

W=frdx=-kloge20.1=-1.5 J

W=K

12×1×v2-12×1×4=-1.5

v=1 m s-1

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