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Question: Answered & Verified by Expert
A block of mass m = 1 kg, moving on a horizontal surface with speed vi = 2 m s–1 enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force Fr on the block in this range is inversely proportional to x over this range, F= -k/x for 0.1 < x < 2.01 m = 0 for x < 0.1m and x > 2.01 m where k = 0.5 J. What is the final kinetic energy and speed vf of the block as it crosses this patch ?
ChemistryWork Power EnergyNEET
Options:
  • A 0.5J , 1m/s
  • B 0.5J , 2m/s
  • C 1J, 0.5m/s
  • D 2J , 1m/s
Solution:
2765 Upvotes Verified Answer
The correct answer is: 0.5J , 1m/s
Correct Option is : (A)
0.5J , 1m/s

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