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Question: Answered & Verified by Expert
A conveyor belt is moving at a constant speed of 2 m s-1. A box is gently dropped on it. The coefficient of friction between them is μ=0.5. The distance that the box will move relative to the belt before coming to rest on it taking g=10 m s-2, is
PhysicsLaws of MotionNEET
Options:
  • A 1.2 m
     
  • B 0.6 m
     
  • C Zero
  • D 0.4 m
     
Solution:
1997 Upvotes Verified Answer
The correct answer is: 0.4 m
 
Frictional force on the box f=μmg

    Acceleration in the box

a=μg=5 m s-2

v2=u2+2as

   0=22+2×(5)s

   s=-25 m (with respect to the belt)

  distance=0.4 m

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