Q36 / 181 · NTA Abhyas (JEE Main) 2020 · Physics · Laws of Motion

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 belt before coming to rest on it taking  g=10m s−2, is

  • A
    1.2 m
  • B
    0.6 m
  • C
    Zero
  • Correct answer
    D
    0.4 m

Solution

The frictional force on the box   f=μmg

  ∴   Acceleration in the box

                 a=μg=5 m  s −2

v2=u2+2as

4=0+2×5×s

s = 0.4 m

Hence, the correct answer is option: 'D'.

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