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JEE Main 2022 (24 Jun, Shift 1) · Friction Basics

Friction Basics question from JEE Main 2022 (24 Jun, Shift 1)

Q11JEE Main 2022 · 24 JUN · SHIFT 1Easy

Friction Basics

A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms−{}^{-}1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is:

[use g = 9.8 ms−{}^{-}2]

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Answer: (b)

  1. Retardation a=μg=0.5×9.8=4.9 m/s2a = \mu g = 0.5 \times 9.8 = 4.9\text{ m/s}^2
  2. s=v22a=(9.8)22×4.9s = \dfrac{v^2}{2a} = \dfrac{(9.8)^2}{2 \times 4.9}
  3. s=9.8s = 9.8 m, so the answer is (b).
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