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MHT CET 2026 (12 May, Shift 2) · Friction Basics

Friction Basics question from MHT CET 2026 (12 May, Shift 2)

Q2MHT CET 2026 · 12 MAY · SHIFT 2Easy

Friction Basics

A block of mass 11 kg is kept on ice surface. When velocity of 44 m/s is given to it, it stops by friction in 55 second. The coefficient of friction is (acceleration due to gravity, g=10g = 10 m/s2{}^{2})

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Answer: (c)

  1. Retardation a=vt=45=0.8 m/s2a = \dfrac{v}{t} = \dfrac{4}{5} = 0.8\text{ m/s}^2
  2. Friction gives a=μg⇒μ=0.810a = \mu g \Rightarrow \mu = \dfrac{0.8}{10}
  3. μ=0.08\mu = 0.08, so the answer is (c).
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