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JEE Main 2021 (17 Mar, Shift 2) · Friction Basics

Friction Basics question from JEE Main 2021 (17 Mar, Shift 2)

Q19JEE Main 2021 · 17 MAR · SHIFT 2Difficult

Friction Basics

A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is __________ N. (Round off to the Nearest Integer) [Take g = 10 ms-2 ]

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Answer: 30

  1. The boy pulls the rope with force TT; the rope pulls the boy up by TT and pulls the wood horizontally by TT.
  2. Normal force from floor on (boy + wood): R=(4+5)g−T=90−TR = (4+5)g - T = 90 - T.
  3. No sliding: T≤μR=0.5(90−T)T \le \mu R = 0.5(90 - T)
  4. 1.5T≤45⇒Tmax=30 N1.5T \le 45 \Rightarrow T_{max} = 30\text{ N}
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