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JEE Main 2023 (25 Jan, Shift 2) · Friction on an Incline

Friction on an Incline question from JEE Main 2023 (25 Jan, Shift 2)

Q14JEE Main 2023 · 25 JAN · SHIFT 2Medium

Friction on an Incline

Consider a block kept on an inclined plane (incline at 45∘{}^\circ) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane(μ\mu) is equal to:

Figure

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Answer: (b)

  1. Push up: Fup=mg(sin⁡45∘+μcos⁡45∘)F_{up} = mg(\sin 45^\circ + \mu\cos 45^\circ); prevent sliding: Fdown=mg(sin⁡45∘−μcos⁡45∘)F_{down} = mg(\sin 45^\circ - \mu\cos 45^\circ).
  2. Fup=2Fdown⇒1+μ=2(1−μ)F_{up} = 2F_{down} \Rightarrow 1 + \mu = 2(1 - \mu)
  3. μ=13≈0.33\mu = \dfrac{1}{3} \approx 0.33
  4. So the answer is (b) 0.33.
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