A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F₁ is the force required to just move the block up the inclined plane and F₂ is the force required to just prevent the block from sliding down, then the value of is: [Use g = 10 m/s²] (a) 5√(3) N (b) (5√(3))/(2) N (c) 10 N (d) 25√(3) N

Answer: (a). Just moving up: F₁ = mgsin θ + μ mgcos θ. Just preventing sliding down: F₂ = mgsin θ - μ mgcos θ. |F₁| - |F₂| = 2μ mgcos θ = 2(0.1)(50)(√(3)/2) So the answer is (a) 5√(3) N.

A block of mass 5 kg is placed on a rough inclined surface as shown… | JEE Main 2024 (31 Jan, Shift 2)

Q9JEE Main 2024 · 31 JAN · SHIFT 2Medium

Friction on an Incline

A block of mass $5\text{ kg}$ is placed on a rough inclined surface as shown in the figure. If $\vec{F_1}$ is the force required to just move the block up the inclined plane and $\vec{F_2}$ is the force required to just prevent the block from sliding down, then the value of is: [Use $g = 10\text{ m/s}^{2}]$

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Answer: (a)

Just moving up: $F_1 = mg\sin\theta + \mu mg\cos\theta$ .
Just preventing sliding down: $F_2 = mg\sin\theta - \mu mg\cos\theta$ .
$|F_1| - |F_2| = 2\mu mg\cos\theta = 2(0.1)(50)\left(\dfrac{\sqrt{3}}{2}\right)$
So the answer is (a) $5\sqrt{3}\text{ N}$ .

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