As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30°, with horizontal. For μs = 0.25, the block will just start to move for the value of F: [Given g = 10 ms⁻² ] (a) 25.2 N (b) 35.7 N (c) 20 N (d) 33.3 N

Answer: (a). F30°NmgμN Normal force: N = mg - Fsin 30° = 100 - 0.5F Just to start motion: Fcos 30° = μs N = 0.25(100 - 0.5F) F(0.866 + 0.125) = 25 ⇒ F = (25)/(0.991) F ≈ 25.2 N, so the answer is (a).

As shown in the figure a block of mass 10 kg lying on a horizontal… | JEE Main 2023 (1 Feb, Shift 2)

Q9JEE Main 2023 · 1 FEB · SHIFT 2Medium

Friction Basics

As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle $30^\circ$ , with horizontal. For $\mu_s = 0.25$ , the block will just start to move for the value of F: [Given $g = 10\text{ ms}^{-2}$ ]

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Answer: (a)

Normal force: $N = mg - F\sin 30^\circ = 100 - 0.5F$
Just to start motion: $F\cos 30^\circ = \mu_s N = 0.25(100 - 0.5F)$
$F(0.866 + 0.125) = 25 \Rightarrow F = \dfrac{25}{0.991}$
$F \approx 25.2$ N, so the answer is (a).

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