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JEE Main 2023 (12 Apr, Shift 1) · Equilibrium, Third Law & Frames

Equilibrium, Third Law & Frames question from JEE Main 2023 (12 Apr, Shift 1)

Q10JEE Main 2023 · 12 APR · SHIFT 1Easy

Equilibrium, Third Law & Frames

Three forces $F_1 = 10\text{ N},\ F_2 = 8\text{ N},\ F_3 = 6\text{ N}$ are acting on a particle of mass $5\text{ kg}$ . The forces $F_2$ and $F_3$ are applied perpendicularly so that particle remains at rest. If the force $F_1$ is removed, then the acceleration of the particle is:

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Answer: (c)

At rest, $F_1$ balances the resultant of $F_2$ and $F_3$ : $\sqrt{8^2 + 6^2} = 10\text{ N} = F_1$ .
Removing $F_1$ leaves a net force of $10\text{ N}$ .
$a = \frac{10}{5} = 2\text{ m s}^{-2}$
So the answer is (c) $2\text{ m s}^{-2}$ .

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