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Physics · LAWS OF MOTION
हिन्दीA particle of mass \( m = 2\text{ kg} \) is kept in equilibrium under the action of three concurrent forces: \( \vec{F}_1 = (2\hat{i} + 3\hat{j} - \hat{k})\text{ N} \), \( \vec{F}_2 = (-\hat{i} - 4\hat{j} + 2\hat{k})\text{ N} \), and \( \vec{F}_3 \). If the force \( \vec{F}_2 \) is suddenly removed, what will be the magnitude of the acceleration of the particle?
Correct answer · Option 2
\( \frac{\sqrt{21}}{2}\text{ m/s}^2 \)
Explanation
In equilibrium, \( \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0 \Rightarrow \vec{F}_1 + \vec{F}_3 = -\vec{F}_2 \). When \( \vec{F}_2 \) is removed, net force is \( \vec{F}_{net} = \vec{F}_1 + \vec{F}_3 = -\vec{F}_2 = (\hat{i} + 4\hat{j} - 2\hat{k})\text{ N} \). Magnitude \( |\vec{F}_{net}| = \sqrt{1^2 + 4^2 + (-2)^2} = \sqrt{21}\text{ N} \). Acceleration \( a = \frac{\sqrt{21}}{2}\text{ m/s}^2 \).
