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Physics · LAWS OF MOTION
हिन्दीA block placed on an inclined plane just begins to slide down when the angle of inclination is \( 30^\circ \). If the angle of inclination of the plane is increased to \( 45^\circ \), assuming \( \mu_k \approx \mu_s \), the acceleration of the block down the incline will be (take \( g = 10\text{ m/s}^2 \)):
Correct answer · Option 3
\( 5\sqrt{2}(1 - \frac{1}{\sqrt{3}})\text{ m/s}^2 \)
Explanation
Angle of repose \( \theta = 30^\circ \Rightarrow \mu_s = \tan 30^\circ = \frac{1}{\sqrt{3}} \). At \( \theta = 45^\circ \), acceleration \( a = g(\sin 45^\circ - \mu_k \cos 45^\circ) = g(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}}) = 5\sqrt{2}(1 - \frac{1}{\sqrt{3}})\text{ m/s}^2 \).
