Motion On An Inclined Plane

On a rough incline the component of gravity along the slope drives the block downward while kinetic friction opposes the motion up the slope. Resolving gravity, the driving component is $mg\sin \theta$ and the normal force is $mg\cos \theta$, so friction is $\mu mg\cos \theta$. Newton's Second Law along the incline gives $a=g(\sin \theta -\mu \cos \theta )$. Substituting, $a=10(\sin 30^{\circ }-0.2\cos 30^{\circ })=10(0.5-0.2\times 0.866)=10(0.5-0.173)=10(0.327)=3.27{\text{ m/s}}^2$. The 5.00 m/s^2 value ignores friction entirely and uses only $g\sin \theta$. The 1.73 m/s^2 value mistakenly uses $g\cos \theta$ scaled by friction alone. The 2.30 m/s^2 value comes from swapping sine and cosine in the two gravity components. A key conceptual point is that the acceleration here is independent of the block's mass, because both the driving gravity term and the friction term scale with mass, so it cancels throughout. Notice also that the block keeps sliding only because $\tan \theta =0.577$ exceeds $\mu =0.2$; had friction been large enough that $\mu \ge \tan \theta$, the block would not accelerate at all. This is the standard NCERT inclined-plane analysis with friction. As a check, the answer is positive and smaller than the frictionless 5 m/s^2, confirming that friction reduces but does not reverse the downhill acceleration.