A block sliding down a rough inclined plane is a standard Newton's-second-law problem. Two forces act along the plane: the component of gravity pulling the block down the slope, and kinetic friction opposing the motion (therefore acting up the slope, since the block moves down).

Given: incline angle θ = 30°, coefficient of friction μ = 0.2. Take g as the acceleration due to gravity and m the block mass.

Step 1 — Resolve the weight.

• Component along the plane (driving the block down): mg·sin θ = mg·sin30° = 0.5 mg.
• Component perpendicular to the plane: mg·cos θ = mg·cos30° = 0.866 mg. This equals the normal reaction N.

Step 2 — Friction force (opposing, up the plane):

• f = μN = μ·mg·cos θ = 0.2 × 0.866 mg = 0.1732 mg.

Step 3 — Net force and acceleration along the plane:

• Net force = mg·sin θ − μ·mg·cos θ = mg(sin30° − 0.2·cos30°) = mg(0.5 − 0.1732) = 0.3268 mg.
• Since F = ma, acceleration a = g(sin θ − μ·cos θ) = 0.3268 g ≈ 0.326 g, directed down the incline.

Notice the mass cancels — the acceleration depends only on θ and μ.

Why the other choices are wrong: 0.5 g is what you would get by ignoring friction altogether (a = g·sin30°), which is not permitted on a rough plane. Zero acceleration would require the block to be on the verge of sliding, i.e., tan θ = μ (μ = tan30° = 0.577); here μ = 0.2 is much smaller, so the block does accelerate. 0.676 g is unphysically large — it exceeds even the frictionless value of 0.5 g, which is impossible when friction is retarding the motion.