Kinetics on an Incline — Component of Weight For a smooth (frictionless) inclined plane at angle θ to the horizontal, what is the linear acceleration of a body sliding down the plane?

Introduction / Context:
Motion on an incline is one of the first applications of Newton’s second law. Decomposing weight into components parallel and perpendicular to the plane gives the acceleration for a frictionless slide.

Given Data / Assumptions:

• Incline angle θ.
• Frictionless contact (smooth plane).
• Neglect air resistance; body treated as a particle.

Concept / Approach:
Resolve weight mg into a component parallel to the plane mg sin θ (down the slope) and a normal component mg cos θ. With no friction, the only unbalanced force along the plane is mg sin θ, producing acceleration a.

Step-by-Step Solution:

Verification / Alternative check:
Boundary checks: θ = 0° ⇒ a = 0 (horizontal surface, no sliding force). θ = 90° ⇒ a = g (free fall vertically). The expression a = g sin θ satisfies both limits.

Why Other Options Are Wrong:
“g cos θ” is the normal reaction divided by m; “g tan θ” does not represent net acceleration; “Zero” ignores the nonzero parallel component for θ > 0.

Common Pitfalls:
Mixing up sine and cosine with the chosen angle reference; forgetting to remove friction only from the parallel direction while keeping the normal reaction.

Final Answer:
g sin θ.