A frictionless incline with θ = 30° accelerates a block downward. Which expression gives the acceleration along the plane?

A frictionless incline with θ = 30° accelerates a block downward. Which expression gives the acceleration along the plane?

• A. a = g sin θ ≈ 4.9 m/s^2 down the plane
• B. a = g cos θ ≈ 8.5 m/s^2 down the plane
• C. a = g tan θ ≈ 5.66 m/s^2 down the plane
• D. a = g ≈ 9.8 m/s^2 down the plane

Answer: (A)

The motion down a frictionless incline is driven by the component of gravity along the plane. Project gravity onto the plane: the unbalanced force along the plane is m g sin θ, while the normal force cancels the perpendicular component m g cos θ. With no friction, the acceleration along the plane is a = F/m = g sin θ. For θ = 30°, sin 30° = 0.5, so a = 9.8 × 0.5 ≈ 4.9 m/s^2 down the plane. The cos component is the perpendicular (normal) component, not the driving one, and tan θ would mix components and isn’t the correct expression for the along-plane acceleration. Free-fall acceleration g would occur only if the motion were vertical, not constrained to the plane.

The motion down a frictionless incline is driven by the component of gravity along the plane. Project gravity onto the plane: the unbalanced force along the plane is m g sin θ, while the normal force cancels the perpendicular component m g cos θ. With no friction, the acceleration along the plane is a = F/m = g sin θ. For θ = 30°, sin 30° = 0.5, so a = 9.8 × 0.5 ≈ 4.9 m/s^2 down the plane. The cos component is the perpendicular (normal) component, not the driving one, and tan θ would mix components and isn’t the correct expression for the along-plane acceleration. Free-fall acceleration g would occur only if the motion were vertical, not constrained to the plane.