A 2-kg block on a frictionless ramp with θ = 30° is released. What is its acceleration down the ramp?

A 2-kg block on a frictionless ramp with θ = 30° is released. What is its acceleration down the ramp?

• A. a = g sin θ ≈ 2.45 m/s^2 down the ramp
• B. a = g cos θ ≈ 8.5 m/s^2 down the ramp
• C. a = g tan θ ≈ 5.66 m/s^2 down the ramp
• D. a ≈ 4.9 m/s^2 down the ramp

Answer: (D)

On a frictionless incline, only the component of gravity along the ramp drives the motion. That component is mg sin theta, while the normal force cancels the perpendicular part. Applying F = ma along the ramp gives m a = m g sin theta, so the mass cancels and the acceleration along the ramp is a = g sin theta. With theta = 30°, sin 30° = 0.5, so a = 9.8 m/s^2 × 0.5 ≈ 4.9 m/s^2 down the ramp. This matches because the full gravitational acceleration is reduced to the along-surface component by the angle. The other expressions (g cos theta or g tan theta) refer to different directions or combinations and don’t describe the motion along the incline.

On a frictionless incline, only the component of gravity along the ramp drives the motion. That component is mg sin theta, while the normal force cancels the perpendicular part. Applying F = ma along the ramp gives m a = m g sin theta, so the mass cancels and the acceleration along the ramp is a = g sin theta.

With theta = 30°, sin 30° = 0.5, so a = 9.8 m/s^2 × 0.5 ≈ 4.9 m/s^2 down the ramp. This matches because the full gravitational acceleration is reduced to the along-surface component by the angle. The other expressions (g cos theta or g tan theta) refer to different directions or combinations and don’t describe the motion along the incline.