For a block on a frictionless incline, the acceleration along the incline is a = g sin θ. If θ is used in the expression, which statement is correct?

For a block on a frictionless incline, the acceleration along the incline is a = g sin θ. If θ is used in the expression, which statement is correct?

• A. a = g cos θ
• B. a = g sin θ
• C. a = g tan θ
• D. a = g

Answer: (B)

The key idea is that motion along a frictionless incline is driven only by the component of gravity that lies along the surface. Gravity of magnitude g can be split into a component parallel to the incline, mg sin θ, and a component perpendicular to the incline, mg cos θ. The normal force cancels the perpendicular component, so the net force along the plane is mg sin θ. Using Newton’s second law along that direction gives a = F/m = g sin θ. That’s why this expression is correct. The cos θ term is the perpendicular component, not the driving force along the plane; tan θ would relate the two components, not the acceleration itself; and g would apply if there were no constraint of the incline.

The key idea is that motion along a frictionless incline is driven only by the component of gravity that lies along the surface. Gravity of magnitude g can be split into a component parallel to the incline, mg sin θ, and a component perpendicular to the incline, mg cos θ. The normal force cancels the perpendicular component, so the net force along the plane is mg sin θ. Using Newton’s second law along that direction gives a = F/m = g sin θ. That’s why this expression is correct. The cos θ term is the perpendicular component, not the driving force along the plane; tan θ would relate the two components, not the acceleration itself; and g would apply if there were no constraint of the incline.