Component of gravitational force parallel to the slope

Steps:

• Resolve forces on M: gravitational component parallel to slope is Mg sin θ, perpendicular is Mg cos θ (balanced by normal force).
• For m: tension T = mg + ma (assuming m accelerates upward).
• Net force on system: Mg sin θ - mg = (M + m)a.
• Since M accelerates down slope, Mg sin θ > mg, so M sin θ > m (g cancels).

Why C is correct:

• M sin θ represents the ratio of parallel gravitational force on M to g, which drives the acceleration down the slope per Newton's second law.

Why the others are wrong:

• A: m sin θ is irrelevant; m hangs vertically, no incline.
• B: m cos θ is perpendicular to m's motion, not a driving force.
• D: M cos θ is perpendicular component on slope, balanced by normal force, causes no acceleration.

Final answer: C