Force balance for constant speed on incline

Steps:

• Resolve weight into components: parallel down slope (mg sinθ) and perpendicular (mg cosθ).
• Normal force equals mg cosθ, perpendicular to slope, upward.
• Friction acts up slope, equal to mg sinθ, for net zero force parallel to slope.
• Correct diagram vectors: weight vertical down, normal perpendicular out, friction parallel up.

Why D is correct:

• Newton's first law requires net force zero for constant speed; D shows friction up slope balancing weight's parallel component.

Why the others are wrong:

• A: Friction shown down slope, which would accelerate the ball.
• B: Normal force not perpendicular to slope.
• C: Weight vector not vertical.

Final answer: D