A body of mass 2kg is released from rest on a plane inclined at an angle of 60° to the horizontal. Calculate…

Analyze the question: A 2 kg body slides from rest down a smooth inclined plane (no friction is mentioned) at 60° to the horizontal. We need its acceleration down the plane.

Concept: On an incline, the weight mg splits into two components:

• Along the plane: mg sinθ (this makes the body accelerate)
• Perpendicular to the plane: mg cosθ (balanced by the normal reaction)

With no friction, the net force down the plane is mg sinθ. By Newton's second law:

F = ma, so mg sinθ = ma, which gives a = g sinθ.

The mass cancels out, so the 2 kg is extra information.

Solution:

a = g sinθ = 10 × sin 60°

sin 60° = √3/2 ≈ 0.866

a = 10 × 0.866 = 8.66 ≈ 8.7 m/s²

Answer: The acceleration is about 8.7 m/s², which is option D.

Tip: For a frictionless incline, always use a = g sinθ. The mass does not matter. Remember these values:

• sin 30° = 0.5, so a = 5 m/s²
• sin 45° ≈ 0.707
• sin 60° ≈ 0.866

If the angle is larger, the acceleration is larger, and it can never exceed g (at 90°, the body is in free fall). Here 8.7 is less than 10, so the answer is sensible.