Question

The acceleration of a body sliding down an inclined surface is:

• A. g sin θ
• B. g cos θ
• C. g tan θ
• D. none of these

Hint:

If the surface has friction, the acceleration decreases to $g(\sin \theta - \mu \cos \theta)$. However, in standard ideal cases where friction isn't mentioned, we consider only the gravity component.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When an object is on an inclined plane, gravity acts vertically downward. This force is resolved into two components: one perpendicular to the plane and one parallel to the plane.

Step 2: Key Formula or Approach:
According to Newton's Second Law, $F = ma$. The net force acting along the direction of motion (down the plane) determines the acceleration.

Step 3: Detailed Explanation:
Consider a body of mass $m$ on a frictionless plane inclined at an angle $\theta$. 1. The weight $mg$ acts straight down. 2. The component of weight perpendicular to the plane is $mg \cos \theta$ (balanced by the Normal force). 3. The component of weight parallel to the plane (acting downwards) is $mg \sin \theta$. Using $F = ma$: \[ mg \sin \theta = ma \] Dividing both sides by $m$: \[ a = g \sin \theta \]

Step 4: Final Answer:
The acceleration is g sin θ.

Answer 2

This question asks for the acceleration of a body sliding down a frictionless inclined surface, in terms of the angle of incline \( \theta \) and gravitational acceleration \( g \). Instead of resolving forces again, we can check each option against two limiting cases we already know from experience: a completely vertical surface (\( \theta = 90^\circ \), which is just free fall) and a completely flat surface (\( \theta = 0^\circ \), where nothing slides at all).

1. \( g\sin\theta \): At \( \theta = 90^\circ \), \( \sin 90^\circ = 1 \), so this gives \( a = g \), exactly matching free fall along a vertical wall. At \( \theta = 0^\circ \), \( \sin 0^\circ = 0 \), so \( a = 0 \), matching a flat, horizontal surface where nothing accelerates. Both limiting cases check out.
2. \( g\cos\theta \): At \( \theta = 90^\circ \), \( \cos 90^\circ = 0 \), which would mean a body falling down a vertical surface has zero acceleration. That contradicts free fall, so this expression cannot be right.
3. \( g\tan\theta \): At \( \theta = 90^\circ \), \( \tan\theta \) grows without bound, implying an infinite acceleration on a vertical surface. Since acceleration on any incline can never exceed \( g \) (free fall is the fastest a body can accelerate under gravity alone), this expression is physically impossible.
4. None of these: Since one of the listed expressions already survives both limiting-case checks, this option is ruled out.

Only \( g\sin\theta \) behaves correctly in both limiting situations, confirming it as the acceleration of the body on the incline.

Therefore, the correct answer is \( g\sin\theta \).