Question

A block of mass 5 kg is placed on a frictionless incline of angle $30^\circ$. What is the acceleration of the block down the incline? (Take $g = 9.8\, m/s^2$)

• A. \(9.8\, m/s^2\)
• B. \(4.9\, m/s^2\)
• C. \(5.6\, m/s^2\)
• D. \(3.2\, m/s^2\)

Hint:

On an incline, acceleration of an object without friction is \( a = g \sin \theta \).

Answer 1

The Correct Option is B

To find the acceleration of the block on a frictionless incline, we must consider the forces acting on the block. The primary force is gravity, which acts vertically downward. The component of gravity acting along the incline causes the block to accelerate. This is determined by the equation for acceleration on an incline:

\(a = g \sin \theta\)

where \(a\) is the acceleration, \(g\) is the gravitational acceleration (\(9.8\, m/s^2\)), and \(\theta\) is the angle of the incline. Here, \(\theta = 30^\circ\). Substituting the values, we get:

\(a = 9.8 \times \sin 30^\circ\)

Since \(\sin 30^\circ = 0.5\), we find:

\(a = 9.8 \times 0.5 = 4.9\, m/s^2\)

Therefore, the acceleration of the block down the incline is \(4.9\, m/s^2\).

Answer 2

Step 1: Write the force component causing acceleration
The component of gravitational force along the incline is \[ F = mg \sin \theta \] Step 2: Calculate acceleration
Using Newton's second law \(F = ma\): \[ ma = mg \sin \theta \Rightarrow a = g \sin \theta \] Step 3: Substitute values \[ a = 9.8 \times \sin 30^\circ = 9.8 \times 0.5 = 4.9\, m/s^2 \]