Question

A block slides down a smooth inclined plane, and its acceleration is found to be \( \frac{g}{8} \). If \( g \) is the acceleration due to gravity, what is the angle of inclination \( \theta \) of the plane?

• A. \( \tan^{-1} \left( \frac{1}{8} \right) \)
• B. \( \tan^{-1} \left( \frac{1}{2} \right) \)
• C. \( \tan^{-1} \left( \frac{1}{4} \right) \)
• D. \( \tan^{-1} \left( \frac{1}{16} \right) \)

Hint:

For inclined plane problems involving acceleration, use the formula \( a = g \sin \theta \) to find the angle of inclination when acceleration is known.

Answer 1

The Correct Option is A

The acceleration of a block descending a frictionless inclined plane is determined by: \[ a = g \sin \theta \] In this equation: - \( a \) represents the block's acceleration. - \( g \) denotes the acceleration due to gravity. - \( \theta \) signifies the angle of the plane's inclination. Given that the block's acceleration is \( \frac{g}{8} \), we have: \[ \frac{g}{8} = g \sin \theta \] Dividing both sides by \( g \): \[ \frac{1}{8} = \sin \theta \] Consequently, the angle \( \theta \) is: \[ \theta = \sin^{-1} \left( \frac{1}{8} \right) \] Therefore, the angle of inclination \( \theta \) is equivalent to \( \tan^{-1} \left( \frac{1}{8} \right) \), corresponding to option (a).