Question

A body of mass \( 5 \, \text{kg} \) is placed on a frictionless inclined plane of angle \( 30^\circ \). What is the component of the weight of the body along the plane?

• A. \( 25 \, \text{N} \)
• B. \( 50 \, \text{N} \)
• C. \( 45 \, \text{N} \)
• D. \( 75 \, \text{N} \)

Hint:

Remember: The component of the weight along the plane depends on both the mass of the body and the angle of inclination. For an inclined plane, \( \sin \theta \) gives the projection of the weight along the surface.

Answer 1

The Correct Option is A

Step 1: Use the formula for the component of weight along an inclined plane The component of the weight along the inclined plane is given by: \[ W_{\parallel} = mg \sin \theta \] where: - \( m \) is the mass of the body, - \( g \) is the acceleration due to gravity, - \( \theta \) is the angle of inclination. Step 2: Substitute the given values Given: - Mass \( m = 5 \, \text{kg} \), - \( g = 10 \, \text{m/s}^2 \), - Angle \( \theta = 30^\circ \). Substitute these values into the formula: \[ W_{\parallel} = 5 \times 10 \times \sin(30^\circ) = 50 \times \frac{1}{2} = 25 \, \text{N} \] Answer: Therefore, the component of the weight of the body along the plane is \( 25 \, \text{N} \). So, the correct answer is option (1).