Question

A block of mass 2 kg is placed on a rough inclined plane at an angle of 30 degrees. The coefficient of friction between the block and the plane is 0.2. If the block is released from rest, will it slide down the plane? Justify your answer with calculations.

• A. Yes, it will slide down.
• B. No, it will not slide down.
• C. It will slide down after some time.
• D. Cannot be determined.

Hint:

If the force due to gravity exceeds the frictional force, the object will slide down the inclined plane.

Answer 1

The Correct Option is A

We can calculate the force acting on the block and compare the forces: The gravitational force parallel to the inclined plane is: \[ F_{\text{gravity}} = mg \sin(\theta) \] Where: - \( m = 2 \, \text{kg} \), - \( g = 9.8 \, \text{m/s}^2 \), - \( \theta = 30^\circ \). The frictional force is: \[ F_{\text{friction}} = \mu mg \cos(\theta) \] Where: - \( \mu = 0.2 \) (coefficient of friction). Now, calculating the forces: \[ F_{\text{gravity}} = 2 \times 9.8 \times \sin(30^\circ) = 9.8 \, \text{N} \] \[ F_{\text{friction}} = 0.2 \times 2 \times 9.8 \times \cos(30^\circ) \approx 3.4 \, \text{N} \] Since the force due to gravity is greater than the frictional force, the block will slide down the plane.