**Problem Statement:**Consider the following diagram:A block is placed on an inclined plane with a slope angle of 60° to the horizontal. The inclined plane is smooth with a grid background for reference.If the mass (m) of the block sitting on the incline is 20 kg, and the coefficient of kinetic friction between the block and the incline is 0.30, find the acceleration of the block down the incline after it is released.**Diagram Explanation:**- The diagram shows a block positioned on a sloped surface.- The incline is labeled with a 60° angle.- The block is represented as a simple rectangle on the incline, indicating its position.**Approach:**To solve the problem, use the following physics concepts:1. **Forces on the Inclined Plane:** - Component of gravitational force parallel to the incline: \( mg \sin(\theta) \). - Component of gravitational force perpendicular to the incline: \( mg \cos(\theta) \).2. **Frictional Force:** - Frictional force, \( f_k = \mu_k \cdot N \), where \( N = mg \cos(\theta) \).3. **Net Force:** - The net force \( F \) down the incline is given by: \[ F = mg \sin(\theta) - f_k \]4. **Acceleration:** - Using Newton's second law: \( F = ma \), \[ a = \frac{mg \sin(\theta) - \mu_k \cdot mg \cos(\theta)}{m} \] - Simplifying, the acceleration \( a \) is: \[ a = g (\sin(\theta) - \mu_k \cos(\theta)) \]By substituting the given values:- \( m = 20 \) kg- \( \mu_k = 0.30 \)- \( \theta = 60° \)- \( g = 9.8 \text{ m/s}^2 \)Calculate the acceleration \( a \).

Name: **Problem Statement:**Consider the following diagram:A block is place
Uploaded: 2024-03-20T06:57:57.000Z
Channel: Bartleby
Description: **Problem Statement:**Consider the following diagram:A block is placed on an inclined plane with a slope angle of 60° to the horizontal. The inclined plane is smooth with a grid background for reference.If the mass (m) of the block sitting on the in