**Understanding the Coefficient of Friction: An Example Problem**A block is pushed across a horizontal surface by the forces \( F_1 \) and \( F_2 \) as shown in the accompanying diagram. If the block moves with an acceleration of \( 2 \, \text{m/s}^2 \), \( F_1 = 15 \, \text{N} \), \( F_2 = 10 \, \text{N} \), and \( M = 2.5 \, \text{kg} \), which of the following is an approximation of the coefficient of kinetic friction between the block and the surface?### Diagram Description:The diagram includes:- A rectangular block labeled \( M \).- \( F_1 \) is a horizontal force acting to the right.- \( F_2 \) is a vertical force acting downward on the block.- A horizontal surface on which the block rests.### Answer Choices:- 0.60- 0.10- 0.29- 1.2**Calculating the Coefficient of Kinetic Friction:**1. **Calculate the Net Horizontal Force:** Using Newton's second law, \( F_{\text{net}} = Ma \): \[ F_{\text{net}} = M \times a = 2.5 \, \text{kg} \times 2 \, \text{m/s}^2 = 5 \, \text{N} \]2. **Determine the Friction Force (\( F_f \)):** The net horizontal force is the applied force minus the friction force: \[ F_{\text{net}} = F_1 - F_f \] Solving for \( F_f \): \[ F_f = F_1 - F_{\text{net}} = 15 \, \text{N} - 5 \, \text{N} = 10 \, \text{N} \]3. **Calculate the Normal Force (\( F_N \)):** The normal force is increased by \( F_2 \): \[ F_N = Mg + F_2 \] Calculating \( Mg \): \[ Mg = 2.5 \, \text{kg} \times 9.

Name: **Understanding the Coefficient of Friction: An Example Problem**A bl
Uploaded: 2024-03-20T06:57:57.000Z
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Description: **Understanding the Coefficient of Friction: An Example Problem**A block is pushed across a horizontal surface by the forces \( F_1 \) and \( F_2 \) as shown in the accompanying diagram. If the block moves with an acceleration of \( 2 \, \text{m/s}^