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We are asked to find the **work done** by the \( 131 \, \text{N} \) force as it drags a \( 16.2 \, \text{kg} \) block over a rough surface with a displacement of \( 76.7 \, \text{m} \). The force acts at an angle of \( 25.9^\circ \) above the horizontal, and the coefficient of kinetic friction is given as \( \mu_k = 0.157 \). The acceleration due to gravity is \( 9.8 \, \text{m/s}^2 \).

### Step 1: Breakdown of Forces
The applied force of \( 131 \, \text{N} \) is at an angle, so we need to decompose it into horizontal and vertical components:
- Horizontal component: 
  \[
  F_{\text{horizontal}} = 131 \cos(25.9^\circ)
  \]
- Vertical component:
  \[
  F_{\text{vertical}} = 131 \sin(25.9^\circ)
  \]

### Step 2: Normal Force
The normal force is affected by the weight of the block and the vertical component of the applied force:
- Weight of the block: 
  \[
  W = mg = 16.2 \times 9.8 = 158.76 \, \text{N}
  \]
- The effective normal force \( F_N \) is given by:
  \[
  F_N = W - F_{\text{vertical}} = 158.76 - 131 \sin(25.9^\circ)
  \]

### Step 3: Frictional Force
The kinetic friction force is:
\[
F_{\text{friction}} = \mu_k \times F_N = 0.157 \times F_N
\]

### Step 4: Net Horizontal Force
The net horizontal force is the difference between the horizontal component of the applied force and the friction force:
\[
F_{\text{net}} = F_{\text{horizontal}} - F_{\text{friction}}
\]

### Step 5: Work Done
The work done by the \( 131 \, \text{N} \) force is:
\[
W = F_{\text{horizontal}} \times \text{displacement}
\]
This is the total work done by the applied force over a distance of \( 76.7 \, \text{m} \).

Let me compute these values step by step.Here are the detailed calculations:

1. **Horizontal component of the applied force**:  
   \[
   F_{\text{horizontal}} = 117.84 \, \text{N}
   \]

2. **Vertical component of the applied force**:  
   \[
   F_{\text{vertical}} = 57.22 \, \text{N}
   \]

3. **Normal force** (after accounting for the vertical component of the applied force):  
   \[
   F_{\text{normal}} = 101.54 \, \text{N}
   \]

4. **Frictional force**:  
   \[
   F_{\text{friction}} = 15.94 \, \text{N}
   \]

5. **Work done by the applied force** over the distance of \( 76.7 \, \text{m} \):  
   \[
   W = 9038.49 \, \text{J}
   \]

Would you like more details on any of these calculations or additional clarifications?

Here are 5 questions related to this:
1. How does the angle of the applied force affect the work done?
2. What happens to the normal force if the applied force angle increases?
3. How would the work done change if the coefficient of friction were higher?
4. How do we calculate work done by friction?
5. What is the relationship between displacement and work done?

**Tip**: The horizontal component of force is the one that contributes to the work done, so always decompose forces at angles!

A 16.2 kg block is dragged over a rough horizontal surface by a constant force of 131 N acting at an angle of 25.9° above the horizontal. The block is displaced 76.7 m, and the coefficient of kinetic friction is 0.157. Find the work done by the 131 N force.

This problem involves calculating the work done by a 131 N force applied to a 16.2 kg block over a rough surface. Using trigonometry and physics concepts such as forces, friction, and the work-energy theorem, we determine the work done over a 76.7 m displacement. The coefficient of kinetic friction is 0.157.

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