To solve the problem, we need to find the tension in the rope at a point 6 meters away from the point of application of the force. Let's break it down step by step.
### Step 1: Identify the given data
- Length of the rope, \( L = 10 \, \text{m} \)
- Linear density of the rope, \( \mu = 0.5 \, \text{kg/m} \)
- Force applied, \( F = 25 \, \text{N} \)
- Distance from the point of application to where we need to find the tension, \( d = 6 \, \text{m} \)
### Step 2: Calculate the mass of the rope
The total mass of the rope can be calculated using the linear density:
\[
\text{Total mass} = \mu \times L = 0.5 \, \text{kg/m} \times 10 \, \text{m} = 5 \, \text{kg}
\]
### Step 3: Calculate the acceleration of the rope
Using Newton's second law, the acceleration \( a \) of the entire rope can be calculated as:
\[
F = m \cdot a \implies a = \frac{F}{m} = \frac{25 \, \text{N}}{5 \, \text{kg}} = 5 \, \text{m/s}^2
\]
### Step 4: Calculate the mass of the segment of the rope
We need to find the tension at a point 6 meters from the pulling end. The mass of the 6-meter segment of the rope is:
\[
\text{Mass of 6 m segment} = \mu \times 6 \, \text{m} = 0.5 \, \text{kg/m} \times 6 \, \text{m} = 3 \, \text{kg}
\]
### Step 5: Calculate the mass of the remaining segment of the rope
The remaining length of the rope is \( 10 - 6 = 4 \, \text{m} \). The mass of this segment is:
\[
\text{Mass of 4 m segment} = \mu \times 4 \, \text{m} = 0.5 \, \text{kg/m} \times 4 \, \text{m} = 2 \, \text{kg}
\]
### Step 6: Set up the equation for tension
The tension \( T \) at the 6-meter point must support the mass of the remaining 4-meter segment (2 kg) and provide the necessary acceleration. Using Newton's second law for the 2 kg mass:
\[
T - (2 \, \text{kg} \cdot 5 \, \text{m/s}^2) = 0
\]
This can be rearranged to find \( T \):
\[
T = 2 \, \text{kg} \cdot 5 \, \text{m/s}^2 = 10 \, \text{N}
\]
### Conclusion
The tension in the rope at a point 6 meters away from the point of application of the force is \( \boxed{10 \, \text{N}} \).
