A uniform thick string of length 5 m is resting on a horizontal frictionless surface. It is pulled by a horizontal force of 5 N from one end. The tension in the string at 1m from the force applied is :

Options

A. zero
B. 5N
C. 4N
D. 1N

Explanation

To solve the problem, we need to determine the tension in the string at a point 1 meter from the end where a 5 N force is applied. We can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
We have a uniform thick string of length 5 m resting on a frictionless surface. A force of 5 N is applied horizontally at one end.

2. **Define Mass of the String**:
Let the total mass of the string be \( m \). Since the string is uniform, the mass per unit length is \( \frac{m}{5} \) kg/m.

3. **Calculate Acceleration**:
The net force acting on the entire string is equal to the applied force. According to Newton's second law, we have:
\[ F_{\text{net}} = m \cdot a \]
Here, \( F_{\text{net}} = 5 \, \text{N} \). Therefore,
\[ 5 = m \cdot a \implies a = \frac{5}{m} \]

4. **Consider the Segment of the String**:
We will analyze the segment of the string that is 1 m long, which is the part of the string at a distance of 1 m from the point where the force is applied. The mass of this 1 m segment is:
\[ m_{\text{segment}} = \frac{m}{5} \, \text{kg} \]

5. **Apply Newton's Second Law to the Segment**:
For the 1 m segment, the forces acting on it are the tension \( T \) at the end of the segment and the applied force of 5 N. According to Newton's second law:
\[ 5 - T = m_{\text{segment}} \cdot a \]
Substituting the values we have:
\[ 5 - T = \left(\frac{m}{5}\right) \cdot \left(\frac{5}{m}\right) \]

6. **Simplify the Equation**:
The right side simplifies to:
\[ 5 - T = 1 \]
Therefore, we can solve for \( T \):
\[ T = 5 - 1 = 4 \, \text{N} \]

7. **Conclusion**:
The tension in the string at 1 m from the force applied is \( T = 4 \, \text{N} \).

### Final Answer:
The tension in the string at 1 m from the force applied is **4 N**.