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To solve this problem, let's break down each part step-by-step.

**Problem Analysis:**
We have a mass \( m \) attached to a string of length \( L \), which is initially held horizontally and then released from rest. It swings down to a position where the string makes an angle of 23° with the vertical.

We are asked to:
1. Derive an expression for the tension in the string when it makes a 23° angle with the vertical.
2. Determine the total work done to bring the string to this position.
3. Find the work done by the tension in the string.

### Part (a): Tension in the String at 23° with the Vertical
1. **Energy Conservation**:
   Since the mass is released from rest, we can use the conservation of mechanical energy. At the starting position (horizontal), the potential energy is maximum and kinetic energy is zero. At 23°, some of the potential energy will have converted into kinetic energy.

2. **Height Calculation**:
   The vertical height \( h \) from the initial position (horizontal) to the 23° position can be found as:
   \[
   h = L - L \cos(23^\circ) = L(1 - \cos(23^\circ))
   \]

3. **Potential Energy and Kinetic Energy**:
   - Initial potential energy \( U_i = mgL \).
   - Potential energy at 23° position \( U_f = mg \cdot h = mg \cdot L(1 - \cos(23^\circ)) \).
   
   By energy conservation, the difference in potential energy will be the kinetic energy at 23°:
   \[
   K = U_i - U_f = mgL - mgL(1 - \cos(23^\circ)) = mgL \cos(23^\circ)
   \]
   Since \( K = \frac{1}{2}mv^2 \), we get:
   \[
   \frac{1}{2}mv^2 = mgL \cos(23^\circ)
   \]
   \[
   v^2 = 2gL \cos(23^\circ)
   \]

4. **Forces and Tension**:
   At this position, the forces acting on the mass are:
   - The gravitational force \( mg \) acting downward.
   - The tension \( T \) in the string acting along the string.

   Using Newton's second law in the radial direction:
   \[
   T - mg \cos(23^\circ) = \frac{mv^2}{L}
   \]
   Substituting \( v^2 \) from earlier:
   \[
   T - mg \cos(23^\circ) = \frac{m(2gL \cos(23^\circ))}{L}
   \]
   \[
   T = mg \cos(23^\circ) + 2mg \cos(23^\circ) = 3mg \cos(23^\circ)
   \]

   So, the tension \( T \) when the string makes a 23° angle with the vertical is:
   \[
   T = 3mg \cos(23^\circ)
   \]

### Part (b): Total Work Done to Bring the String to the 23° Position
The total work done is equal to the change in potential energy of the mass from the initial position to the 23° position, as the only forces doing work are conservative (gravity).

\[
W = U_i - U_f = mgL - mgL(1 - \cos(23^\circ)) = mgL \cos(23^\circ)
\]

Thus, the total work done is:
\[
W = mgL \cos(23^\circ)
\]

### Part (c): Work Done by the Tension
The work done by the tension force is zero. This is because the tension always acts perpendicular to the displacement of the mass along the arc of the circle, meaning there is no component of tension in the direction of the displacement.

### Summary of Answers
1. **Tension in the string**: \( T = 3mg \cos(23^\circ) \).
2. **Total work done**: \( W = mgL \cos(23^\circ) \).
3. **Work done by the tension**: 0.

Would you like more details on any part of this solution?

### Further Related Questions
1. How would the answer change if the angle were different, say 30° instead of 23°?
2. What would happen to the tension if the mass was increased?
3. How does the conservation of angular momentum apply in this scenario?
4. Could this problem be solved using a different method, such as integrating force over the path?
5. What would be the effect on the work done by gravity if friction was introduced?

### Tip
Always analyze forces in the direction perpendicular to the motion path when considering work done, as forces perpendicular to displacement do not contribute to work.

The string in Figure 6-44 is wound around the nail as shown. The end of the string is attached to a mass m, that is released from the horizontal position, starting from rest. Derive an expression for (a) the tension in the string when it makes an angle of 23° with the vertical, (b) the total work done to bring the string to the position in part (a), (c) the work done by the tension.

Learn how to calculate the tension in a string attached to a mass when it makes an angle with the vertical, and determine the work done to bring it to this position. This problem covers principles of circular motion and conservation of energy.

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