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Let's go through this problem step-by-step. Here’s a breakdown of each part:

### Given:
- A mass \( m \) attached to a string of length \( L \).
- The mass is released from a horizontal position and swings down.
- We need to find values related to the tension in the string and the work done as the mass moves to make an angle of \( 23^\circ \) with the vertical.

### Part (a): Tension in the string at \( 23^\circ \)
1. **Determine the speed of the mass at \( 23^\circ \)**:
   - Use conservation of energy.
   - Initial height \( h_i = L \) (when the mass is horizontal).
   - Final height \( h_f = L \cos(23^\circ) \).
   - Change in height: \( \Delta h = h_i - h_f = L(1 - \cos(23^\circ)) \).
   - The potential energy lost converts to kinetic energy:
     \[
     m g \Delta h = \frac{1}{2} m v^2
     \]
     Solve for \( v \) (speed at \( 23^\circ \)).

2. **Use centripetal force to find tension**:
   - The forces acting on the mass at \( 23^\circ \) are gravity and tension.
   - The component of gravitational force along the radial direction: \( mg \cos(23^\circ) \).
   - Apply Newton's second law in the radial direction:
     \[
     T - mg \cos(23^\circ) = \frac{m v^2}{L}
     \]
     Substitute \( v \) from the energy equation to find \( T \).

### Part (b): Total work done to bring the string to \( 23^\circ \)
- Work done by gravity as the mass falls from \( L \) to \( L \cos(23^\circ) \).
- This work is equal to the change in gravitational potential energy:
  \[
  W_{\text{total}} = m g \Delta h = m g L (1 - \cos(23^\circ))
  \]

### Part (c): Work done by the tension
- Tension is always perpendicular to the direction of motion, so it does no work on the mass.
- Thus, the work done by the tension is zero.

Would you like to see the detailed calculations for each step?

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Here are some related questions to explore further:
1. How would the tension change if the mass were increased?
2. What if the angle were different, say \( 30^\circ \) instead of \( 23^\circ \)?
3. How does the length \( L \) affect the speed of the mass at \( 23^\circ \)?
4. What would happen if there were no gravity in this scenario?
5. Can you calculate the period of this pendulum-like motion?

**Tip:** In circular motion problems, always remember that tension and gravity work together to provide the required centripetal force.

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), and (c) the work done by the tension.

Explore the physics problem involving a mass m attached to a string, released from a horizontal position. Derive expressions for the tension in the string when the mass reaches a 23° angle with the vertical, the total work done, and the work done by tension. This detailed solution covers concepts in circular motion, work, energy, and tension.

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