To solve the problem (assuming it asks for the tension at the lowest point of a pendulum released from rest at angle θ and the maximum angle before the string breaks), here's the step-by-step solution:

Problem Context

A simple pendulum of length (L) and mass (m) is released from rest at an angle (\theta) from the vertical.

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Part 1: Tension at the Lowest Point

We use conservation of mechanical energy and Newton’s second law for centripetal force.

Step 1: Energy Conservation

• Initial kinetic energy ((K_i)) = 0 (released from rest).
• Initial potential energy ((U_i)) = (mgL(1-\cos\theta)) (height above lowest point: (h = L(1-\cos\theta))).
• Final kinetic energy ((K_f)) = (\frac{1}{2}mv^2) (at lowest point, potential energy (U_f = 0)).

Conservation equation:
[K_i + U_i = K_f + U_f]
[mgL(1-\cos\theta) = \frac{1}{2}mv^2]
[mv^2 = 2mgL(1-\cos\theta)]

Step 2: Centripetal Force at Lowest Point

Forces on the mass: tension ((T) upward) and weight ((mg) downward). Net force provides centripetal acceleration ((\frac{v^2}{L})):
[T - mg = \frac{mv^2}{L}]

Substitute (mv^2) from energy:
[T = mg + \frac{2mgL(1-\cos\theta)}{L}]
[T = mg(3 - 2\cos\theta)]

Part 2: Maximum Angle Before String Breaks

Let (T{\text{max}}) be the breaking tension. Set (T = T{\text{max}}):
[T{\text{max}} = mg(3 - 2\cos\theta{\text{max}})]

Solve for (\theta{\text{max}}):
[cos\theta{\text{max}} = \frac{3mg - T{\text{max}}}{2mg}]
[θ{\text{max}} = \arccos\left(\frac{3mg - T_{\text{max}}}{2mg}\right)]

Final Results

1. Tension at lowest point: (\boxed{T = mg(3 - 2\cos\theta)})
2. Maximum angle before breaking: (\boxed{θ{\text{max}} = \arccos\left(\frac{3mg - T{\text{max}}}{2mg}\right)})

If the problem refers to a pendulum in a moving car (uniform circular motion), the solutions are:

• Speed of car: (\boxed{v = \sqrt{Rg\tanθ}})
• Tension: (\boxed{T = \frac{mg}{\cosθ}})

(Note: Adjust based on the exact problem statement, but these are the two common versions of Problem 10.15.)

Answer:
Assuming the first version (release from angle), the key result is (\boxed{mg(3 - 2\cos\theta)}) for tension at the lowest point. For maximum angle, it's (\boxed{\arccos\left(\frac{3mg - T_{\text{max}}}{2mg}\right)}).

If you provide more details from the image, I can refine the answer further!
(\boxed{mg(3 - 2\cos\theta)}) (default for tension at lowest point)

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