A body of m kg slides from rest along the curve of vertical circle from point A to B in friction less path. The velocity of the body at B is :(Given, \( R = 14 \, \text{m}, \, g = 10 \, \text{m/s}^2 \, \text{and} \, \sqrt{2} = 1.4 \))
Using the Work-Energy Theorem (W.E.T.) from point A to point B:
1. Apply W.E.T. from A to B: \[ W_{mg} = K_B - K_A. \] Since the body starts from rest at A, \( K_A = 0 \). Thus, \[ mg \times \left( \frac{R}{\sqrt{2}} + R \right) = \frac{1}{2} mv_B^2. \]
2. Substitute Values and Solve for \( v_B \): Simplifying, we get: \[ mgR \frac{(\sqrt{2} + 1)}{\sqrt{2}} = \frac{1}{2} mv_B^2. \] Solving for \( v_B \), we find: \[ v_B = \sqrt{\frac{2gR(\sqrt{2} + 1)}{\sqrt{2}}}. \] Substitute \( g = 10 \, \text{m/s}^2 \), \( R = 14 \, \text{m} \), and \( \sqrt{2} = 1.4 \): \[ v_B = \sqrt{\frac{2 \times 10 \times 14 \times 2.4}{1.4}} = 21.9 \, \text{m/s}. \] Answer: 21.9 m/s
