Question

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, $R = 14 \, \text{m}, \, g = 10 \, \text{m/s}^2 \, \text{and} \, \sqrt{2} = 1.4$)

• A. 19.8 m/s
• B. 21.9 m/s
• C. 16.7 m/s
• D. 10.6 m/s

Answer 1

Answer: (B)

21.9 m/s

Answer 2

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