Question

A body slides down a fixed curved track that is one quadrant of a circle of radius $R$, as in the figure. If there is no friction and the body starts from rest, its speed at the bottom of the track is:

• A. $5gR$
• B. $\sqrt{5gR}$
• C. $\sqrt{2gR}$
• D. $\sqrt{gR}$

Answer 1

Answer: (C)

$\sqrt{2gR}$

According to the law of conservation of energy.
Kinetic energy at top point $+$ Potential energy at top point $=$ Kinetic energy at lowest point $+$ Potential energy at lowest point
$0+$ $mgR$ = $\frac{1}{2}m{v}^2$$+0$

$\Rightarrow v=\sqrt{2gR}$