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:

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:

Toppr Admin · Accepted Answer

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 point0- mgR - 12mv2-0

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:

$\sqrt{5 g R}$
$\sqrt{2 g R}$
$5 g R$
$\sqrt{g R}$

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
$0 +$ $m g R$ = $\frac{1}{2} m v^{2}$ $+ 0$

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:√5gR√2gR5gR√gR

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 point0+ mgR = 12mv2+0

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:

$\sqrt{5 g R}$
$\sqrt{2 g R}$
$5 g R$
$\sqrt{g R}$

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
$0 +$ $m g R$ = $\frac{1}{2} m v^{2}$ $+ 0$