A body of mass 2 kg slides down a curved track which is quadrant of a circle of radius 1 m (figure) All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is

Question

A body of mass 2 kg slides down a curved track which is quadrant of a circle of radius 1 m (figure) All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is (A) 4.43 m/s (B) 2 m/s (C) 0.5 m/s (D) 19.6 m/s

Tardigrade Admin · Accepted Answer

By applying law of conservation of energy mgR=(1/2)mv2 ⇒ v=√2Rg =√2× 1× 10 =4.43 m/s

Q. A body of mass $2 k g$ slides down a curved track which is quadrant of a circle of radius $1 m$ (figure) All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is

4.43 m/s

2 m/s

0.5 m/s

19.6 m/s

By applying law of conservation of energy $m g R = \frac{1}{2} m v^{2}$
$\Rightarrow$ $v = 2 R g$
$= 2 \times 1 \times 10$
$= 4.43 m / s$

Q. A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1m (figure) All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is