Question

A block is released from rest at point P and slides along the frictionless track shown. At point Q, its speed is

Answer 1

Let mass of object is $m$

At point $P$

Kinetic energy$=0$

Potential energy $=mg{h}_1$

Total energy , $T_1=mg{h}_1$

At point $Q$

Let $v$ be the velocity of object

Kinetic energy$=\frac{1}{2}m{v}^2$

Potential energy $=mg{h}_2$

Total energy , $T_2=mg{h}_2+\frac{1}{2}m{v}^2$

Using conservation of energy

$⟹T_1=T_2$

$⟹mg{h}_1=mg{h}_2+\frac{1}{2}m{v}^2$

$⟹v=\sqrt{2g(h_1-h_2)}$