Question

A block of mass m is released from rest at a height R above a horizontal surface. The acceleration due to gravity is g. The block slides along the inside of a frictionless circular hoop of radius R. Which one of the following expressions gives the speed of the block at the bottom of the hoop?

Answer 1

Answer:

$v=\sqrt{2gR}\ m/s$

Explanation:

Given that

Mass = m

Height h=R

acceleration due to gravity = g m/s²

Initially the speed of the mass ,u=0  m/s

The final speed of the mass at bottom = v m/s

Now from work power energy theorem

Work done by all forces=Change in the kinetic energy

Given that surface is friction less that is why work done by the friction force is zero.

$mgh=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$mgR=\dfrac{1}{2}mv^2$

$v=\sqrt{2gR}\ m/s$

Therefore the speed at the bottom of the circular loop is $\sqrt{2gR}$