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blagie [28]
3 years ago
6

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 bloc

k 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?
Physics
1 answer:
astraxan [27]3 years ago
5 0

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}

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