Question

A small block with mass 0.0500 kg slides in a vertical circle of radius 0.0800 m on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.85 N.

What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

Express your answer with the appropriate units.

Answer 1

Given is :-

The mass of the block m = 0.05kg

The raius of the circular path R=0.08m

The normal force on the block at the bottom of the path N=3.85N

Now,

At the bottom the forces are

By plugging all the values in above equation, we get

Now in the y direction(vertical ) there is only gravitational force is working thus by applying third equation of motion in y direction we get

by plugging all the values we get

now by equating force at the top of the path