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# A block of mass m is released from rest at a height R above

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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.

A.Which expression would give the speed of the block at the bottom of the hoop and why?

1.v=mgR
2.v=mg/2R
3.v^2=g^2/R
4.v^=2gR.

B. What is the magnitude of the normal force exerted on the block by the hoop when the block has reached the bottom?

See the attached file.

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#### Solution Preview

From position 1 to 2, potential energy has changed into kinetic energy.
Position 1: Potential energy = mgR
Position 2 (bottom of hoop): Kinetic ...

#### Solution Summary

The solution provides detailed discussions and explanations for the problem.

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