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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.

https://brainmass.com/physics/energy/block-mass-released-rest-height-104608

#### 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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## A block suspended above the ground by a light rope wrapped around a pulley wheel, is released allowing it to rotate freely, letting the block fall to the ground. To determine the rotational speed, speed, time to reach the ground.

A 2 Kg block is suspended 5 m above the ground by a light rope wrapped around a 3 kg solid pulley wheel, as shown in figure. The pulley wheel, which was initially at rest, is then released allowing it to rotate freely, letting the block fall to the ground.

Ignoring friction effects the rope mass:

(a) Briefly state where, and in what forms, energy is stored in this system:

(Note - numerical calculations are not required here, just clear comments about the forms and locations of energy storage. State any assumptions you make)

(1) Before the pulley wheel is released.

(2) Just before block hits ground.

(b) What is the rotational speed of the pulley wheel (in revolutions per sec) as the block hits the ground?

(c) At what speed will the block hit the ground?

(d) How long does it take (after release) for the block to reach the ground?

See attached file for full problem description.

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