Answer:
<u><em>note:</em></u>
<u><em>find the attached solution:</em></u>
The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.
<h3>What is Potential and Kinetic energy?</h3>
Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.
mass of rod, mab = 2.4kg
mass of rod, mbc = 4kg
conservation of energy


potential energy at position 1,

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18
V1 = 11.30112
kinetic energy T1 at position 1 is zero
potential energy at position 2 is zero
K.E at position 2,


= 1/3 *4 * (0.36)²
=0.10368kg m²

= 1/12 *4 * (0.6)²
=0.12kg m²
on putting the values in above equation we get,
T₂ = 1.0667vb²
0 + 11.30112 = 1.0667vb² + 0
vb = 3.2549 m/s
to learn more about Kinetic and potential energy go to - brainly.com/question/18963960
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Answer:

Explanation:
In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

The initial potential energy of the spring is given by the equation:

the Kinetic energy of the block is then given by the equation:

so we can now set them both equal to each other, so we get:

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

so now we can solve this for the final velocity, so we get:
