A spring and block is set up on frictionless tabletop. The 30N/m spring is compressed 16 centimeters and released. What is the maximum speed the 2.0 kilogram block will reach?

To find the maximum speed the block will reach, we can use the principle of conservation of mechanical energy.

The potential energy stored in the spring when it is compressed is given by the formula:

Potential energy (PE) = (1/2) * k * (x^2)

Where k is the spring constant and x is the displacement of the spring.

In this case, the spring constant (k) is 30 N/m and the displacement (x) is 16 centimeters, which is equivalent to 0.16 meters. Substituting these values into the formula, we get:

PE = (1/2) * 30 * (0.16^2)
= 0.768 Joules

According to the conservation of mechanical energy, this potential energy is converted into kinetic energy when the spring is released. The kinetic energy (KE) is given by the formula:

Kinetic energy (KE) = (1/2) * m * (v^2)

Where m is the mass of the block and v is its velocity.

In this case, the mass (m) of the block is 2.0 kilograms. We want to find the velocity (v) at the point where all potential energy is converted to kinetic energy. Equating the potential energy and kinetic energy, we get:

0.768 = (1/2) * 2.0 * (v^2)
= v^2

Simplifying the equation, we have:

0.768 = v^2

To find the maximum speed, we need to take the square root of both sides of the equation:

v = √0.768

Using a calculator, we find:

v ≈ 0.88 m/s

Therefore, the maximum speed the 2.0 kilogram block will reach is approximately 0.88 m/s.