A 1.26 kg block is held against a spring of force constant 2.0 103 N/m, compressing it a distance of 0.18 m. How fast is the block moving after it is released and the spring pushes it away?

To find the speed of the block after it is released and the spring pushes it away, we can use the principle of conservation of energy. The potential energy stored in the compressed spring is converted into kinetic energy of the block as it moves away.

First, let's calculate the potential energy stored in the compressed spring:

Potential Energy (PE) = 0.5 * k * x^2

where k is the force constant of the spring and x is the compression distance.

PE = 0.5 * 2.0 * 10^3 N/m * (0.18 m)^2
= 0.5 * 2.0 * 10^3 N/m * 0.0324 m^2
= 648 N·m

Next, we can equate the potential energy to the kinetic energy of the block:

PE = KE

0.5 * m * v^2 = PE

where m is the mass of the block and v is the speed of the block.

Substituting the known values:

0.5 * 1.26 kg * v^2 = 648 N·m

Simplifying the equation:

0.63 kg * v^2 = 648 N·m

Dividing both sides by 0.63 kg:

v^2 = 1032 N·m / 0.63 kg

v^2 ≈ 1638.1 m^2/s^2

Taking the square root of both sides:

v ≈ √(1638.1 m^2/s^2)

v ≈ 40.5 m/s

Therefore, the block is moving at approximately 40.5 m/s after it is released and the spring pushes it away.