A 0.150 kg block is placed on a light vertical spring (k = 4.60 103 N/m) and pushed downward, compressing the spring 0.100 m. After the block is released, it leaves the spring and continues to travel upward. What height above the point of release will the block reach if air resistance is negligible?

To find the height above the point of release that the block will reach, we need to first find the energy stored in the spring when it is compressed.

The potential energy stored in a spring is given by the equation:

PE = (1/2) k x^2

where PE is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

In this case, the spring constant is given as k = 4.60 * 10^3 N/m and the displacement of the spring is x = 0.100 m.

Substituting these values into the equation, we have:

PE = (1/2) * (4.60 * 10^3 N/m) * (0.100 m)^2
= 2.30 * 10^1 J

Next, we need to find the initial potential energy of the block when it is released from the spring. This can be calculated as the sum of the potential energy stored in the spring and the gravitational potential energy of the block.

The potential energy due to gravity is given by the equation:

PE = m * g * h

where m is the mass of the block, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the point of release.

In this case, the mass of the block is given as m = 0.150 kg. Substituting the values into the equation, we have:

PE = (0.150 kg) * (9.8 m/s^2) * h
= 1.47 * h J

Now, since the initial potential energy of the block is equal to the energy stored in the spring, we can equate the two expressions:

1.47 * h = 2.30 * 10^1

Solving for h, we find:

h = (2.30 * 10^1) / (1.47)
= 15.65 m

Therefore, the block will reach a height of approximately 15.65 meters above the point of release.