a 500-g mass is attached to the end of an initially unstretched vertical spring for which k=30 n/m. the mass is then released, so that it falls and stretches the spring. how far will it fall before stopping? (the PEg lost by the mass must appear as PEe)

To find the distance the mass will fall before stopping, we need to consider the equilibrium between the gravitational potential energy (PEg) lost by the mass and the elastic potential energy (PEe) stored in the spring.

The potential energy lost by the mass as it falls is given by the equation: PEg = mgh, where m is the mass (500 g = 0.5 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).

PEg = (0.5 kg) * (9.8 m/s²) * h

Now, we need to find the amount of elastic potential energy stored in the spring when the mass is at equilibrium. The elastic potential energy of a spring is given by the equation: PEe = (1/2) * k * x², where k is the spring constant and x is the displacement of the spring from its equilibrium position.

Since the mass is attaching to the end of the spring and stretching it, the displacement of the spring is the same as the distance the mass will fall before stopping.

We want to find the distance x. Rearranging the equation for the elastic potential energy, we have:

x = sqrt((2 * PEe) / k)

Substituting the values, k = 30 N/m and PEe = PEg, we can find the value of x.

x = sqrt((2 * PEg) / k)

Now we can substitute the values of PEg and k into the equation and find the distance:

x = sqrt((2 * (0.5 kg * 9.8 m/s² * h)) / 30 N/m)

Simplifying further:

x = sqrt((0.98 * h) / 0.03)

Therefore, the distance the mass will fall before stopping depends on the height (h) from where it was released. Plug in the value of h to get the final answer.