A spring id vertically, as shown below, and a 500g mass is hung from its lower end. The EQUILIBRIUM extension of the spring is observed to be 10.0cm.

(1) What is the spring constant of the spring?
A: 49N/m
(B) If the is pulled down by 5.00cm and released, what is the frequency of the subsequent oscillations.
A: 1.58
(C) What ist he velocity of the mass at time t=2.5 sec?
(D) What is the acceleration of the mass at time t=4.75 sec?
(E) What is the net force acting on the mass at time t=5.0sec?
(F) What are the potential and kinetic energies of the mass when the displacement 7 =-1.5cm?

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

(1) First, let's convert the mass of the object to kilograms: 500g = 0.5kg.

The equilibrium extension of the spring is given as 10.0cm, which is equal to 0.1m.

Using Hooke's Law and the formula F = kx, where F is the force, k is the spring constant, and x is the displacement, we can solve for k.

F = m * g, where m is the mass and g is the acceleration due to gravity (9.8 m/s²).
F = 0.5kg * 9.8 m/s² = 4.9 N.

Since the force exerted by the spring is equal to the weight of the object when the spring is in equilibrium, we have:
4.9 N = k * 0.1m.

Solving for k: k = 4.9 N / 0.1m = 49 N/m.

So, the spring constant of the spring is 49 N/m (option A).

(B) To find the frequency of subsequent oscillations, we can use the formula for frequency:

f = (1 / 2π) * (k / m)^(1/2),

where f is the frequency, k is the spring constant, and m is the mass.

Using the given values of k = 49 N/m and m = 0.5 kg, we can substitute them into the formula:
f = (1 / (2π)) * (49 N/m / 0.5 kg)^(1/2) ≈ 1.58 Hz.

So, the frequency of the subsequent oscillations is approximately 1.58 Hz (option B).

For parts (C), (D), and (E), we need more information about the motion of the mass, such as the amplitude or the equation of motion. Could you provide any additional details?

For part (F), we can calculate the potential and kinetic energies using the equations:

Potential Energy (PE) = 1/2 * k * x^2,
Kinetic Energy (KE) = 1/2 * m * v^2,

where k is the spring constant, x is the displacement, m is the mass, and v is the velocity.

Given that the displacement is 7 = -1.5 cm (which is equivalent to -0.015 m), we can calculate the potential and kinetic energies at this position.

PE = 1/2 * 49 N/m * (-0.015 m)^2 ≈ 0.00547 J,
KE = 1/2 * 0.5 kg * v^2.

Unfortunately, we don't have enough information to determine the velocity v at this position. Can you provide any information about the velocity or further equations of motion?