We can use Newton's second law to solve this problem:

Where the only force is from the spring, so we can say:

Rearranging for mass, we get:

Plugging in our values, we get:

We can use the expression for conservation of energy to solve this problem:

There is no initial kinetic (starts at rest) or final potential (at equilibrium), so we can say:

Where work is done by friction. Inserting expressions for each of these, we get:

Multiplying both sides of the equation by 2 and rearranging for velocity, we get:

Plugging in values for each of these variables, we get:

The radius of the circle will be . This is the rest length plus the stretch of the spring.

Thus, the circumference will be

Since the angular velocity is

Thus, the linear velocity is

The force of the spring will be equal to the centripetal force.

Solving for the spring constant:

Plugging in values:

Using

Solving for

Converting to and plugging in values:

Using

Solving for

Converting to and plugging in values:

Using

Solving for 

Converting to

The spring force is going to add to the gravitational force to equal zero.

Plugging in values:

Solving for

The spring force is going to add to the gravitational force to equal zero.

Plugging in values:

Solving for

Once again, using

Solving for

Since we are never asked about anything pertaining to an individual spring, their individual constants and arrangement are irrelevant. We can simply use an equivalent resistance to solve this problem. Beginning with Hooke's Law:

Rearranging for the spring constant:

Plugging in our values, we get:

Now we have a value for our equivalent resistance. At this point, the system has set into a new equilibrium. This is very important to remember when performing further calculations.

Let's move on to the next relevant information: we manually pull down the mass an additional distance. We are ultimately looking for the maximum velocity of the spring. This occurs when the mass passes through our new equilibrium. Furthermore, we have enough information to calculate the potential energy of the spring when it is at it's maximum stretched length. Therefore, we can use the expression for conservation of energy:

Where the initial condition is when the spring is full stretched, and the final condition is when the mass passes through equilibrium. Therefore, we can eliminate initial kinetic and final potential energy to get:

Plugging in expressions for each of these, we get:

Rearranging for velocity, we get:

We have all of these values, so time to solve the problem:

When in equilibrium, the tension in the rope is simply the weight of the mass:

The maximum tension will occur at the maximum extension of the spring, when 

We can then use Hooke's law to calculate the addition tension in the rope:

Adding the two together, we get:

The tension in the rope is a result of the weight of the mass and the force of the spring:

Substituting expressions in for the two forces:

Rearranging for displacement, we get:

Plugging in our values, we get: