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:

This question is essentially a reworded form of how much the spring is stretched when the mass is attached. When the mass is attached, it reaches a new equilibrium, and we are asked to find the distance between this new equilibrium and the original equilibrium prior to the mass being attached.

This distance can be calculated by setting Hooke's Law equal to the weight of the mass (this is the point where the force of the spring is equal and opposite to the weight of the mass:

Rearranging for displacement, we get:

Plugging in our values, we get:

Since the springs are attached in a linear fashion, the weight of the mass is applied equally to both springs. Therefore, we can determine how far each spring will stretch and then add the two together. For a spring in general:

Rearranging:

Applying this expression to each spring, we get:

Plugging in our values, we get:

Much like resistors in a circuit, spring constants attached in parallel and linear fashion can be combined to a single constant. Since the springs are attached in a linear fashion, we will use the following expression:

Which becomes:

From the problem statement, we know that:

So let's substitute that into the problem:

Now we that we have a "total" spring constant, we can use Hooke's Law:

Plugging in our values, we get:

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. Let's move on to the next relevant information: we manually pull down the mass an additional distance. Once again we can use Hooke's Law, but this time we will be calculating the force required to do this:

It's necessary to clarify what this force is. This force is the additional force on top of gravity that requires to get the mass to this point. When the mass is attached and the springs stretch, the force of gravity and the force of the spring cancel each other out. Furthermore, since spring forces are linear, we only need this additional force to calculate the instantaneous acceleration of the mass when released: