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:

Much like resistors in a circuit, spring constants can be added together to get one equivalent constant. Since these springs are added in parallel, we can simply add their constants together:

Furthermore, we know that:

Substituting this in, we get:

Now we can use Hooke's Law to determine what this equivalent resistance is:

Rearranging for the equivalent constant:

Much like resistors in a circuit, spring constants can be combined to obtain a single equivalent constant. When springs are in parallel, we simply add their constants, and when they are added linearly, we add their inverses. Just like a circuit, let's begin with the two springs in parallel:

Now we can combine this equivalent constant with the third spring:

Since the springs are held in constant, we can determine a total equivalent constant by simply adding the two individual constants together:

From the problem statement, we know:

Substituting this in, we get:

The problem statement tells us that a mass is attached and the springs stretch an unknown distance. At this point, the system has reached a new equilibrium. The problem statement tells us that another mass is attached and the springs stretch more. Since spring forces are linear, the new mass and distance stretched is the only information we need to solve the problem. Using Hooke's Law:

Rearranging for the spring constant:

Plugging in our values:

Since the springs are added in parallel, we can simply add their constants together to get an equivalent constant.

From the statement, we know:

Substituting this in, we get:

Now we can use Hooke's Law along with the information of how far the block causes the springs to stretch:

Rearranging for the equivalent constant:

Plugging in our values:

Now we can use the expression for the potential energy stored in a spring:

The problem statement is asking for the total potential energy stored, so we need to use the total distance stretched (which includes both from the mass and manually). Plugging in our values, we get:

The force applied to a spring can be represented by

where F is force, k is spring constant, and x is the distance compressed.

Solving this equation for k

and plugging in our values

Luckily enough, the angles of the two ropes are the same. Therefore, the tension in each will be the same. This immediately eliminates two of the five answers. Now we just need to calculate what that force is.

We know that together, the vertical components of the tension must equal the weight of the block. Therefore we can write:

Since we know that the two tension forces are equal, we can rewrite:

Rearranging for T, we get: