"the gravitational potential energy should DECREASE as it is converted into kinetic energy, but I guess I'm not really sure why it's negative at position 3."
To try and explain this I'll use an analogy. Imagine for a second you have a cat, and you throw the cat up into the air. If you choose to use the ground as a reference point, then the cat starts off with some vertical displacement, and will arc its way downwards until it hits zero, when it hits the ground. Instead, lets try taking the moment the cat leaves our hands as a reference point. Instead, the cat's distance away from you will increase for a while, come to a standstill, drop PAST your hands, and become negative. In a sense, that's the theory behind the statement you've made. Gravitational potential energy really doesn't have any meaning unless you state from which reference point your'e taking the measurement from, just as the cat's height has no meaning unless you say whether you are measuring from your hands or from your feet.
"but why is −mgxsinθ negative and why isn't kx2 negative?"
Because xsinθ is a distance, it will be positive. But that's not what we really want here. Since the block is BELOW the reference point we've taken (point 2), its distance is actually negative. Take the cat example I used, if we take our hands as the reference point, and the cat is below them, it follows that the distance in the upward direction is negative.
kx^2 isn't negative, because the distance in this case is assumed to be inwards toward the spring, as it represents the elastic potential energy. Let's take an analogy. Say the cat that we have thrown arcs, and instead of landing on the floor lands on a trampoline. Logically, the more that the trampoline bends inward, the greater force the cat will feel, and the greater potential the cat will have to fly off into space. This means that the cat's Elastic potential energy will have some correlation to the distance travelled into the trampoline. It's the same principle for the spring, the greater value of x, the more you travel into the spring, the more potential energy you have.
"And finally, how come one side is equal to the other if the right side involves a completely different aspect to it (the spring) that isn't found in the initial conditions. "
Oh, but it IS found in the initial conditions. Here's what the full equation would look like.
Gravitational Potential energy + Kinetic energy + Elastic potential energy (reference point 1)
= Gravitational Potential energy + Kinetic energy + Elastic potential energy(Reference point 3)
But then where did it go on the left hand side? Simple, the block was just not touching the spring yet, so the value of the Elastic potential energy at point one was ZERO, (similar to how the value of the kinetic energy at point 3 is zero).
Hope I answered your question, if something needs clarifying reply, it's my first answer here so it might not be very clear.
You need to add the energies together with some care.
Assume that the system is a spring, spring constant $k$ and a mass $m$ (and the Earth).
Assume that the gravitational potential energy (GPE) and the spring potential energy (SPE) are both zero at the level of the bottom of the unextended spring.
The equilibrium position for the spring-mass system occurs when the condition $mg = kB$ is satisfied where $B$ the extension of the spring.
Let the amplitude of motion be $A$.
Now working out the energy of the system $E(x)$ when the kinetic energy is zero.
$E(B-A) = \frac 1 2 k (B-A)^2 - mg (B-A) = \frac 1 2 kA^2 - \frac 12 k B^2$
$E(B+A) = \frac 1 2 k (B+A)^2 - mg (B+A) = \frac 1 2 kA^2 - \frac 12 k B^2$
The energy of the system is the same at both positions.
A similar analysis could be done by assuming that the gravitational potential energy was zero at a different position, for example at $x=B$ where the energy of the system is now $\frac 1 2 kA^2 + \frac 12 k B^2$
Best Answer
The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it and at some time the maximum compression is reached.
At maximum compression the block is momentarily stationary (between moving down and back up again) and the spring should be stationary as well (though I expect there would be some vibrations). My point is initially and at maximum compression there is no kinetic energy. Overall the initial gravitational energy is converted to elastic potential energy.
With a massive spring it becomes a bit more involved. The spring will compress to some extent under its on weight and as it is compressed its centre of mass will move down reducing its gravitational potential energy.