$\Delta K=\Delta U$ vs $\Delta K = -\Delta U$ While doing physics homework, I noticed that in some problems, the change in kinetic energy(KE) is equal to the change in potential energy(PE), even though I learned that conservation of energy shows that the change in K.E. is equal to the negative of the change in P.E. My question is, when are they equal and when are they equal to the negative of the other? 
For example, a problem with a 2 block system in which one block is on a frictionless table and the other block is attached to the first block by a rope, hanging off the table. The problem gives me both masses and asks for the speed of the blocks after moving 2 meters. In the solution, it states "The only force is gravity (a conservative force), so the change in potential energy of the system is equal to the change in total kinetic energy ΔU=ΔK."
However, in another problem that involves pushing a block against a spring attached to the wall, mass, compression distance, and spring constant are given. The solution states
"ΔU = 0 - .5k(x^2) - 0 = -.5k(x^2) and that energy conservation implies ΔK=−ΔU."
 A: First of all, note that the total energy of a mechanical system is always conserved for this system, i.e., the change in the total energy should always be zero: $$\Delta E=\Delta K+\Delta U=0$$
So, the correct expression is $\Delta K=-\Delta U$ and when only magnitudes are considered, we have $|\Delta K|=|\Delta U|$.
Physical Interpretation:
Consider a pulley system with two unequal masses hanging on it, a larger mass $M$ and a smaller mass $m$. Let the system is released from rest when both the masses are at the same level.
Now, the larger mass $M$ comes down and the smaller mass $m$ goes up and let their common velocity be $V$ and their heights be $H_M$ and $H_m$ respectively.


*

*Total energy for mass $M$: $E_M=\frac{1}{2}MV^2+MgH_M$

*Total energy for mass $m$: $E_m=\frac{1}{2}mV^2+mgH_m$
Since, $M>m$ and $H_M<H_m$, we have the following conclusions:
For the mass $M$, the kinetic energy $K$ increases while the potential energy $U$ decreases as the mass comes down, i.e., $\Delta K>0$ and $\Delta U<0$.
For the mass $m$, the kinetic energy $K$ decreases while the potential energy $U$ increases as the mass goes up, i.e., $\Delta K<0$ and $\Delta U>0$.
A: By the work-kinetic energy theorem for a point particle the total work on the particle changes its kinetic energy, 
$W = \Delta K$. 
Potential energy is defined as the opposite of the work done by a conservative force, 
$\Delta U = -W_c$. 
Thus, if there are no nonconservative forces, 
$-\Delta U = \Delta K$ (or $\Delta E = 0$, where $E=U+K$ is the total mechanical energy). 
This sign is not something that can be chosen, sometimes plus, sometimes minus. 
If this sign is wrong, very strange and wrong results will be found. 
The wrong sign is not physical. 
However, it is the case that 
$|\Delta U| = |\Delta K|$. 
It is possible that your confusion is with what is the initial state and what is the final state. 
Defined these carefully for yourself at the start of every conservation of energy problem. 
(1) Think of the modified Atwood machine problem you describe. 
The potential energy of the block on the table does not change, so the total change in potential energy is due to the falling block which loses potential energy. Thus, $\Delta U < 0$. 
This implies that $\Delta K > 0$, as we expect. 
Initially the blocks were at rest, then they are moving. 
(2) For the second example involving the spring it appears that a block collides with a spring and comes to a stop. 
The potential energy stored by the spring increased, $\Delta U > 0$. 
But the block came to a stop, so $\Delta K < 0$ as expected. 
A: In the first case it is talking about the magnitude only. You can just think yourself, both KE and PE can't increase; as one block moves down, the speed of both blocks increase. The decrease in PE is increase in KE. So there is no difference in two cases. Energy conservation implies $\Delta KE+\Delta PE=0$
