In an Atwood machine, the heavier block has a mass twice that of...

Let the mass of the lighter block be $m$ and the heavier block be $2m$.

The acceleration of the system is $a = \dfrac{2m - m}{2m + m}g = \dfrac{g}{3}$.

The tension in the string is $T = \dfrac{2(2m)(m)}{2m + m}g = \dfrac{4mg}{3}$.

Given $T = 16.0 \text{ N}$, we have $\dfrac{4mg}{3} = 16.0 \Rightarrow mg = 12.0 \text{ N}$.

The distance moved by the blocks in the first second ($t = 1 \text{ s}$) is $s = \dfrac{1}{2}at^2 = \dfrac{1}{2}\left(\dfrac{g}{3}\right)(1)^2 = \dfrac{g}{6}$.

The decrease in gravitational potential energy is $\Delta U = (2m)gs - mgs = mgs$.

Substituting the values, $\Delta U = (12.0)\left(\dfrac{g}{6}\right) = 2.0g$.

Taking $g = 9.8 \text{ m/s}^2$, $\Delta U = 2.0 \times 9.8 = 19.6 \text{ J}$.