An Atwoods machine is set up by suspending two blocks connected by astring of negligible mass over a pulley as shown above The blocks are initiallyheld at rest and then released at time to = 0 s The s

We can use the conservation of energy to solve this problem. At time t=0, the potential energy of the system is entirely gravitational potential energy: $U_{initial}=mgh$, where $m$ is the total mass of the system (5 kg), $g$ is the acceleration due to gravity, and $h$ is the initial height difference between the blocks.

As the blocks fall, potential energy is converted to kinetic energy. At time t=2.0 s, the potential energy has decreased to $U_{final}=mgh-2mgd$, where $d$ is the distance that the blocks have fallen (which is also the length of the string that connects them). The kinetic energy at this time is entirely in the 3 kg block, since the 2 kg block is still moving downward. Therefore,

$$\frac{1}{2}mv^2=U_{initial}-U_{final}$$

Plugging in the values, we get:

$$\frac{1}{2}(3 kg)v^2=(5 kg)(9.8 m/s^2)(h-d)-(5 kg)(9.8 m/s^2)h$$

Simplifying and solving for v, we get:

$$v=\sqrt{\frac{2(5 kg)(9.8 m/s^2)d}{3 kg}}\approx \boxed{7.7 m/s}$$