An Atwood machine is set up by suspending two blocks connected by astring of negligible mass over a pulley as shown in the figure above Thepulley has negligible mass but there is friction as it rotate

We can use the equations of motion to solve for the acceleration of the system and then use that to find the frictional force. Let's choose downward as the positive direction for the 1 kg block and upward as the positive direction for the 2 kg block. Then we have:

For the 1 kg block: $F_{net} = T - mg = ma$

For the 2 kg block: $F_{net} = mg - T - f = ma$

where $T$ is the tension in the string, $f$ is the frictional force, $m$ is the mass of each block, and $a$ is the acceleration of the system.

We can eliminate $T$ by adding the two equations:

$mg - mg - f = 2a$

$f = -2a$

The negative sign indicates that the frictional force acts in the opposite direction to the motion of the system.

We can find the acceleration using the speed of the 2 kg block after 1.0 s:

$v = at$

$1.8 \text{ m/s} = a \cdot 1.0 \text{ s}$

$a = 1.8 \text{ m/s}^2$

Substituting into the equation for the frictional force:

$f = -2a = -2 \cdot 1.8 \text{ m/s}^2 = -3.6 \text{ N}$

The best estimate of the external frictional force acting on the two-block system is therefore 3.6 N, but this is not one of the answer choices. We need to choose the closest one, which is 1.3 N, or D.