Answer
$$W_{Cue}=\left| W_{frictotal}\right| =4.32J$$
Work Step by Step
$$\dfrac {mv^{2}}{2}=W_{fric}\Rightarrow W_{fric}=\dfrac {0.42\times \left( 4.2\right) ^{2}}{2}\approx 3.7J=F_{fric}\times 12\Rightarrow F_{fric}=\dfrac {3.7}{12}\approx 0.3083N\Rightarrow W_{frictotal}=F_{fric}\times 14\approx 4.32J$$
The total work done by friction will be equal in magnitute to the work done by the cue becouse the disk will stop at the end.
$W_{Cue}=\left| W_{frictotal}\right| =4.32J$
