Answer
The translational kinetic energy is $~~0.019~J$
Work Step by Step
In part (c), we found that the linear speed as it reaches the end of the string is $~~v = 0.56~m/s$
We can find the translational kinetic energy:
$K_{trans} = \frac{1}{2}Mv^2$
$K_{trans} = \frac{1}{2}(0.12~kg)(0.56~m/s)^2$
$K_{trans} = 0.019~J$
The translational kinetic energy is $~~0.019~J$.
