Answer
The translational kinetic energy is $~~0.12~J$
Work Step by Step
In part (c), we found that the linear speed as it reaches the end of the string is $~~1.4~m/s$
We can find the translational kinetic energy:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(0.12~kg)(1.4~m/s)^2$
$K = 0.12~J$
The translational kinetic energy is $~~0.12~J$.
