Answer
The skater does $69.7~J$ of work to pull in her arms while spinning.
Work Step by Step
We can use conservation of angular momentum to find the final angular velocity:
$L_f = L_0$
$I_f~\omega_f = I_0~\omega_0$
$\omega_f = \frac{I_0~\omega_0}{I_f}$
$\omega_f = \frac{(2.50~kg~m^2)(10.0~rad/s)}{1.60~kg~m^2}$
$\omega_f = 15.6~rad/s$
The new angular velocity is $15.6~rad/s$
We can find the change in rotational kinetic energy:
$\Delta KE_{rot} = \frac{1}{2}I_f~\omega_f^2 - \frac{1}{2}I_0~\omega_0^2$
$\Delta KE_{rot} = \frac{1}{2}(1.60~kg~m^2)(15.6~rad/s)^2 - \frac{1}{2}(2.50~kg~m^2)(10.0~rad/s)^2$
$\Delta KE_{rot} = 69.7~J$
Since the rotational kinetic energy increases by $69.7~J$, the skater does $69.7~J$ of work to pull in her arms while spinning.
