A flywheel with a moment of inertia of 5.0 kg-m^2 is rotated at a...

Initial kinetic energy of the flywheel is given by $K_i = \dfrac{1}{2} I \omega_i^2$.

Substituting the given values, $I = 5.0 \text{ kg-m}^2$ and $\omega_i = 60 \text{ rad/s}$:

$K_i = \dfrac{1}{2} \times 5.0 \times (60)^2$

$K_i = \dfrac{1}{2} \times 5.0 \times 3600 = 9000 \text{ J}$

Final kinetic energy $K_f = 0$ since the flywheel comes to rest.

According to the work-energy theorem, the total work done by friction is equal to the change in kinetic energy:

$W = K_f - K_i = 0 - 9000 \text{ J} = -9000 \text{ J}$

The magnitude of the total work done by friction is $9.0 \text{ kJ}$.