A flywheel with a moment of inertia of 5.0 kg-m ^2 is rotated at a speed of 60 rad/s . Due to friction at the axle, it comes to rest in 5.0 minutes . Determine the total work done by the friction.

Step-by-step solution

Initial kinetic energy of the flywheel is given by K_i = 1 2 I _i^2 . Substituting the given values, I = 5.0 kg-m ^2 and _i = 60 rad/s : K_i = 1 2 5.0 (60)^2 K_i = 1 2 5.0 3600 = 9000 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 J = -9000 J The magnitude of the total work done by friction is 9.0 kJ .