Given that, moment of inertia, I=10‌kg−m2 Initial angular velocity, ω0=50‌rad∕s Final angular velocity, ω=0 Time taken to stop wheel, t=10s We know that, work done by torque = change of rotational KE W=Δ‌KE=Kf−Ki W=
1
2
Iω2−12Iωa2 W=0−
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2
×10×(50)2 = -12500 J So, amount of work done to bring the flywheel to rest is 12500J.