A wheel rotating at an angular speed of 20 rads^–1, is brought to rest by a constant torque in 4s. If the M.I is 0.2 kg m², the work done in first 2s is

${\mathrm{\omega }}_1 = 20 \mathrm{rad}/\mathrm{s}, {\mathrm{\omega }}_2 = 0; \mathrm{t} = 4\sec$
$\mathrm{\alpha }=\frac{{\mathrm{\omega }}_2-{\mathrm{\omega }}_1}{\mathrm{t}}=\frac{0-20}{4}=-5\mathrm{rad}/{\mathrm{s}}^2$

Given I = 0.2kgm².

When t₁ = 2sec, 

${\mathrm{\omega }}_2={\mathrm{\omega }}_1+{\mathrm{\alpha t}}_1 = 20+(–5)\left(2\right) = 10\mathrm{rad}/\mathrm{s}$

$\mathrm{W}=\frac{1}{2}\mathrm{I}\left({{\mathrm{\omega }}_2}^2-{{\mathrm{\omega }}_1}^2\right)=\frac{1}{2} \mathrm{x} 0.2({\left(10\right)}^2-{\left(20\right)}^2)=-30 \mathrm{J}$