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A sphere of diameter 0.2 m and mass 2 kg is rolling on an inclined plane with velocity $v=0.5 \mathrm{~m} / \mathrm{s}$. The kinetic energy of the sphere is
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0.42 J
The kinetic energy of sphere while rolling on an inclined plane is given by
$E_K=\text { Translational kinetic energy }+ ext{ Rotational kinetic energy}$
$\begin{aligned}= & \frac{1}{2} m v^2+\frac{1}{2} I \omega^2 \\ E_K= & \frac{1}{2} \times 2 \times(0.5)^2+\frac{1}{2} \times \frac{2}{3} \times 2 \times(0.1)^2 \times\left(\frac{0.5}{0.1}\right)^2 \\ = & 0.25+0.17=0.42 \mathrm{~J}\end{aligned}$
$E_K=\text { Translational kinetic energy }+ ext{ Rotational kinetic energy}$
$\begin{aligned}= & \frac{1}{2} m v^2+\frac{1}{2} I \omega^2 \\ E_K= & \frac{1}{2} \times 2 \times(0.5)^2+\frac{1}{2} \times \frac{2}{3} \times 2 \times(0.1)^2 \times\left(\frac{0.5}{0.1}\right)^2 \\ = & 0.25+0.17=0.42 \mathrm{~J}\end{aligned}$
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