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If a solid sphere of mass \(1 \mathrm{~kg}\) and radius \(0.1 \mathrm{~m}\) rolls without slipping at a uniform velocity of \(1 \mathrm{~m} / \mathrm{s}\) along a straight line on a horizontal floor, the kinetic energy is
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The correct answer is:
\(\frac{7}{10} \mathrm{~J}\)
\(\begin{gathered}
\mathrm{K} . \mathrm{E} \text { of rolling }=\left(\frac{1}{2} m v^2+\frac{1}{2} I \cdot \frac{v^2}{R^2}\right) \\
\Rightarrow \mathrm{K} \cdot \mathrm{E}=\frac{1}{2} m v^2+\frac{1}{2} \cdot \frac{2}{5} \cdot \frac{m R^2 \cdot v^2}{R^2} \\
\mathrm{~K} . \mathrm{E}=\frac{1}{2} m v^2 \times \frac{7}{5} \\
m=1 \mathrm{~kg}, \quad v = 1 ~m/s \\
\mathrm{~K} \cdot \mathrm{E}=\frac{1}{2} \times 1 \times 1 \times \frac{7}{5}=\frac{7}{10} \mathrm{~J}
\end{gathered}\)
\mathrm{K} . \mathrm{E} \text { of rolling }=\left(\frac{1}{2} m v^2+\frac{1}{2} I \cdot \frac{v^2}{R^2}\right) \\
\Rightarrow \mathrm{K} \cdot \mathrm{E}=\frac{1}{2} m v^2+\frac{1}{2} \cdot \frac{2}{5} \cdot \frac{m R^2 \cdot v^2}{R^2} \\
\mathrm{~K} . \mathrm{E}=\frac{1}{2} m v^2 \times \frac{7}{5} \\
m=1 \mathrm{~kg}, \quad v = 1 ~m/s \\
\mathrm{~K} \cdot \mathrm{E}=\frac{1}{2} \times 1 \times 1 \times \frac{7}{5}=\frac{7}{10} \mathrm{~J}
\end{gathered}\)
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