Question:
A solid sphere of mass \(2\,kg\) rolls on a horizontal surface at \(10\,m/s\) and then rolls up a \(30^\circ\) incline. The maximum height reached is:
A solid sphere of mass \(2\,kg\) rolls on a horizontal surface at \(10\,m/s\) and then rolls up a \(30^\circ\) incline. The maximum height reached is:
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Rolling body \(\Rightarrow\) include rotational KE.
Updated On: Apr 16, 2026
- \(10\,m\)
- \(4.9\,m\)
- \(14.2\,m\)
- \(7.1\,m\)
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The Correct Option is D
Solution and Explanation
Concept:
Total energy = translational + rotational:
\[
E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
\]
For solid sphere:
\[
I = \frac{2}{5}mr^2,\quad \omega = \frac{v}{r}
\]
Step 1: Total energy.
\[ E = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \]
Step 2: Convert to potential energy.
\[ mgh = \frac{7}{10}mv^2 \] \[ h = \frac{7v^2}{10g} = \frac{7 \cdot 100}{10 \cdot 9.8} = 7.1\,m \]
Step 1: Total energy.
\[ E = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \]
Step 2: Convert to potential energy.
\[ mgh = \frac{7}{10}mv^2 \] \[ h = \frac{7v^2}{10g} = \frac{7 \cdot 100}{10 \cdot 9.8} = 7.1\,m \]
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