A solid sphere of mass 2 kg is rolling on a | vedclass

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Question

(C) The total kinetic energy of a rolling solid sphere is the sum of its translational and rotational kinetic energies.$KE_{total} = \frac{1}{2} m V^2

Vedclass · Accepted Answer

$0.7$

Vedclass · Answer

(C) The total kinetic energy of a rolling solid sphere is the sum of its translational and rotational kinetic energies.$KE_{total} = \frac{1}{2} m V^2 + \frac{1}{2} I \omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5} m r^2$ and the rolling condition is $V = r \omega$.$KE_{total} = \frac{1}{2} m V^2 + \frac{1}{2} (\frac{2}{5} m r^2) (\frac{V}{r})^2 = \frac{1}{2} m V^2 + \frac{1}{5} m V^2 = \frac{7}{10} m V^2$.When the sphere compresses the spring by a maximum distance $x$,all its kinetic energy is converted into the potential energy of the spring $(U = \frac{1}{2} k x^2)$.$\frac{1}{2} k x^2 = \frac{7}{10} m V^2$$x^2 = \frac{14}{10} \frac{m V^2}{k} = \frac{1.4 	imes 2 	imes 6^2}{36} = \frac{1.4 	imes 2 	imes 36}{36} = 2.8$.$x = \sqrt{2.8} \approx 1.67 \ m$.Note: Given the options provided,there appears to be a discrepancy in the calculation or the provided options. Based on the standard physics approach,the result is $\sqrt{2.8} \ m$.

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