A solid sphere of mass 2 ,kg rolls on a smoot | vedclass

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(C) Let $h$ be the height attained by the sphere before it stops. According to the law of conservation of energy, the total initial kinetic energy (tr

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$7.0 \,m$

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(C) Let $h$ be the height attained by the sphere before it stops. According to the law of conservation of energy, the total initial kinetic energy (translational + rotational) is converted into potential energy at the maximum height.$K_{	ext{rot}} + K_{	ext{trans}} = mgh$$\frac{1}{2} I \omega^2 + \frac{1}{2} m v_{CM}^2 = mgh$Since the sphere is rolling, $v_{CM} = R\omega$ and for a solid sphere, the moment of inertia $I = \frac{2}{5} mR^2$.Substituting these values:$\frac{1}{2} \left( \frac{2}{5} mR^2 ight) \left( \frac{v_{CM}}{R} ight)^2 + \frac{1}{2} m v_{CM}^2 = mgh$$\frac{1}{5} m v_{CM}^2 + \frac{1}{2} m v_{CM}^2 = mgh$Dividing by $m$ and simplifying:$v_{CM}^2 \left( \frac{1}{5} + \frac{1}{2} ight) = gh$$v_{CM}^2 \left( \frac{2+5}{10} ight) = gh$$\frac{7}{10} v_{CM}^2 = gh$Given $v_{CM} = 10 \,m/s$ and $g = 10 \,m/s^2$:$\frac{7}{10} 	imes (10)^2 = 10 	imes h$$\frac{7}{10} 	imes 100 = 10h$$70 = 10h$$h = 7 \,m$

$A$ solid sphere of mass $2 \,kg$ rolls on a smooth horizontal surface at $10 \,m/s$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g=10 \,m/s^2$].

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