A spherical ball of mass 20 kg is stationary | vedclass

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(A) According to the law of conservation of mechanical energy,the loss in potential energy $(PE)$ is equal to the gain in total kinetic energy $(KE_{t

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$40 \sqrt{\frac{5}{7}} \ m/s$

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(A) According to the law of conservation of mechanical energy,the loss in potential energy $(PE)$ is equal to the gain in total kinetic energy $(KE_{total})$,which includes both translational and rotational kinetic energy.Initial height $h_1 = 100 \ m$,final height $h_2 = 20 \ m$.Change in height $\Delta h = h_1 - h_2 = 100 \ m - 20 \ m = 80 \ m$.Loss in $PE = mg \Delta h = m 	imes 10 	imes 80 = 800m \ J$.Gain in $KE_{total} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$.For a solid spherical ball,the moment of inertia $I = \frac{2}{5} mr^2$ and $\omega = \frac{v}{r}$.Substituting these values: $KE_{total} = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v}{r})^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2$.Equating loss in $PE$ to gain in $KE_{total}$:$800m = \frac{7}{10} mv^2$.$v^2 = \frac{8000}{7}$.$v = \sqrt{\frac{8000}{7}} = \sqrt{\frac{1600 	imes 5}{7}} = 40 \sqrt{\frac{5}{7}} \ m/s$.

$A$ spherical ball of mass $20 \ kg$ is stationary at the top of a hill of height $100 \ m$. It rolls down a surface to the ground,then climbs up another hill of height $30 \ m$ and finally rolls down to a horizontal base at a height of $20 \ m$ above the ground. The velocity attained by the ball is

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