A solid spherical ball is rolled up an inclin | vedclass

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(C) By the law of conservation of energy,the total initial mechanical energy equals the total final mechanical energy at the highest point.Initial ene

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$224$

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(C) By the law of conservation of energy,the total initial mechanical energy equals the total final mechanical energy at the highest point.Initial energy: $E_i = K.E_{trans} + K.E_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,$I = \frac{2}{5}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.$E_i = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$At the highest point,the final kinetic energy is zero,and the potential energy is $mgh$.So,$\frac{7}{10}mv^2 = mgh \Rightarrow h = \frac{7v^2}{10g}$Given $v = 4 \ m/s$ and $g = 10 \ m/s^2$,$h = \frac{7 	imes 4^2}{10 	imes 10} = \frac{7 	imes 16}{100} = 1.12 \ m$.The distance $\ell$ along the plane is given by $\ell = \frac{h}{\sin 30^{\circ}} = \frac{1.12}{0.5} = 2.24 \ m = 224 \ cm$.

$A$ solid spherical ball is rolled up an inclined plane of angle of inclination $30^{\circ}$ with an initial speed of $4 \ m/s$ at the bottom of the inclination. How far will the ball go up the plane (in $cm$)? (Use $g=10 \ m/s^2$)

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