A solid circular disc of mass 50 ,kg rolls al | vedclass

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(D) According to the work-energy theorem, the work done $W$ is equal to the change in kinetic energy $\Delta KE$.Since the disc is rolling, its total

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$6$

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(D) According to the work-energy theorem, the work done $W$ is equal to the change in kinetic energy $\Delta KE$.Since the disc is rolling, its total kinetic energy $KE$ is the sum of translational and rotational kinetic energy: $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a solid disc, the moment of inertia $I = \frac{1}{2}mR^2$ and for rolling without slipping, $\omega = \frac{v}{R}$.Substituting these, $KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.Given $m = 50 \,kg$ and $v = 0.4 \,m/s$, the initial kinetic energy is $KE_i = \frac{3}{4} 	imes 50 	imes (0.4)^2 = \frac{3}{4} 	imes 50 	imes 0.16 = 0.75 	imes 8 = 6 \,J$.To stop the disc, the final kinetic energy $KE_f = 0$.Thus, $W = KE_f - KE_i = 0 - 6 = -6 \,J$.The absolute value of the work done is $|W| = 6 \,J$.

$A$ solid circular disc of mass $50 \,kg$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 \,m/s$. The absolute value of work done on the disc to stop it is . . . . . . $J$.

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