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(D) According to the Work-Energy Theorem,the work done by the frictional force is equal to the change in kinetic energy of the cylinder.$W = \Delta K$

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$10$

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(D) According to the Work-Energy Theorem,the work done by the frictional force is equal to the change in kinetic energy of the cylinder.$W = \Delta K$$-f_k \cdot s = K_f - K_i$Since the final velocity is $0$,the final kinetic energy $K_f = 0$.$-\mu_k \cdot m \cdot g \cdot s = 0 - \frac{1}{2} m v^2$$\mu_k \cdot m \cdot g \cdot s = \frac{1}{2} m v^2$$s = \frac{v^2}{2 \mu_k g}$Substituting the given values: $v = 10 \, m/s$,$\mu_k = 0.5$,and $g = 10 \, m/s^2$.$s = \frac{(10)^2}{2 	imes 0.5 	imes 10} = \frac{100}{10} = 10 \, m$.

$A$ cylinder of $10 \, kg$ is sliding on a plane with an initial velocity of $10 \, m/s$. If the coefficient of friction between the surface and the cylinder is $0.5$,then the distance it will cover before stopping is ........ $m$. $(g = 10 \, m/s^2)$

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