A block of mass 4 text{ kg} is placed on a ho | vedclass

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(D) Given: Mass $m = 4 	ext{ kg}$, coefficient of static friction $\mu_s = 0.5$, coefficient of kinetic friction $\mu_k = 0.4$, applied force $F = 4

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$4$

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(D) Given: Mass $m = 4 	ext{ kg}$, coefficient of static friction $\mu_s = 0.5$, coefficient of kinetic friction $\mu_k = 0.4$, applied force $F = 4 	ext{ N}$, and acceleration due to gravity $g = 10 	ext{ m/s}^2$.First, calculate the maximum static friction (limiting friction) $f_{s, 	ext{max}} = \mu_s N$, where $N = mg = 4 	imes 10 = 40 	ext{ N}$.$f_{s, 	ext{max}} = 0.5 	imes 40 = 20 	ext{ N}$.Since the applied force $F = 4 	ext{ N}$ is less than the limiting friction $f_{s, 	ext{max}} = 20 	ext{ N}$, the block will remain at rest.According to the laws of static friction, if the applied force is less than the limiting friction, the static friction force $f_s$ is equal to the applied force $F$.Therefore, $f_s = F = 4 	ext{ N}$.

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