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(A) Given: Weight of the body,$W = 64 \ N$. Coefficient of static friction,$\mu_s = 0.8$. Coefficient of dynamic (kinetic) friction,$\mu_d = 0.6$. The

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$0.2 \ g$

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(A) Given: Weight of the body,$W = 64 \ N$. Coefficient of static friction,$\mu_s = 0.8$. Coefficient of dynamic (kinetic) friction,$\mu_d = 0.6$. The force $F$ required to just start the motion is equal to the limiting friction: $F = \mu_s 	imes W = 0.8 	imes 64 \ N$. Once the body starts moving,the kinetic friction acting on it is $f_k = \mu_d 	imes W = 0.6 	imes 64 \ N$. The net force acting on the body is $F_{net} = F - f_k = (0.8 	imes 64) - (0.6 	imes 64) = 64(0.8 - 0.6) = 64 	imes 0.2 \ N$. Using Newton's second law,$F_{net} = m 	imes a$,where mass $m = \frac{W}{g} = \frac{64}{g}$. Substituting the values: $\frac{64}{g} 	imes a = 64 	imes 0.2$. Therefore,$a = 0.2 \ g$.

$A$ body of weight $64 \ N$ is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are $0.8$ and $0.6$ respectively,then the acceleration of the body will be

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