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(C) The maximum static frictional force (limiting friction) is given by $f_{L} = \mu_{s} N$,where $N = mg$.Given: $m = 2\,kg$,$\mu_{s} = 0.54$,and $g

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(C) The maximum static frictional force (limiting friction) is given by $f_{L} = \mu_{s} N$,where $N = mg$.Given: $m = 2\,kg$,$\mu_{s} = 0.54$,and $g = 10\,m/s^2$.$f_{L} = 0.54 	imes 2 	imes 10 = 10.8\,N$.The applied horizontal force is $F_{applied} = 2.8\,N$.Since the applied force is less than the limiting friction $(F_{applied} According to the law of static friction,the frictional force adjusts itself to be equal to the applied force when the object is at rest.Therefore,the frictional force is $2.8\,N$.

$A$ block of mass $2\,kg$ is at rest on a floor. The coefficient of static friction between the block and the floor is $0.54$. $A$ horizontal force of $2.8\,N$ is applied to the block. What will be the frictional force between the block and the floor (in $.8$)? $(g = 10\,m/s^2)$

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