A block of mass 1, kg lies on a horizontal su | vedclass

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(B) The limiting friction $f_{L}$ is given by $f_{L} = \mu_{s} N = \mu_{s} mg$.Given $\mu_{s} = 0.6$,$m = 1\, kg$,and taking $g = 10\, m\, s^{-2}$ for

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$5$

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(B) The limiting friction $f_{L}$ is given by $f_{L} = \mu_{s} N = \mu_{s} mg$.Given $\mu_{s} = 0.6$,$m = 1\, kg$,and taking $g = 10\, m\, s^{-2}$ for simplicity (or $9.8\, m\, s^{-2}$ as per standard convention).Using $g = 10\, m\, s^{-2}$,$f_{L} = 0.6 	imes 1 	imes 10 = 6\, N$.The pseudo force acting on the block due to the acceleration of the truck is $F_{p} = ma = 1 	imes 5 = 5\, N$.Since the applied pseudo force $(5\, N)$ is less than the limiting friction $(6\, N)$,the block does not slide relative to the truck.Therefore,the static frictional force adjusts itself to balance the pseudo force.Thus,the frictional force is $5\, N$.

$A$ block of mass $1\, kg$ lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is $0.6$. If the acceleration of the truck is $5\, m\, s^{-2}$,the frictional force acting on the block is ........ $N$.

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