A block of mass 2 kg is placed on a rough ho | vedclass

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(B) Given: Mass $m = 2 \ kg$,Applied force $F = 20 \ N$,Acceleration $a = 7 \ m \ s^{-2}$,Acceleration due to gravity $g = 10 \ m \ s^{-2}$.According

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$0.3$

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(B) Given: Mass $m = 2 \ kg$,Applied force $F = 20 \ N$,Acceleration $a = 7 \ m \ s^{-2}$,Acceleration due to gravity $g = 10 \ m \ s^{-2}$.According to Newton's second law of motion,the net force acting on the block is $F_{net} = F - f_k = ma$,where $f_k$ is the kinetic frictional force.The kinetic frictional force is given by $f_k = \mu_k N$,where $N$ is the normal reaction force. On a horizontal surface,$N = mg = 2 \ kg 	imes 10 \ m \ s^{-2} = 20 \ N$.Substituting the values into the equation $F - f_k = ma$:$20 - f_k = 2 	imes 7$$20 - f_k = 14$$f_k = 20 - 14 = 6 \ N$.Now,using $f_k = \mu_k N$:$6 = \mu_k 	imes 20$$\mu_k = 6 / 20 = 0.3$.Therefore,the coefficient of kinetic friction is $0.3$.

$A$ block of mass $2 \ kg$ is placed on a rough horizontal surface. If a horizontal force of $20 \ N$ acting on the block produces an acceleration of $7 \ m \ s^{-2}$ in it,then the coefficient of kinetic friction between the block and the surface is $(g = 10 \ m \ s^{-2})$.

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