A horizontal force of 40,N is applied to a 5, | vedclass

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(A) Given:Mass of the block,$m = 5\,kg$Applied horizontal force,$F = 40\,N$Coefficient of kinetic friction,$\mu_k = 0.4$Acceleration due to gravity,$g

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$4$

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(A) Given:Mass of the block,$m = 5\,kg$Applied horizontal force,$F = 40\,N$Coefficient of kinetic friction,$\mu_k = 0.4$Acceleration due to gravity,$g = 10\,m/s^2$Step $1$: Calculate the kinetic frictional force $(f_k)$:The normal force on the horizontal surface is $N = mg = 5 	imes 10 = 50\,N$.The kinetic frictional force is given by $f_k = \mu_k N = 0.4 	imes 50 = 20\,N$.Step $2$: Calculate the net force $(F_{net})$:The net force acting on the block is $F_{net} = F - f_k = 40\,N - 20\,N = 20\,N$.Step $3$: Calculate the acceleration $(a)$:Using Newton's second law,$F_{net} = ma$.$a = \frac{F_{net}}{m} = \frac{20\,N}{5\,kg} = 4\,m/s^2$.Therefore,the acceleration of the block is $4\,m/s^2$.

$A$ horizontal force of $40\,N$ is applied to a $5\,kg$ block which is at rest on a horizontal surface. If the coefficient of kinetic friction is $0.4$,then the acceleration of the block is ........ $m/s^2$ $(g = 10\,m/s^2)$.

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