A block of mass 5, kg is on a rough horizonta | vedclass

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(D) Given: Mass $m = 5\, kg$,Applied force $F = 24\, N$,Coefficient of kinetic friction $\mu_k = 0.4$,Acceleration due to gravity $g = 9.8\, m/s^2$.Fi

Vedclass · Accepted Answer

$0.88$

Vedclass · Answer

(D) Given: Mass $m = 5\, kg$,Applied force $F = 24\, N$,Coefficient of kinetic friction $\mu_k = 0.4$,Acceleration due to gravity $g = 9.8\, m/s^2$.First,calculate the kinetic frictional force: $f_k = \mu_k \cdot m \cdot g = 0.4 	imes 5 	imes 9.8 = 19.6\, N$.Since the applied force $F = 24\, N$ is greater than the frictional force $f_k = 19.6\, N$,the block will accelerate.The net force acting on the block is $F_{net} = F - f_k = 24 - 19.6 = 4.4\, N$.Using Newton's second law,$F_{net} = ma$,the acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{4.4}{5} = 0.88\, m/s^2$.

$A$ block of mass $5\, kg$ is on a rough horizontal surface and is at rest. Now a force of $24\, N$ is imparted to it with negligible impulse. If the coefficient of kinetic friction is $0.4$ and $g = 9.8\, m/s^2$,then the acceleration of the block is ........ $m/s^2$.

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