A heavy box of mass 50 ,kg is moving on a hor | vedclass

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(B) The force of kinetic friction $f_k$ is given by the formula:$f_k = \mu_k N$where $\mu_k$ is the coefficient of kinetic friction and $N$ is the nor

Vedclass · Accepted Answer

$147$

Vedclass · Answer

(B) The force of kinetic friction $f_k$ is given by the formula:$f_k = \mu_k N$where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force.For a box on a horizontal surface, the normal force $N$ is equal to the weight of the box, $mg$.Given:Mass $m = 50 \,kg$Coefficient of kinetic friction $\mu_k = 0.3$Acceleration due to gravity $g = 9.8 \,m/s^2$Calculating the normal force:$N = mg = 50 	imes 9.8 = 490 \,N$Calculating the kinetic friction force:$f_k = 0.3 	imes 490 = 147 \,N$

$A$ heavy box of mass $50 \,kg$ is moving on a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is $0.3$, then the force of kinetic friction is: (in $\,N$)

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