A block of mass m is moving on a rough horizo | vedclass

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Question

(A) The surface exerts two forces on the block: the normal force $(N)$ acting vertically upwards and the kinetic frictional force $(f_{k})$ acting hor

Vedclass · Accepted Answer

$mg(1+\mu_{k}^{2})^{1/2}$

Vedclass · Answer

(A) The surface exerts two forces on the block: the normal force $(N)$ acting vertically upwards and the kinetic frictional force $(f_{k})$ acting horizontally opposite to the direction of motion.$1$. The normal force is $N = mg$.$2$. The kinetic frictional force is $f_{k} = \mu_{k} N = \mu_{k} mg$.$3$. The net force $(F_{net})$ exerted by the surface is the vector sum of the normal force and the frictional force:$F_{net} = \sqrt{N^{2} + f_{k}^{2}}$$F_{net} = \sqrt{(mg)^{2} + (\mu_{k} mg)^{2}}$$F_{net} = \sqrt{(mg)^{2} (1 + \mu_{k}^{2})}$$F_{net} = mg \sqrt{1 + \mu_{k}^{2}}$$F_{net} = mg(1 + \mu_{k}^{2})^{1/2}$Therefore,option $A$ is correct.

$A$ block of mass $m$ is moving on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is $\mu_{k}$. The net force exerted by the surface on the block is (where $g$ is the acceleration due to gravity).

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