A block of mass 8, kg is at rest on a rough i | vedclass

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Question

(D) Since the block is at rest,the net force acting on it is zero.The forces acting on the block are its weight $(mg)$,the normal force $(N)$ exerted

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) Since the block is at rest,the net force acting on it is zero.The forces acting on the block are its weight $(mg)$,the normal force $(N)$ exerted by the surface,and the static friction force $(f)$ exerted by the surface.The net force exerted by the surface on the block is the vector sum of the normal force and the friction force,which is given by $F_{surface} = \sqrt{N^2 + f^2}$.For a block at rest on an inclined plane:$N = mg \cos 	heta$$f = mg \sin 	heta$Therefore,the net force exerted by the surface is:$F_{surface} = \sqrt{(mg \cos 	heta)^2 + (mg \sin 	heta)^2}$$F_{surface} = \sqrt{m^2g^2(\cos^2 	heta + \sin^2 	heta)}$$F_{surface} = mg$Given $m = 8\, kg$ and $g = 10\, m/s^2$:$F_{surface} = 8 	imes 10 = 80\, N$.Thus,the magnitude of the net force exerted by the surface on the block is $80\, N$.

$A$ block of mass $8\, kg$ is at rest on a rough inclined plane as shown in the figure below. The magnitude of the net force exerted by the surface on the block will be ........ $N$ $(g = 10\, m/s^2)$.

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