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

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(D) When a body is at rest on an inclined plane,the net force acting on it is zero.Therefore,the net force exerted by the surface on the block must ba

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$80$

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(D) When a body is at rest on an inclined plane,the net force acting on it is zero.Therefore,the net force exerted by the surface on the block must balance the weight of the block $(mg)$ to keep it in equilibrium.The force exerted by the surface consists of two components: the normal force $(N)$ perpendicular to the surface and the static frictional force $(f_s)$ parallel to the surface.$N = mg \cos 	heta$$f_s = mg \sin 	heta$The magnitude of the net force $(F)$ exerted by the surface is given by:$F = \sqrt{N^2 + f_s^2} = \sqrt{(mg \cos 	heta)^2 + (mg \sin 	heta)^2}$$F = \sqrt{m^2g^2(\cos^2 	heta + \sin^2 	heta)} = mg$Given $m = 8\, kg$ and taking $g = 10\, m/s^2$:$F = 8 	imes 10 = 80\, N$.

$A$ block of mass $8\, kg$ is at rest on a rough inclined plane as shown in the figure. The magnitude of the net force exerted by the surface on the block will be ......... $N$.

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