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(D) The block is in equilibrium on the inclined plane. The forces acting on the block are its weight $(mg)$,the normal force $(N)$ exerted by the surf

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) The block is in equilibrium on the inclined plane. The forces acting on the block are its weight $(mg)$,the normal force $(N)$ exerted by the surface,and the static friction force $(f_s)$ acting up the incline.For equilibrium,the net force on the block is zero.The normal force is $N = mg \cos 	heta$.The static friction force is $f_s = mg \sin 	heta$.The net force exerted by the surface on the block is the resultant of the normal force and the friction force,given by $F_{net} = \sqrt{N^2 + f_s^2}$.Substituting the expressions: $F_{net} = \sqrt{(mg \cos 	heta)^2 + (mg \sin 	heta)^2} = \sqrt{m^2g^2(\cos^2 	heta + \sin^2 	heta)} = mg$.Given $m = 8 \ kg$ and $g = 10 \ m/s^2$,the net force is $F_{net} = 8 	imes 10 = 80 \ N$.

$A$ 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 $..... \ N$. (Take $g = 10 \ m/s^2$)

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