A block of mass 2 kg slides down an inclined | vedclass

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(A) Given: Mass $m = 2 \ kg$,angle $	heta = 30^{\circ}$,coefficient of friction $\mu = 0.5$,acceleration due to gravity $g = 10 \ m/s^2$.The contact

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$5\sqrt{15} \ N$

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(A) Given: Mass $m = 2 \ kg$,angle $	heta = 30^{\circ}$,coefficient of friction $\mu = 0.5$,acceleration due to gravity $g = 10 \ m/s^2$.The contact force $F_c$ is the resultant of the normal force $N$ and the frictional force $f$.Normal force $N = mg \cos 	heta = 2 	imes 10 	imes \cos 30^{\circ} = 20 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3} \ N$.Since the block is sliding,the frictional force is kinetic friction: $f = \mu N = 0.5 	imes 10\sqrt{3} = 5\sqrt{3} \ N$.The contact force $F_c$ is given by $F_c = \sqrt{N^2 + f^2}$.Substituting the values: $F_c = \sqrt{(10\sqrt{3})^2 + (5\sqrt{3})^2} = \sqrt{300 + 75} = \sqrt{375} = \sqrt{25 	imes 15} = 5\sqrt{15} \ N$.

$A$ block of mass $2 \ kg$ slides down an inclined plane of inclination $30^{\circ}$. The coefficient of friction between the block and the plane is $0.5$. The contact force between the block and the plane is:

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