A cube of mass m slides down an inclined righ | vedclass

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Question

(D) The component of the gravitational force perpendicular to the inclined plane is $mg \cos 	heta$. This force is balanced by the normal forces $N$

Vedclass · Accepted Answer

$g(\sin 	heta - \sqrt{2} \mu_k \cos 	heta)$

Vedclass · Answer

(D) The component of the gravitational force perpendicular to the inclined plane is $mg \cos 	heta$. This force is balanced by the normal forces $N$ exerted by the two sides of the trough.Since the trough is a right-angle trough,the angle between the normal force from each side and the vertical is $45^{\circ}$.Thus,$2N \cos 45^{\circ} = mg \cos 	heta$.$2N \left(\frac{1}{\sqrt{2}}ight) = mg \cos 	heta \implies N = \frac{mg \cos 	heta}{\sqrt{2}}$.The total kinetic friction force is $f_k = 2 \mu_k N = 2 \mu_k \left(\frac{mg \cos 	heta}{\sqrt{2}}ight) = \sqrt{2} \mu_k mg \cos 	heta$.The net force along the incline is $F_{net} = mg \sin 	heta - f_k = ma$.$ma = mg \sin 	heta - \sqrt{2} \mu_k mg \cos 	heta$.Therefore,the acceleration $a = g(\sin 	heta - \sqrt{2} \mu_k \cos 	heta)$.

$A$ cube of mass $m$ slides down an inclined right-angle trough. If the coefficient of kinetic friction between the cube and the trough is $\mu_k$,then the acceleration of the block is:

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