A block of mass 5 kg is moving on an incline | vedclass

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(B) For the block to move down without acceleration, the net force along the inclined plane must be zero.Let $F$ be the applied force directed up the

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(B) For the block to move down without acceleration, the net force along the inclined plane must be zero.Let $F$ be the applied force directed up the incline.The forces acting down the incline are the component of gravity $mg \sin 30^{\circ}$.The forces acting up the incline are the applied force $F$ and the kinetic friction $f_k = \mu N = \mu mg \cos 30^{\circ}$.Equating the forces: $mg \sin 30^{\circ} = F + \mu mg \cos 30^{\circ}$.Substituting the values: $5 	imes 10 	imes \frac{1}{2} = F + \frac{\sqrt{3}}{2} 	imes 5 	imes 10 	imes \frac{\sqrt{3}}{2}$.$25 = F + \frac{\sqrt{3}}{2} 	imes 50 	imes \frac{\sqrt{3}}{2} = F + \frac{3}{4} 	imes 50 = F + 37.5$.$F = 25 - 37.5 = -12.5 \ N$.The negative sign indicates that the force $F$ must be applied in the opposite direction (i.e.,down the incline) to maintain constant velocity.Therefore, a force of $12.5 \ N$ must be applied down the incline.

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