A block of mass m slides down an inclined pla | vedclass

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Question

(B) The forces acting along the inclined plane are the component of gravity $mg \sin 30^{\circ}$ acting downwards and the kinetic friction force $f_k

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{3}}$

Vedclass · Answer

(B) The forces acting along the inclined plane are the component of gravity $mg \sin 30^{\circ}$ acting downwards and the kinetic friction force $f_k = \mu N = \mu mg \cos 30^{\circ}$ acting upwards.Applying Newton's second law along the incline:$mg \sin 30^{\circ} - \mu mg \cos 30^{\circ} = ma$Given $a = \frac{g}{4}$,we substitute the values:$mg \sin 30^{\circ} - \mu mg \cos 30^{\circ} = m \left( \frac{g}{4} ight)$Dividing by $mg$:$\sin 30^{\circ} - \mu \cos 30^{\circ} = \frac{1}{4}$$\frac{1}{2} - \mu \left( \frac{\sqrt{3}}{2} ight) = \frac{1}{4}$$\mu \left( \frac{\sqrt{3}}{2} ight) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$\mu = \frac{1}{4} 	imes \frac{2}{\sqrt{3}} = \frac{1}{2 \sqrt{3}}$

$A$ block of mass $m$ slides down an inclined plane at an angle of $30^{\circ}$ with an acceleration of $\frac{g}{4}$. The value of the coefficient of kinetic friction will be:

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