A block slides down an incline of angle 30^o | vedclass

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(C) The forces acting on the block along the incline are the component of gravity $mg \sin 	heta$ acting downwards and the kinetic friction force $f_

Vedclass · Accepted Answer

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

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(C) The forces acting on the block along the incline are the component of gravity $mg \sin 	heta$ acting downwards and the kinetic friction force $f_k = \mu_k N = \mu_k mg \cos 	heta$ acting upwards.According to Newton's second law,the net force is $F_{net} = ma$.$mg \sin 30^o - \mu_k mg \cos 30^o = m \left( \frac{g}{4} ight)$.Dividing both sides by $mg$,we get $\sin 30^o - \mu_k \cos 30^o = \frac{1}{4}$.Substituting the values $\sin 30^o = \frac{1}{2}$ and $\cos 30^o = \frac{\sqrt{3}}{2}$,we have $\frac{1}{2} - \mu_k \frac{\sqrt{3}}{2} = \frac{1}{4}$.Rearranging the terms: $\mu_k \frac{\sqrt{3}}{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.Solving for $\mu_k$: $\mu_k = \frac{1}{4} 	imes \frac{2}{\sqrt{3}} = \frac{1}{2\sqrt{3}}$.

$A$ block slides down an incline of angle $30^o$ with an acceleration $\frac{g}{4}$. Find the coefficient of kinetic friction.

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