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(B) The forces acting on the block along the inclined plane are the component of gravity $mg \sin 	heta$ acting downwards and the kinetic friction fo

Vedclass · Accepted Answer

$\sqrt{3} - 1$

Vedclass · Answer

(B) The forces acting on the block along the inclined plane 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.Applying Newton's second law along the incline:$mg \sin 	heta - f_k = ma$$mg \sin 	heta - \mu_k mg \cos 	heta = ma$Dividing by $m$:$g \sin 	heta - \mu_k g \cos 	heta = a$Given $	heta = 60^{\circ}$ and $a = g/2$:$g \sin 60^{\circ} - \mu_k g \cos 60^{\circ} = g/2$$g \left( \frac{\sqrt{3}}{2} ight) - \mu_k g \left( \frac{1}{2} ight) = \frac{g}{2}$Dividing by $g$:$\frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2}$$\sqrt{3} - \mu_k = 1$$\mu_k = \sqrt{3} - 1$

$A$ block slides down an inclined plane with an acceleration $g/2$ as shown in the figure. Then the coefficient of kinetic friction is

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