A body is made to move up along an inclined plane of inclination $30^{\circ}$ and the coefficient of friction is 0.5 , then its retardation is ( $\mathrm{g}$ - acceleration due to gravity)

Question

Aicharya · Accepted Answer

Retardation, $a=g(\sin 	heta+\mu \cos 	heta)$ $=\mathrm{g}\left[\sin 30^{\circ}+0.5\left(\cos 30^{\circ}ight)ight]$ $=g\left(\frac{1}{2}+\frac{1}{2} 	imes \frac{\sqrt{3}}{2}ight)=\left(\frac{1}{2}+\frac{\sqrt{3}}{4}ight) \mathrm{g}=\left(\frac{2+\sqrt{3}}{4}ight) \mathrm{g}$

A body is made to move up along an inclined plane of inclination $30^{\circ}$ and the coefficient of…

Solution