An inclined plane makes an angle $30^{\circ}$ with horizontal. A solid sphere rolls down from the top of the inclined plane from rest without slipping has a linear acceleration along the plane equal to (where g is acceleration due to gravity) (given $\sin 30^{\circ}=0.5$ )

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Aicharya · Accepted Answer

$\begin{aligned} & a=\frac{g \sin 	heta}{\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}ight)}=\frac{g \sin 30^{\circ}}{\left(1+\frac{2}{5}ight)} \ 	herefore \quad & a=\frac{5 \mathrm{~g}}{7} 	imes\left(\frac{1}{2}ight)=\frac{5 \mathrm{~g}}{14}\end{aligned}$ .

An inclined plane makes an angle $30^{\circ}$ with horizontal. A solid sphere rolls down from the top of the…

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