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A particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is 0.5 , the acceleration is ( $g=$ Acceleration due to gravity $)$
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$\frac{3 g}{2 \sqrt{2}}$
Acceleration of an object up a rough inclined plane
$\begin{aligned} a & =g(\sin \theta+\mu \cos \theta)=g\left(\sin 45^{\circ}+0.5 \cos 45^{\circ}\right) \\ & =g\left(\frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right)=\frac{g}{\sqrt{2}}\left(1+\frac{1}{2}\right)=\frac{3 g}{2 \sqrt{2}}\end{aligned}$
$\begin{aligned} a & =g(\sin \theta+\mu \cos \theta)=g\left(\sin 45^{\circ}+0.5 \cos 45^{\circ}\right) \\ & =g\left(\frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right)=\frac{g}{\sqrt{2}}\left(1+\frac{1}{2}\right)=\frac{3 g}{2 \sqrt{2}}\end{aligned}$
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