A body is sliding down an inclined plane (angle of inclination $45^o$). If the coefficient of friction is $0.5$ and $g = 9.8\, m/s^2$. then the acceleration of the body downwards in $m/s^2$ is
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$ a =g \sin \theta-\mu g \cos \theta$
$=g\left[\sin 45^{\circ}-0.5 \times \cos 45^{\circ}\right]$
$=\frac{4.9}{\sqrt{2}} \mathrm{m} / \mathrm{s}^{2} 1 $
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