The coefficient of friction between a body and the surface of an inclined plane at $45^°$ is $0.5.$ If $g = 9.8\,m/{s^2}$, the acceleration of the body downwards in $m/{s^2}$ is
The coefficient of friction between a body and the surface of an inclined plane at $45^°$ is $0.5.$ If $g = 9.8\,m/{s^2}$, the acceleration of the body downwards in $m/{s^2}$ is
$\frac{{4.9}}{{\sqrt 2 }}$
$4.9\sqrt 2 $
$19.6\sqrt 2 $
$4.9$
The coefficient of friction between a body and the surface of an inclined plane at $45^°$ is $0.5.$ If $g = 9.8\,m/{s^2}$, the acceleration of the body downwards in $m/{s^2}$ is
$a = g(\sin \theta - \mu \cos \theta ) = 9.8(\sin 4{5^o} - 0.5\cos {45^o})$
$ = \frac{{4.9}}{{\sqrt 2 }}\;m/{\sec ^2}$
Other Language