$A$ body weighing $20 \ kg$ just slides down a rough inclined plane that rises $5$ in $12$. The coefficient of friction is

An insect is at the bottom of a hemispherical ditch of radius $R = 1\, m$. It crawls up the ditch but starts slipping after it is at height $h$ from the bottom. If the coefficient of friction between the ground and the insect is $\mu = 0.75$,then $h$ is $.......\, m$. $(g = 10\, m s^{-2})$

The coefficient of friction between a body and the surface of an inclined plane at $45^o$ is $0.5$. If $g = 9.8\,m/s^2$,the acceleration of the body downwards in $m/s^2$ is

$A$ block of a certain mass is placed on a rough inclined plane. The angle between the plane and the horizontal is $30^{\circ}$. The coefficients of static and kinetic friction between the block and the inclined plane are $0.6$ and $0.5$ respectively. Then,the magnitude of the acceleration of the block is [Take $g = 10 \ ms^{-2}$]

The upper portion of an inclined plane of inclination $\alpha$ is smooth and the lower portion is rough. $A$ particle slides down from rest from the top and just comes to rest at the foot. If the ratio of the smooth length to rough length is $m : n$,the coefficient of friction is

$A$ block of mass $15\,kg$ is resting on a rough inclined plane as shown in the figure. The block is tied up by a horizontal string which has a tension of $50\,N$. The coefficient of friction between the surface of contact is: $(g = 10\,m/s^2)$