$A$ $2 \ kg$ brick begins to slide over a surface which is inclined at an angle of $45^{\circ}$ with respect to the horizontal axis. The coefficient of static friction between their surfaces is:

$A$ block of mass $5 \text{ kg}$ is placed on a rough inclined surface as shown in the figure. If $\vec{F}_1$ is the force required to just move the block up the inclined plane and $\vec{F}_2$ is the force required to just prevent the block from sliding down,then the value of $|\vec{F}_1|-|\vec{F}_2|$ is: [Use $g=10 \text{ m/s}^2$]

Consider a block kept on an inclined plane (inclined at $45^{\circ}$) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down,the coefficient of friction between the block and inclined plane $(\mu)$ is equal to

$A$ box of mass $2 \ kg$ is placed on an inclined plane that makes $30^{\circ}$ with the horizontal. The coefficient of friction between the box and the inclined plane is $0.2$. $A$ force $F$ is applied on the box perpendicular to the incline to prevent the box from sliding down. The minimum value of $F$ is (acceleration due to gravity $= 10 \ ms^{-2}$) (in $N$)

$A$ homogeneous rectangular brick lies on a rough inclined plane of very small inclination. Which half part of the brick exerts a greater contact force on the plane?

$A$ block moves down a smooth inclined plane of inclination $\theta$. Its velocity on reaching the bottom is $v$. If it slides down a rough inclined plane of the same inclination,its velocity on reaching the bottom is $v/n$,where $n$ is a number greater than one. The coefficient of friction $\mu$ is given by: