$A$ block of mass $M$ is placed on a rough inclined plane with an angle of inclination $\theta$ and coefficient of friction $\mu$. $A$ force $F$ is applied parallel to the inclined plane as shown in the figure,such that the block just starts moving upward. The value of $F$ is:

Consider a small block sliding down an inclined plane of inclination $30^{\circ}$ with the horizontal. The coefficient of friction is $\mu = \frac{2}{3} x$,where $x$ is the distance (in meters) through which the mass slides down. The distance covered by the mass before it stops is

Two inclined planes are placed as shown in the figure. $A$ block is projected from point $A$ of the inclined plane $AB$ along its surface with a velocity just sufficient to carry it to the top point $B$ at a height of $10 \ m$. After reaching point $B$, the block slides down the inclined plane $BC$. The time it takes to reach point $C$ from point $A$ is $t(\sqrt{2}+1) \ s$. The value of $t$ is........ (use $g = 10 \ m/s^2$)

$A$ board is balanced on a rough horizontal semicircular log. Equilibrium is obtained with the help of the addition of a weight to one of the ends of the board when the board makes an angle $\theta$ with the horizontal. The coefficient of friction between the log and the board is:

$A$ block of mass $m$ is at rest on an inclined plane with an angle of inclination $\theta$ and coefficient of friction $\mu$,as shown in the figure. The frictional force acting on the block is:

The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If $\mu$ is the coefficient of friction,the inclination of the plane to the horizontal is: