$A$ $2 \ kg$ 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:

When a body is placed on a rough plane (coefficient of friction $= \mu$) inclined at an angle $\theta$ to the horizontal,its acceleration is (acceleration due to gravity $= g$)

$A$ block of mass $10 \,kg$ is kept on a fixed rough $(\mu=0.8)$ inclined plane of angle of inclination $30^{\circ}$. The frictional force acting on the block is ........... $N$.

$A$ block of mass $10 \ kg$,initially at rest,makes a downward motion on a $45^{\circ}$ inclined plane. The distance travelled by the block after $2 \ s$ is (Assume the coefficient of kinetic friction to be $0.3$ and $g=10 \ m/s^2$)

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

$A$ disc arranged in a vertical plane has two grooves of the same length directed along the vertical chord $AB$ and the chord $CD$ as shown in the figure. Particles slide down along $AB$ and $CD$ starting from rest. The ratio of the time $t_{AB}/t_{CD}$ is