$A$ block slides down an inclined plane with an acceleration $g/2$ as shown in the figure. Then the coefficient of kinetic friction is

If $300\,J$ of work is done to move a $2\,kg$ block up a rough inclined plane to a height of $10\,m$,find the work done against friction in $J$. (Take $g = 10\,m/s^2$)

The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is $F_{1}$,while the minimum force needed to prevent it from sliding down is $F_{2}$. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta = 2\mu$,then the ratio $\frac{F_{1}}{F_{2}}$ is:

$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \, N$,the mass of the block (in $kg$) is (take $g = 10 \, m/s^2$).

$A$ block is lying on an inclined plane which makes $60^\circ$ with the horizontal. If the coefficient of friction between the block and the plane is $0.25$ and $g = 10\,m/s^2$,then the acceleration of the block when it moves along the plane will be ........ $m/s^2$.

The acceleration of a body of mass $m$ sliding down an inclined plane with an angle of inclination $\theta$ and a coefficient of kinetic friction $\mu$ is: