$A$ block of $10 \, kg$ mass is placed on a rough inclined surface as shown in the figure. The acceleration of the block will be

$A$ block of mass $2 \, kg$ slides down an inclined plane of inclination $30^{\circ}$. The coefficient of friction between the block and the plane is $0.5$. The contact force between the block and the plane is:

$A$ particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is $0.5$,the acceleration is ($g=$ Acceleration due to gravity).

$A$ 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:

$A$ block of mass $1 \,kg$ slides down on a rough inclined plane of inclination $60^\circ$ starting from its top. If the coefficient of kinetic friction is $0.5$ and the length of the plane is $1 \,m$,then the work done against friction is ........ $J$.

$A$ particle of mass $m$ slides from rest down a plane inclined at $30^{\circ}$ to the horizontal. The force of resistance acting on the particle during motion is $ms^2$,where $s$ is the displacement of the particle from its initial position. The velocity (in $m/s$) of the particle when $s = 1\,m$ is $v$. The value of $\frac{3v^2}{14}$ is: