$A$ given object takes $n$ times the time to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down an identical perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is:

Two touching blocks $1$ and $2$ are placed on an inclined plane forming an angle $60^{\circ}$ with the horizontal. The masses are $m_1$ and $m_2$ and the coefficients of friction between the inclined plane and the two blocks are $1.5 \mu$ and $1.0 \mu$, respectively. The force of reaction between the blocks during the motion is ($g=$ acceleration due to gravity).

$A$ block of wood resting on an inclined plane of angle $30^o$ just starts moving down. If the coefficient of friction is $0.2$,its velocity (in $ms^{-1}$) after $5\, s$ is: $(g = 10\, ms^{-2})$

$A$ small block starts slipping down from a point $B$ on an inclined plane $AB$,which is making an angle $\theta$ with the horizontal. The section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$) of the inclined plane. If $BC = 2AC$,the coefficient of friction is given by $\mu = k \tan \theta$. The value of $k$ is

$A$ 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 ........ $J$.

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