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

$A$ uniform chain of length $L$ hangs partly from a table and is kept in equilibrium by friction. If the maximum length of the chain that can hang without slipping is $l$,then the coefficient of friction between the table and the chain is:

$A$ body of mass $2 \ kg$ is at rest at the bottom of an inclined plane of length $8 \ m$ and height $1 \ m$ as shown in the figure. If the coefficient of friction is $0.2$,what is the work done in moving the body from the bottom to the top of the incline (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:

To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at $45^{\circ}$ and the block is released from rest. The block takes a time $t$ in moving a distance $d$. The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time $t/2$ in moving down the same distance $d$. The coefficient of friction is

Two blocks $A$ and $B$ of equal mass are initially in contact when released from rest on an inclined plane. The coefficients of friction between the inclined plane and blocks $A$ and $B$ are $\mu_1$ and $\mu_2$ respectively.