What percentage of the total length of a chain can hang off the edge of a table without sliding,given that the coefficient of friction between the chain and the table is $0.25$?

$A$ uniform rope of length $l$ lies on a table. If the coefficient of friction is $\mu$,then the maximum length $l_1$ of the part of this rope which can overhang from the edge of the table without sliding down is

$A$ 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: $(g = 10\, ms^{-2})$

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F_1)$ required to move the block up the inclined plane will be

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

Two blocks $A$ and $B$ of equal masses are sliding down along straight parallel lines on an inclined plane of $45^{\circ}$. Their coefficients of kinetic friction are $\mu_A = 0.2$ and $\mu_B = 0.3$ respectively. At $t = 0$,both the blocks are at rest and block $A$ is $\sqrt{2} \text{ m}$ behind block $B$. Calculate the time and distance from the initial position where the front faces of the blocks come in line on the inclined plane as shown in the figure. (Use $g = 10 \text{ m/s}^2$)