$A$ given object takes $n$ times more time to slide down a $45^o$ rough inclined plane as it takes to slide down a perfectly smooth $45^o$ incline. The coefficient of kinetic friction between the object and the incline is

$A$ block of mass $5 \ kg$ is kept against an accelerating wedge with a wedge angle of $45^{\circ}$ to the horizontal. The coefficient of friction between the block and the wedge is $\mu = 0.4$. What is the minimum absolute value of the acceleration of the wedge to keep the block steady? Assume $g = 10 \ m \ s^{-2}$.

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 $g = 10 \text{ m/s}^2$)

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 upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If the coefficient of friction is $0.25$,the angle of inclination of the plane is ...... $^o$.