$A$ 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:

$A$ girl holds a book of mass $m$ against a vertical wall with a horizontal force $F$ using her finger,so that the book does not move. The frictional force on the book by the wall is

$A$ body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is $0.5$. The ratio of the net force required for the body to slide down and the normal reaction on the body is $1:2$. Then the angle of the inclined plane is (in $^{\circ}$)

For the equilibrium of a body on an inclined plane of inclination $45^{\circ}$, the coefficient of static friction must be:

$A$ smooth block is released at rest on a $45^\circ$ incline and then slides a distance $d$. The time taken to slide is $n$ times as much to slide on a rough incline than on a smooth incline. The coefficient of kinetic friction is: