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

$A$ rocket is fired vertically from the earth with an acceleration of $2g,$ where $g$ is the gravitational acceleration. On an inclined plane inside the rocket,making an angle $\theta$ with the horizontal,a point object of mass $m$ is kept. The minimum coefficient of friction $\mu_{min}$ between the mass and the inclined surface such that the mass does not move is

The coefficient of friction between a body and the surface of an inclined plane at $45^o$ is $0.5$. If $g = 9.8\,m/s^2$,the acceleration of the body downwards in $m/s^2$ is

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:

$A$ body is moving down a long inclined plane of slope $37^{\circ}$. The coefficient of friction between the body and the plane varies as $\mu = 0.3x$,where $x$ is the distance travelled down the plane. The body will have maximum speed at:

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