Consider a block kept on an inclined plane (inclined at $45^{\circ}$) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down,the coefficient of friction between the block and inclined plane $(\mu)$ is equal to

$A$ 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

$A$ body of $5 \,kg$ weight kept on a rough inclined plane of angle $30^{\circ}$ starts sliding with a constant velocity. Then the coefficient of friction is (assume $g=10 \,ms^{-2}$)

$A$ block of mass $4\, kg$ rests on an inclined plane. The inclination of the plane is gradually increased. It is found that when the inclination is $3$ in $5$ $\left( \sin \theta = \frac{3}{5} \right)$,the block just begins to slide down the plane. The coefficient of friction between the block and the plane is

$A$ particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is $0.5$,the acceleration is ($g=$ Acceleration due to gravity).

$A$ block of a certain mass is placed on a rough inclined plane. The angle between the plane and the horizontal is $30^{\circ}$. The coefficients of static and kinetic friction between the block and the inclined plane are $0.6$ and $0.5$ respectively. Then,the magnitude of the acceleration of the block is [Take $g = 10 \ ms^{-2}$]