$A$ 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).

$A$ body is projected up along a rough inclined plane of inclination $45^{\circ}$. The coefficient of friction is $0.5$. Then the retardation of the block is

$A$ block of weight $1 \,N$ rests on an inclined plane of inclination $\theta$ with the horizontal. The coefficient of friction is $\mu$. The minimum force that has to be applied parallel to the inclined plane to make the body just move up the plane is

$A$ block is at rest on an inclined plane making an angle $\alpha$ with the horizontal. As the angle $\alpha$ of the incline is increased,the block starts slipping when the angle of inclination becomes $\theta$. The coefficient of static friction between the block and the surface of the inclined plane is:

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

$A$ 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