$A$ body is sliding down an inclined plane (angle of inclination $45^{\circ}$). If the coefficient of friction is $0.5$ and $g = 9.8\, m/s^2$,then the acceleration of the body downwards in $m/s^2$ is

$A$ wooden box lying at rest on an inclined surface of a wet wood is held at static equilibrium by a constant force $F$ applied perpendicular to the incline. If the mass of the box is $1 \ kg$, the angle of inclination is $30^{\circ}$ and the coefficient of static friction between the box and the inclined plane is $0.2$, the minimum magnitude of $F$ is (Use $g=10 \ m/s^2$)

If a block moving up an inclined plane at $30^{\circ}$ with a velocity of $5 \, m/s$ stops after $0.5 \, s$,then the coefficient of friction will be nearly:

$A$ 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

$A$ box is lying on an inclined plane. What is the coefficient of static friction if the box starts sliding when the angle of inclination is $60^\circ$?

$A$ block is projected upwards on an inclined plane of inclination $37^{\circ}$ along the line of greatest slope with a coefficient of friction $\mu = 0.5$ and an initial velocity of $5 \ m/s$. The block first stops at a distance of .......... $m$ from the starting point.