When a body is lying on a rough inclined plane and does not move,the force of friction

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F_1)$ required to move the block up the inclined plane will be

$A$ piece of ice slides down a rough inclined plane at $\theta=45^{\circ}$ inclination in twice the time that it takes to slide down an identical but frictionless inclined plane. What is the coefficient of friction between ice and incline?

Starting from rest,the time taken by a body sliding down on a rough inclined plane at $45^{\circ}$ with the horizontal is twice the time taken to travel on a smooth plane of the same inclination and same distance. Then the coefficient of kinetic friction is

$A$ block of mass $5 \ kg$ is kept against an accelerating wedge with a wedge angle of $45^{\circ}$ to the horizontal. The coefficient of friction between the block and the wedge is $\mu = 0.4$. What is the minimum absolute value of the acceleration of the wedge to keep the block steady? Assume $g = 10 \ m \ s^{-2}$.

$A$ block of mass $M$ is sliding down an inclined plane. An external force $F$ is applied vertically downwards on the block. The coefficient of static friction is $\mu_s$ and the coefficient of kinetic friction is $\mu_k$. The friction force acting on the block is: