The wedge is fixed and the mass of the block is $m$. The minimum value of $F$ so that the block is in equilibrium is (neglect the dimension of the block).

$A$ 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).

$A$ cylindrical vessel filled with water is released on an inclined surface of angle $\theta$ as shown in the figure. The friction coefficient of the surface with the vessel is $\mu (< \tan \theta)$. Then the contact angle made by the surface of water with the incline will be

Starting from rest,a body slides down a $45^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

$A$ block slides down an inclined plane of slope angle $\theta$ with a constant velocity. It is then projected up the plane with an initial velocity $u$. The distance up to which it will rise before coming to rest is

Two blocks of masses $1 \ kg$ and $2 \ kg$ are connected by a light rod and the system is slipping down a rough incline at an angle of $45^{\circ}$ with the horizontal. The coefficient of kinetic friction at both contacts is $0.4$. If the acceleration of the system is $\alpha \sqrt{2} \ m/s^2$,find the value of $\alpha$. (Use $g = 10 \ m/s^2$)