Frictional forces act over the common surfaces of the two bodies to avoid and restrict relative moment between those two surfaces.

When the body is in rest it is under static friction but when it starts moving (neither rolling nor sliding), the static friction slowly chngs to kinetic friction as the coefficient of static friction start decreasing and that of kinetic friction starts increasing. In case it starts rolling motion then the friction is rolling friction & if it slides then sliding fiction.

The impending motion refers to the state of a body when it is on the verge of slipping. In such cases, the static friction has reached its upper limit and is given by the equation, F=μ_sN

Limiting friction, f = μmg = 0.6 × 1 × 9.8 = 5.88 N

Applied force = F = ma = 1 × 5 = 5 N

As F < f, so force of friction = 5 N

We know that centripetal acceleration, c = v²/r
Thus we get r = v²/c
= 36 / 120
= .3 /m

At the time when the block just starts to move, we get that net force acting upon it is 0, thus we get, f - mg.sin a = 0
Where f is friction force and a is angle of incline.
We also know that f = 1√3 x N
= 1√3 x mg.cos a
Thus we get 1√3 x mg.cos a = mg.sin a
Thus we get tan a = 1√3
And a = 30

Mass of the block = 2 kg

Weight of the block = mg = 2 × 9.8 = 19.6 N

The component of the weight along normal = 19.6 Cos 30° = 16.97 N

Hence, the normal force (N) = 16. 97 N

Now, the component of the weight along the inclined plane = 19.6 Sin 30°

Along the inclined plane = 9.8 N

Now, the friction = μ N = 0.7 × 16.97 = 11.87 N

Since, 9.8 N is less than the maximum value of the friction (11.87 N)

Hence, the frictional force = 9.8 N

We know that centripetal acceleration, a = v x v /r
Here v = 6m/s and r = 0.3m
Thus a = 36 / 0.3 m/s²
= 120 m/s²

What you need to remember and apply here is that friction always tries to oppose motion.
So, in case of the wheel, the point P which is in contact with the ground tries to go backwards due to rotation and it's velocity as shown in the figure.
The frictional force F will try to oppose this motion and thus, it will be in the forward direction.
 

The normal force due to the weight of the bloch is 40N, hence the friction force is 4N acting to restrict the motion.
Hence the net force on the body is thus 20 - 4 = 16N
Thus the net acceleration of the block is 16/4 = 4ms²