Explanation:

1. Understanding the given information:
- A box with a mass of 8 kg is placed on a rough inclined plane.
- The inclined plane has an angle of inclination, theta.
- The box can be prevented from moving downwards by applying an upward pull force.
- The box can be made to slide upwards by applying a force of 2f.
- The coefficient of friction between the box and the inclined plane is unknown.

2. Forces acting on the box:
When the box is placed on the inclined plane, several forces come into play:

- Weight (W): The force due to the gravitational pull acting vertically downwards. Its magnitude is given by W = mg, where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Normal Force (N): The force exerted by the inclined plane perpendicular to its surface. Its magnitude is equal to the component of the weight acting perpendicular to the inclined plane, which is N = mg*cos(theta).
- Friction Force (Ff): The force opposing the motion of the box along the inclined plane. Its magnitude is given by Ff = μ*N, where μ is the coefficient of friction between the box and the inclined plane.

3. Preventing downward motion:
To prevent the box from moving downwards, an upward pull force must be applied. This force should be equal to or greater than the force of gravity pulling the box downwards. Mathematically, this can be expressed as 2f ≥ mg.

4. Making the box slide upwards:
To make the box slide upwards, a force of 2f must be applied. This force should be greater than the force of friction acting on the box. Mathematically, this can be expressed as 2f > Ff.

5. Determining the coefficient of friction:
By comparing the inequalities from steps 3 and 4, we can deduce that the force of friction acting on the box is equal to the weight of the box. Therefore, we have Ff = mg.

Substituting this into the equation for friction force, we get mg = μ*N.

Since N = mg*cos(theta), we can rewrite the equation as mg = μ*mg*cos(theta).

Simplifying the equation, we find that the coefficient of friction, μ = 1/cos(theta).

Conclusion:
The coefficient of friction between the box and the inclined plane is given by μ = 1/cos(theta). This equation shows that as the angle of inclination, theta, increases, the coefficient of friction decreases. Therefore, the box will slide more easily on steeper inclines.

N = normal force by the incline on the block
f' = frictional force on the block
m = mass of the block = 8 kg
θ = angle of incline
μ = Coefficient of friction
consider the force diagram to prevent the motion of block in down direction :
force equation perpendicular to incline is given as
N = mg Cosθ
frictional force on the block is given as
f' = μ N
f' = μ mg Cosθ   
parallel to incline , the force equation is given as
f + f' = mg Sinθ
f + μ mg Cosθ = mg Sinθ
f = mg Sinθ - μ mg Cosθ                                           eq-1
consider the force diagram to make the block slide upward:
force equation perpendicular to incline is given as
N = mg Cosθ
frictional force on the block is given as
f' = μ N
f' = μ mg Cosθ   
parallel to incline , the force equation is given as
2 f = mg Sinθ + f'
2 f = mg Sinθ +  μ mg Cosθ
using eq-1
2 (mg Sinθ - μ mg Cosθ) = mg Sinθ +  μ mg Cosθ
2 mg Sinθ - 2 μ mg Cosθ = mg Sinθ +  μ mg Cosθ
mg Sinθ = 3  μ mg Cosθ
tanθ =  3  μ
μ = tanθ/3