Let's denote the coefficient of friction as μ, the gravitational acceleration as g (9.81 m/s² approximately), and the force required to just prevent the box from sliding down as F.
When the box is in equilibrium (not sliding down), the force F balances the component of the gravitational force parallel to the incline and frictional force.
The force due to gravity acting down the incline is m * g * sin(θ), where m is the mass of the box (8 kg in this case). The frictional force acting opposite to gravity force is μ * m * g * cos(θ), where μ is the coefficient of friction.
So for equilibrium (just preventing the box from sliding down), we can write:
F + μ * m * g * cos(θ) = m * g * sin(θ)
F = m * g * sin(θ) - μ * m * g * cos(θ)
When applying a force of 2F to move the box upward, the total force opposing motion is the sum of friction (μ * m * g * cos(θ)) and the force due to gravity down the slope (m * g * sin(θ)). So we can write:
2F = μ * m * g * cos(θ) + m * g * sin(θ)
Now, let's put F from the first equation into the second equation:
2 * (m * g * sin(θ) - μ * m * g * cos(θ)) = μ * m * g * cos(θ) + m * g * sin(θ)
Simplify and solve for μ:
2 * m * g * sin(θ) - 2 * μ * m * g * cos(θ) = μ * m * g * cos(θ) + m * g * sin(θ)
Now, let's collect like terms, with all terms with μ on one side:
2 * m * g * sin(θ) - m * g * sin(θ) = μ * m * g * cos(θ) + 2 * μ * m * g * cos(θ)
Simplify further:
m * g * sin(θ) = 3 * μ * m * g * cos(θ)
Divide both sides by m * g * cos(θ) to isolate μ:
sin(θ) / cos(θ) = 3 * μ
tan(θ) = 3 * μ
Solve for μ:
μ = (1/3) * tan(θ)
Thus, the coefficient of friction between the box and the inclined plane is:
A. 1/3 tan θ
