As per the given conditions, both the magnet and the spring are in equilibrium in the vertical direction, and the system is frictionless with massless spring and carriage. Ignoring the horizontal force of the magnet, we need to determine the horizontal force required to prevent the system from sliding down the inclined plane.
In regards to the given options, the answer would be A, as the force required needs to be determined at any distance along the vertical direction after it is positioned.
To calculate the horizontal force required, we need to consider the forces acting on the system. The normal force due to the spring pushes the carriage system down the inclined plane, while the magnetic force due to the magnet's vertical force becomes stronger as it gets closer to the steel.
To prevent the system from sliding down the inclined plane, we need to apply a force equal and opposite to the force pushing the system down. Therefore, the required horizontal force would be the sum of the magnet's weight (mg) and the magnetic force, multiplied by the sine of the angle of the inclined plane. This is because the forces acting on the system are perpendicular to each other, and the horizontal force is the component of the magnetic and gravitational forces that is parallel to the inclined plane.