In the hover, it must produce a lift force equal to the weight of the helicopter, in order for the helicopter to remain at constant altitude. A helicopter with a mass of 700 kg must produce a lift of 700*g = 6,867 N to hover at constant altitude.
In forward flight at constant altitude, the rotor must produce more thrust with the thrust vector tilted forward at angle α. Vertical thrust component = lift must be equal to weight for the helicopter to fly at constant altitude. As the helicopter picks up speed, less power is required to provide the same amount of thrust due to reduction of induced drag - if power is not reduced, the helicopter starts to climb when transferring from hover to forward speed.
If in the hover the rotor produces more lift than the weight, let's say 710*g,it accelerates upwards, resulting in a vertical velocity component that results in an additional aerdynamic damping force, proportional to vertical speed, caused by change of rotor blade Angle of Attack and fuselage vertical drag. So for a 700 kg helicopter with a rotor that exerts 710*g = 6,965 N, the extra 98 N is used for overcoming aerodynamic drag associated with vertical speed.
Assuming that by "lift" only the vertical component of thrust is meant: more lift than weight results in a constant upwards velocity. Less lift than weight results in a constant downwards velocity, even in a free fall (terminal velocity).