The force exerted by an elevator's floor on the foot of a person...

Let $N$ be the normal force exerted by the elevator floor and $mg$ be the true weight of the person.

Taking the upward direction as positive, the equation of motion for the person is $N - mg = ma$, where $a$ is the acceleration of the elevator.

This gives $N = m(g + a)$.

For the normal force to be greater than the weight ($N > mg$), the acceleration $a$ must be positive, meaning the net acceleration is directed upwards.

An upward acceleration occurs in two scenarios:

The elevator is travelling upward (positive velocity) and its speed is increasing (speeding up, so acceleration is in the direction of velocity).

The elevator is travelling downward (negative velocity) and its speed is decreasing (slowing down, so acceleration is opposite to velocity, which is upwards).

Thus, the force exerted by the floor is greater than the person's weight when the elevator is travelling upward and speeding up, or travelling downward and slowing down.