Abuggy and baby together weigh 140n. center of gravity g shown on diagram.
what is the corresponding upward force the pusher needs to exert to lift the back wheels of the pram up a step whilst moving backwards?

What a delightful little problem !

I think what you have here is a Class-II lever:

-- The fulcrum is the front wheel.

-- The resistance is the 140N, acting downward, 30 cm from the fulcrum.

-- The effort is the required force 'F' acting upward, 110 cm from the fulcrum
and on the same side of it.

In order to rotate the buggy upward around the fulcrum, the effort-moment
must be at least equal to the resistance-moment, in the opposite direction.

     110 F = (30 x 140) = 4,200 newton-cm

           F = (4,200 newton-cm) / (110 cm) = 38.2 Newtons.

Anything greater than that will lift the back end. 

85.25 kg

Explanation:

Force = 31 N

v1 = 0 m/s

v2 = 8 m/s

t = 22 s

Acceleration = (v2 - v1) / t

                    = (8 - 0) / 22

                    = 8 / 22

Force = mass * acceleration

mass = force / acceleration

mass = 31 / (8/22)

mass = (31*22) / 8

mass = 85.25 kg