Moving evenly, a student weighing 40 kg lifted a load weighing 10 kg to the 2nd floor

Moving evenly, a student weighing 40 kg lifted a load weighing 10 kg to the 2nd floor of a school building located at a height of 3m. Moving evenly, another student weighing 45 kg lifted a load weighing 8 kg to the same height. Which of them did the big work?

m1 = 40 kg.
m2 = 10 kg.
m3 = 45 kg.
m4 = 8 kg.
h = 3 m.
g = 9.8 m / s ^ 2.
A1 -?
A2 -?
The work that boy A did is equal to the change in potential energy En: A = ΔEp.
ΔEn = En – Eno, where Eno is the potential energy at the beginning, En is the potential energy at the end of the ascent.
The potential energy is determined by the formula En = m * g * h, where m is the mass, g is the acceleration of gravity, h is the height.
Since at the moment of movement h = 0, then A = En.
The work done by the first student A1 = (m1 + m2) * g * h.
A1 = (40 kg + 10 kg) * 9.8 m / s ^ 2 * 3 m = 1470 J.
The work done by the second student A2 = (m3 + m4) * g * h.
A2 = (45 kg + 8 kg) * 9.8 m / s ^ 2 * 3 m = 1558.2 J.
Answer: The second boy did more work than the first.