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

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 3 m. Moving evenly, another student weighing 45 kg lifted a load weighing 8 kg to the same height. Which of them did the great job?

The work of students can be calculated using the formula:
A = F * S, where F is the applied force (F = Ft = mg, m is the mass of the student with the load, g is the acceleration of gravity (we take g = 10 m / s ^ 2)), S is the movement of the student.
Let’s calculate the work done by the first student:
A = F * S = m * g * S = (40 + 10) * 10 * 3 = 50 * 10 * 3 = 1500 J = 1.5 kJ.
Let’s calculate the work done by the second student:
A = F * S = m * g * S = (45 + 8) * 10 * 3 = 53 * 10 * 3 = 1590 J = 1.59 kJ.
Answer: The second student did a great job.