The athlete, using a movable block, lifted the load to a height of 2.5 m, applying a force of 240 N

The athlete, using a movable block, lifted the load to a height of 2.5 m, applying a force of 240 N. to the free end of the rope. What work did he do? (Do not take into account the weight of the unit and the frictional force.)

The useful work of an athlete can be calculated using the formula:
A = P * h, where P is the acting force, h is the height to which the load was lifted (h = 2.5 m).
Since the movable block gives a 2-fold gain in strength, then P = 2F, where F is the force applied by the athlete (F = 240 N).
Let’s calculate the work:
A = P * h = A = 2F * h = 2 * 240 * 2.5 = 1200 J = 1.2 kJ.
Check: J = N * m = J.
Answer. The work done by the athlete is 1.2 kJ.