With a vertical uniformly accelerated lifting of a load weighing 3 kg to a height of 15 m, a work of 675 J was performed

With a vertical uniformly accelerated lifting of a load weighing 3 kg to a height of 15 m, a work of 675 J was performed, then the acceleration of the body will be (g = 10m / s²)

Initial data: m (cargo weight) = 3 kg; h (the height to which the load was lifted) = 15 m; A (work done while lifting the load) = 675 J; the lifting of the load is uniformly accelerated.
Reference data: by condition g (acceleration of gravity) = 10 m / s ^ 2.
The work that was done with uniformly accelerated lifting of the load: A = F * S = m * (a + g) * h, whence a + g = A / (m * h) and a = A / (m * h) – g.
Calculation: a = 675 / (3 * 15) – 10 = 15 – 10 = 5 m / s ^ 2.
Answer: The load was lifted with a constant acceleration of 5 m / s ^ 2.