# A load weighing 1 kg, under the action of a force of 50 N directed vertically upwards

**A load weighing 1 kg, under the action of a force of 50 N directed vertically upwards, rises to a height of 3 m. The change in the kinetic energy of the load in this case is equal to**

Given:

m = 1 kilogram is the mass of the cargo;

F = 50 Newton – force acting on the load;

g = 10 m / s ^ 2 – acceleration of gravity;

H = 3 meters – the height to which the load was lifted.

It is required to find the change in the kinetic energy Eк (Joule).

Before we started to lift the load, it has zero potential and kinetic energy. After the start of lifting the load, according to the law of conservation of energy, the work expended by the force F will go to change the potential and kinetic energy of the load, that is:

A (F) = Epot + Ekin;

F * H = m * g * H + Ekin;

Ekin = F * H – m * g * H = 50 * 3 – 10 * 1 * 3 = 150 – 30 = 120 Joules.

Answer: the change in kinetic energy is 120 Joules