A load weighing 1 kg under the action of a force of 30 N directed vertically upwards

A load weighing 1 kg under the action of a force of 30 N directed vertically upwards rises to a height of 5 m. What is the change in the kinetic energy of the load?

Given:

m = 1 kilogram (kg) is the mass of the load lifted vertically upward by force F;

F = 30 Newton (N) – force value;

g = 10 Newton / kilogram (N / kg) – acceleration of gravity;

h = 5 meters – the height to which the load is lifted.

It is required to determine dE (Joule) – the change in the kinetic energy of the load.

Since the condition of the problem is not specified, the air resistance is not taken into account.

Then, according to the law of conservation of energy:

A = dE + m * g * h, where A is the work done by force F.

dE = A – m * g * h = F * h – m * g * h = h * (F – m * g) =

= 5 * (30 – 1 * 10) = 5 * (30 – 10) = 5 * 20 = 100 Joules.

Answer: the change in kinetic energy will be equal to 100 Joules.