What is the efficiency of the movable unit, with the help of which a 90 kg load is lifted to a height of 4 m?

What is the efficiency of the movable unit, with the help of which a 90 kg load is lifted to a height of 4 m? It is known that the work done in this case is equal to 4000J.

Let us determine what the work that must be performed in order to lift a load equal to 4 meters to a height equal to 4 meters, which, according to the condition of our problem, has a mass of 90 kilograms, should be equal in the ideal case:

A = m * g * h;

A = 90 * 4 * 10 = 3600.

Let us determine what the efficiency of our moving unit will equal, when from the condition of our problem we know that in fact it took 4000 Joules to lift the load:

Efficiency = 3600/4000 * 100% = 0.9 * 100% = 90%.

Answer: 90%.