With the help of a stationary block, a load weighing 50 kg is lifted to a height of 12 m

With the help of a stationary block, a load weighing 50 kg is lifted to a height of 12 m, while a force of 600N acts. Define useful work, total work and efficiency

Given:

m = 50 kilograms – the mass of the load, which is lifted using a fixed block;

h = 12 meters – the height to which the load is lifted;

F = 600 Newton – the force with which the load is lifted.

It is required to determine A1 (Joule) – useful work, A2 (Joule) – full work and n – engine efficiency.

The useful work will be equal to the potential energy of the load:

A1 = m * g * h, where g = 10 Newton / kilogram (approximate value;

A1 = 50 * 10 * 12 = 500 * 12 = 6000 Joules.

The total work expended will be equal to:

A2 = F * h = 600 * 12 = 7200 Joules.

The efficiency will be equal to:

n = A1 / A2 = 6000/7200 = 0.83 = 83%.

Answer: the useful work is 6000 Joules, the total work is 7200 Joules, the efficiency of the unit is 83%.