What work must be done to raise a cast-iron anchor weighing 100 kg from the bottom of a lake 10 m deep to the surface?

Given:

m = 100 kilograms – the mass of the cast-iron anchor, which must be lifted from the bottom of the lake;

h = 10 meters – the depth at which the anchor is located;

ro = 7200 kg / m3 (kilogram per cubic meter) – density of cast iron;

ro1 = 1000 kg / m3 – water density;

g = 9.8 N / kg (Newton per kilogram) is the acceleration of gravity.

It is required to determine A (Joule) – what work needs to be done to raise the anchor from the bottom of the lake.

Since it is not specified in the problem statement, we assume that the anchor is raised evenly. Then, the force with which the anchor will be raised will be equal to the difference between the force of gravity and the Archimedean force:

F = F gravity – Farchimedes = m * g – ro1 * V * g = m * g – ro1 * m * g / ro =

= m * g * (1 – ro1 / ro) = 100 * 9.8 * (1 – 1000/7200) = 980 * 0.86 = 842.8 Newtons.

Then the work will be equal to:

A = F * h = 842.8 * 10 = 8428 Joules.

Answer: To lift the cast iron anchor from the bottom of the lake, you need to do a work equal to 8428 Joules.