What force must be applied to the end of the cord to evenly lift a 12kg load?

What force must be applied to the end of the cord to evenly lift a 12kg load? How high does it rise if the free end of the cord drops by 10cm

m = 12 kg.

g = 10 N / kg.

l = 10 cm = 0.1 m.

F -?

h -?

1) Fixed block.

According to 1 Newton’s law, with uniform rectilinear motion, the action of forces on it is compensated. The force of gravity m * g is balanced by the force that is applied to the end of the cord F: m * g = F.

F = 12 kg * 10 N / kg = 120 N.

The body rises to a height h equal to the length of the cord l, which came down: h = l.

h = 0.1 m.

2) Moving unit system.

The movable block gives a gain in strength 2 times, but a loss of 2 times the height of lifting.

F = m * g / 2.

F = 12 kg * 10 N / kg / 2 = 60 N.

h = 2 * l.

h = 2 * 0.1 m = 0.2 m.

Answer: 1) F = 120 N, h = 0.1 m; 2) F = 60 N, h = 0.2 m.