Using a fixed block, a load of 120 kg is lifted. What force is applied in this case if the efficiency of the unit is 0.8.

Using a fixed block, a load of 120 kg is lifted. What force is applied in this case if the efficiency of the unit is 0.8. The efficiency of the mobile unit is 75%. How many times is the mass of the load greater than the mass of the block? Consider block friction to be negligible.

m = 120 kg.

g = 9.8 m / s2.

k = 0.8.

F -?

A fixed block only changes the direction of the force. If we assume that the load is lifted evenly, then the weight of the load P = m * g is equal to the force F with which the rope is pulled, multiplied by the coefficient k: F * k = m * g.

F = m * g / k.

F = 120 kg * 9.8 m / s2 / 0.8 = 1470 N.

Answer: the load is lifted with a force of F = 1470 N.

Efficiency = 75%.

mg / mb -?

Efficiency shows what part of the spent work of Azatr turns into useful work of Apol, expressed as a percentage: Efficiency = Apol * 100% / Azatr.

Apol = mg * g * h.

If friction is neglected, then the work on lifting the block is 100% – efficiency = 100% – 75% = 25% of the work expended.

We express the work on lifting the movable block Ab by the formula: Ab = mb * g * h work goes to lifting the movable block.

mg * g * h – is 75%.

mb * g * h – is 25%.

mg / mb = 75% / 25% = 3.

Answer: the mass of the block is 3 times less than the mass of the load: mg / mb = 3.