The walking excavator throws out 14 m3 of soil in one step, lifting it to a height of 20 meters. Bucket weight without soil is 20kN.

The walking excavator throws out 14 m3 of soil in one step, lifting it to a height of 20 meters. Bucket weight without soil is 20kN. Determine the work that the excavator engine does on lifting the soil and bucket. Soil density is 1500 kg * m3.

Vg = 14 m3.

h = 20 m.

g = 10 m / s2.

Pk = 20 kN = 20,000 N.

ρg = 1500 kg / m3.

A – ?

Mechanical work A is expressed by the formula: A = F * S, where F is the force developed by the excavator, S is the movement of the bucket with the ground.

The bucket S moves to the height h: S = h.

We will assume that the bucket rises in a straight line uniformly, therefore, the action of forces on the bucket is compensated. The lifting force of the excavator F is compensated by the gravity of the bucket with soil m * g, where m is the mass of the bucket with soil, g is the free acceleration.

m = mk + mg = Pk / g + Vg * ρg.

А = (Pк / g + Vg * ρg) * g * h.

A = (20,000 N / 10 m / s2 + 14 m3 * 1500 kg / m3) * 10 m / s2 * 20 m = 4600000 J.

Answer: the crane is doing work A = 4600000 J.