A truck crane lifts a load weighing 3000 kg using a movable block to a height of 6 m

| February 7, 2021 | education

A truck crane lifts a load weighing 3000 kg using a movable block to a height of 6 m. Determine the efficiency of the block if the cable is pulled with a constant force of 16 kN.

m = 3000 kg.

g = 10 m / s2.

h = 6 m.

Ftr = 16 kN = 16000 N.

Efficiency -?

Efficiency coefficient The efficiency of a simple mechanism shows what part of the spent work Az, expressed as a percentage, goes into the useful Ap: Efficiency = Ap * 100% / Az.

We will express the useful work Ap by the formula: Ap = m * g * h, where h is the height to which the load was lifted.

Ap = 3000 kg * 10 m / s2 * 6 m = 180,000 J.

The expended work Az is expressed by the formula: Az = Ftr * L, where L is the length of the cable.

For the movable block L = 2 * h.

Az = Ftr * 2 * h.

Az = 16000 N * 2 * 6 m = 192000 J.

Efficiency = 180,000 J * 100% / 192,000 J = 93.75%.

Answer: the truck crane has an efficiency of 93.75%.



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