A weight of 100 kg is fixed on the short arm of the lever. In order to raise the load to a height

| September 28, 2021 | education

A weight of 100 kg is fixed on the short arm of the lever. In order to raise the load to a height of 8 cm, a force equal to 200 N was applied to the long arm of the lever. At the same time, the point of application of this force dropped by 50 cm. Determine the efficiency of the lever.

The short arm is acted upon by the force F1 = mg = 100 kg * 10 m / s2 = 1000 N.

When climbing to a height of h1 = 8 cm, useful work A1 was performed:

A1 = F1 * h1 = 1000 N * 0.08 m = 80 J.

The work A2 of the force F2 = 200 N was spent on lifting the load:

A2 = F2 * h2 = 200 N * 0.5 m = 100 J, where h2 is the displacement of the point of application of the force.

The efficiency of the lever is equal to the ratio of useful work to spent:

η = (A1 / A2) * 100% = (80 J / 100 J) * 100% = 80%.

Answer: 80%.



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