With a uniform movement of a load weighing 40 kg, suspended from the short arm of the lever, a force of 250 N

With a uniform movement of a load weighing 40 kg, suspended from the short arm of the lever, a force of 250 N was applied to the long arm. If the load rose to a height of 50 cm, and the point of application of the force dropped by 1 m, then the efficiency of the lever is equal to.

m1 = 40 kg.

F2 = 250 N.

h1 = 50 cm = 0.5 m.

h2 = 1 m.

g = 10 m / s2.

Efficiency -?

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

Useful work when lifting a load with the lever Ap is expressed by the formula: Ap = m1 * g * h1, where m1 is the mass of the load that was lifted, g is the acceleration of gravity, h1 is the height of the load.

The expended work Az will be expressed by the formula: Az = F2 * h2, where the force F2, with which they act on the larger shoulder, h2 is the distance the applied force has dropped.

The formula for determining the efficiency of the lever will take the form: efficiency = m1 * g * h1 * 100% / F2 * h2.

Efficiency = 40 kg * 10 m / s2 * 0.5 m * 100% / 250 N * 1 m = 80%.

Answer: the lever has an efficiency of 80%.