# Efficiency The height of the inclined plane is 1.2 m, and its height is 10.8 m. To lift a body weighing

**Efficiency The height of the inclined plane is 1.2 m, and its height is 10.8 m. To lift a body weighing 180 kg along this plane, a force of 250 N was applied. Determine the efficiency of the inclined plane and the friction force.**

m = 180 kg.

g = 10 m / s2.

h = 1.2 m.

L = 10.8 m.

F = 250 N.

Efficiency -?

Ftr -?

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

Useful work when lifting a body along an inclined plane An is expressed by the formula: An = m * g * h, where m is the mass of the load, g is the acceleration of gravity, h is the height of the inclined plane.

The expended work Az is expressed by the formula: Az = F * L, where the force F with which the load is pulled, L is the length of the inclined plane.

The formula for determining the efficiency of an inclined plane will take the form: efficiency = m * g * h * 100% / F * L.

Efficiency = 180 kg * 10 m / s2 * 1.2 m * 100% / 250 N * 10.8 m = 80%.

Ftr = (Az – Ap) / L = (F * L – m * g * h) / L.

Ftr = (250 N * 10.8 m – 180 kg * 10 m / s2 * 1.2 m) / 10.8 m = 50 N.

Answer: the inclined plane has an efficiency = 80%, the friction force Ffr = 50 N.