# A load weighing 3.5 kN is lifted on an inclined plane to a height of 1.8 m. Knowing that the efficiency

**A load weighing 3.5 kN is lifted on an inclined plane to a height of 1.8 m. Knowing that the efficiency of an inclined plane is 60%, find the work performed with an active force.**

P = 3.5 kN = 3500 N.

g = 10 m / s2.

h = 1.8 m.

Efficiency = 60%.

Az -?

The efficiency factor The efficiency of the inclined plane shows what part of the spent work Az, expressed as a percentage, goes into the 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.

We will assume that the lifting of the body occurs uniformly, therefore P = m * g.

Ap = P * h.

Efficiency = P * h * 100% / Az.

Az = P * h * 100% / efficiency.

Az = 3500 N * 1.8 m * 100% / 60% = 10500 J.

Answer: when lifting a load along an inclined plane, the active force performs the work Az = 10500 N.