A load weighing 50 kg is evenly pulled along an inclined plane with a force of 50 N.

A load weighing 50 kg is evenly pulled along an inclined plane with a force of 50 N. Determine the efficiency of the inclined plane if its length is 5 m and its height is 40 cm. What is the frictional force when lifting a load?

m = 50 kg.

g = 9.8 m / s2.

F = 50 N.

h = 40 cm = 0.4 m.

S = 5 m.

Efficiency -?

Ftr -?

The efficiency factor determines what part of the spent work Az goes into useful work Ap: Efficiency = Ap * 100% / Az.

We will express the useful work Ap by the formula: Ap = m * g * h.

The expended work Az is expressed by the formula: Az = F * S.

Efficiency = m * g * h * 100% / F * S.

Efficiency = 50 kg * 9.8 m / s2 * 0.4 m * 100% / 50 N * 5 m = 78.4%.

F = Ftr + m * g * sinα.

sinα = h / S.

Ftr = F – m * g * sinα = F – m * g * h / S.

Ftr = 50 N – 50 kg * 9.8 m / s2 * 0.4 m / 5 m = 10.8 N.

Answer: efficiency = 78.4%, Ftr = 10.8 N.