A body weighing 10 kg was lifted along an inclined plane (an angle with the horizontal of 30 degrees)

A body weighing 10 kg was lifted along an inclined plane (an angle with the horizontal of 30 degrees) to a height of 2.5 m. The coefficient of sliding friction on the surface was 0.2. Determine the amount of work done.

To determine the value of the perfect work A = F ∙ S when lifting a body weighing m = 10 kg along an inclined plane (angle with the horizontal α = 30 °) to a height h = 2.5 m, it is necessary to find the distance traveled S = 2 ∙ h = 2 ∙ 2.5 = 5 (m) – by the property of the leg, which lies opposite the angle α = 30 °. According to Newton’s second law, the resultant F = m ∙ a or m ∙ a = Ftr + m ∙ g ∙ sinα, where the proportionality coefficient g = 9.8 N / kg, and the friction force Ffr = μ ∙ N = μ ∙ m ∙ g ∙ cosα. Then m ∙ a = μ ∙ m ∙ g ∙ cosα + m ∙ g ∙ sinα or a = g ∙ (μ ∙ cosα + sinα). We get that F = m ∙ g ∙ (μ ∙ cosα + sinα) or F = 10 ∙ 9.8 ∙ (0.1 ∙ √3 + 0.5) ≈ 65.97 (H), since from the condition of the problem it is known that the coefficient of sliding friction on the surface was μ = 0.2. Substitute the values ​​of physical quantities in the calculation formula and find the work A = 65.97 ∙ 5 ≈ 330 (J).

Answer: 330 J is the amount of perfect work.