Determine the coefficient of friction if a body with a mass of 2 tons is pulled horizontally by a rope

Determine the coefficient of friction if a body with a mass of 2 tons is pulled horizontally by a rope with an acceleration of 0.5 m / s², and the rope with a stiffness of 40,000 N is stretched by 5 cm.

Data: m (weight of the body pulled by the rope) = 2 t (2000 kg); a (body acceleration) = 0.5 m / s2; k (cable stiffness coefficient) = 40,000 N; Δl (cable elongation) = 5 cm (0.05 m).

Constants: g (acceleration due to gravity) = 9.81 m / s2 ≈ 10 m / s2.

The coefficient of friction is determined from the equation of projections of forces: m * a = Ftr + Fcont. = μ * m * g + k * Δl, whence μ = (m * a – k * Δl) / (m * g).

Let’s calculate: μ = (2000 * 0.5 – 40,000 * 0.05) / (2000 * 10) = -0.05.

Answer: The coefficient of friction is 0.05.