A body weighing 5 kg slides evenly on a horizontal surface under the action of a horizontal force applied by means

A body weighing 5 kg slides evenly on a horizontal surface under the action of a horizontal force applied by means of a spring. Determine the elongation of the spring if the coefficient of friction = 0.2 and the stiffness of the spring is k = 500N / m.

Since the speed of a given body during sliding is constant, the equality will be true: Ftr = Fcont and μ * m * g = k * Δx, whence Δx = μ * m * g / k.

Constants and variables: μ – coefficient of friction (μ = 0.2); m is the mass of the sliding body (m = 5 kg); g – acceleration due to gravity (g ≈ 10 m / s2); k is the stiffness of the spring used (k = 500 N / m).

Calculation: Δx = μ * m * g / k = 0.2 * 5 * 10/500 = 0.02 m or 2 cm.

Answer: The spring used should have lengthened by 2 cm.