Help with physics is urgently needed. A body weighing m = 4.0 kg began to move on a horizontal surface.

Help with physics is urgently needed. A body weighing m = 4.0 kg began to move on a horizontal surface. Determine the modulus of the horizontal directional force that was applied to the body if, with the sliding friction coefficient equal to 0.20, after a time interval Δt = 3.0 s after the start of movement, it accelerated to a speed whose modulus v = 0.90 m / s

A body with a mass of m = 4.0 kg moving on a horizontal surface is affected by a traction force Ft and a friction force Ffr directed against the motion. The resultant of these forces according to Newton’s second law is found by the formula: F = m ∙ a, where acceleration a = (v – vо): Δt. On the other hand, F = Ft – Ftr, then the modulus of the horizontally directed force that was applied to the body will be: Ft = F + Ftr. Ftr = μ ∙ m ∙ g, where the coefficient g = 9.8 N / kg, according to the condition, the coefficient of sliding friction μ = 0.20. After a time interval Δt = 3.0 s, after the start of motion (vо = 0 m / s), the body accelerated to a speed whose modulus v = 0.90 m / s, we obtain: Fт = m ∙ a + μ ∙ m ∙ g or Fт = m ∙ (a + μ ∙ g); Fт = m ∙ (v / Δt + μ ∙ g); Ft = 4 ∙ (0.9 / 3 + 0.2 ∙ 9.8); Ft = 9.04 N.