# A force of 3.99 N acts on a body weighing 0.8 kg and lying on a horizontal

A force of 3.99 N acts on a body weighing 0.8 kg and lying on a horizontal surface. The coefficient of friction is 0.5. Find the acceleration of the body.

According to Newton’s second law:

ma = F – Ffr, where m – body mass (m = 0.8 kg), a – body acceleration (m / s²), F – force acting on the body (F = 3.99 N), Ffr – friction force (Ftr. = Μ * N, where μ is the coefficient of friction (μ = 0.5), N is the reaction force of the support (N = Ft = m * g, where g is the acceleration of gravity (g = 9.8 m / s² ))).

Body Acceleration:

a = (F – Ftr.) / m = (F – μ * m * g) / m = (3.99 – 0.5 * 0.8 * 9.8) / 0.8 = 0.07 / 0 , 8 = 0.0875 m / s².

Answer: The acceleration of the body is equal to 0.0875 m / s².

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